92 Remember this Remember this Equation (4.12) tells that the temperature of a system changes in an isobaric process therefore, its internal energy also changes (Eq. 4.11). The heat exchanged (Eq. 4.12) is partly used for increasing the temperature and partly to do some work. The change in the temperature of the system depends on the specific heat at constant pressure C p . p-V diagram of an isochoric process. For an isochoric process, ∆V =0, and we have, from the first law of thermodynamics, ∆U = Q. This means that for an isochoric change, all the energy added in the form of heat remains in the system itself and causes an increase in its internal energy. Heating a gas in a constant volume container or diffusion of a gas in a closed chamber are some examples of Example 4.6: isochoric process. One mole of an ideal gas is initially kept in a cylinder with a movable frictionless and massless piston at pressure of 1.0mPa, and temperature 27°C. It is then expanded till its volume is doubled. How much work is done if the expansion is isobaric? Solution: Work done in isobaric process given by W = p∆V = (Vf - Vi ). Vf = 2Vi ∴W = 2pVi . Vi can be found by using the ideal gas equation for initial state. pi Vi = nRTi for n = 1 mol, V RT p W W i i i fi fi ff ff fi ff ffl fi ff ff ff fi ffi ffi 8 31 300 1 10 24 9 10 2 10 24 9 10 6 4 3 6 4 . . . m 4 9. kJ Always remember for an isobaric process: 1. ∆p = 0. Constant pressure process. 2. Temperature of the system changes, ∆T ≠ 0. 3. Q = ∆U + W. Energy exchanged is used to do work and also to change internal energy, i.e., to increase its temperature. 4. W = p∆V. Volume changes when work is done. 4. Isochoric process: It is a constant volume process. A system does no work on its environment during an isochoric change. Figure 4.17 shows the Fig. 4.17: p-V diagram of isochoric process. Thermodynamics of Isochoric process: For an isochoric process, we have, ∆V =0. The system does not do any work and all the energy supplied to the system is converted into its internal energy. The first law of thermodynamics for isochoric process is Q = ∆U --- (4.13) The change in internal energy is given by fi fi U n ff C T V The work done is given by W = p∆V = 0 (because ∆V = 0). The heat exchanged is given by the first law of thermodynamics, Q U fi ff ffl ffl W U fi fi nC T V ffl --- (4.14) Always remember for an isochoric process: 1. ∆V = 0. Constant volume process. 2. W = 0. No work is done because volume remains constant, ∆V = 0. 3. Q = ∆U. Energy exchanged is used to change internal energy. 4. ∆T ≠ 0. Temperature of the system changes.
93 5. Adiabatic process: It is a process during which there is no transfer of heat from or to the system. Figure 4.18 shows the p-V diagram of an adiabatic process. For an adiabatic change, Q = 0. Heat transfer to or from the system is prevented by either perfectly insulating the system from its environment, or by carrying out the change rapidly so that there is no time for any exchange of heat. Puncturing an inflated balloon or a tyre are some familiar examples of adiabatic changes. For an adiabatic change, ∆U = - W --- (4.15) When a system expands adiabatically, W is positive (work is done by the system) and ∆U is negative, the internal energy of the system decreases. When a system is compressed adiabatically, W is negative (work is done on the system), and ∆U is positive. The internal energy of the system increases in an adiabatic process. It is observed that for many systems, temperature increases when internal energy increases and decreases when the internal energy is decreased. gases and 8 6 4 3 = for polyatomic gases. Equations (3.23), (3.24) and (3.25) can be extended to obtain these values. An adiabatic system is thermally isolated from its environment, therefore, it cannot exchange heat with it. Therefore, when a system undergoes an adiabatic change, its temperature and internal energy both change. The change in internal energy is, ∆U = Cv (∆T) --- (4.17) The work done is, W pdV V V i f fi ff Using Eq. (4.16) we have, W C dV V V V i f fi ff ffl W C V V V i f fi ff ffl ffi fl ffl 1 1 , where V changes from Vi to Vf . W C V V f i fi ffl ffi ff fl ff ffl ffi ff ffl ffi ff fi fi fi fi 1 1 1 1 1 --- (4.18) From (Eq. 4.16) we have, pVγ = C or, pi Vi γ = pf Vf γ Therefore, we can write (Eq. 4.18) as, W p V V p V V f f f i i i fi ffl ffi ff fl ff ffl ffi ff ffl ffi ff fi fi fi fi 1 1 1 1 W p fi f f V pi i V ffl ffi ff fi fi fl ff ffl ffi 1 1 --- (4.19) W nR T T i f p Vf f p Vi i fi ffl ffi ff ffl ffi ff fi ffl ffi ff ffl ffi ff fi fi 1 1 fl fl --- (4.20) Equation (4.20) implies that when work is done by the gas, i.e., when the gas expands, W > 0, and Ti > Tf . This mean that the gas will cool down. Similarly, if the work is done on the gas, i.e., if the gas is compressed W < 0, and Ti < Tf . This means that the gas will warm up. Fig. 4.18: p-V diagram of adiabatic process. Thermodynamics of Adiabatic process: For an adiabatic process we have, pVγ = constant = C --- (4.16) where, γ is the ratio of the specific heat at constant pressure to the specific heat at constant volume, i.e., γ = C C p V γ is also called adiabatic ratio. For moderate temperature changes, the value of γ is 5 3 for monoatomic gases, 7 5 for diatomic
94 Use your brain power Remember this Always remember for an adiabatic process: 1. Equation of state: pVγ = constant. 2. Q = 0. No exchange of heat with the surroundings. The system is perfectly insulated from its environment, or the change is very rapid. 3. ∆U = - W. All the work is utilized to change the internal energy of the system. 4. ∆T ≠ 0. Temperature of the system changes. 5. Adiabatic expansion causes cooling and adiabatic compression causes heating up of the system. 6. W nR T T i f p Vf f p Vi i fi ffl ffi ff ffl ffi ff fi ffl ffi ff ffl ffi ff fi fi 1 1 fl fl 7. Most of the times, an adiabatic change is a sudden change. During a sudden change, the system does not find any time to exchange heat with its environment. Example 4.7: An ideal gas of volume 1.0L is adiabatically compressed to (1/15)th of its initial volume. Its initial pressure and temperature is 1.01 ×105 Pa and 27°C respectively. Given CV for ideal gas = 20.8J/mol.K and γ = 1.4. Calculate (a) final pressure, (b) work done, and (c) final temperature. (d) How would your answers change, if the process were isothermal ? Solution: (a) To calculate the final pressure pf . This can be calculated by using p p V V i f i f fi ff ffl ffi ffi fl fi fi ( . ) ( ) . . Pa P 1 01 10 15 44 8 10 5 1 4 5 a (about 45 atm) . (b) To calculate the work done, W p Vf f p Vi i fi ffl ffi ff ffl ffi 1fffl W P fi i i V Pi f P Pi f P ff ffl ffi ff fi fi ff fl fl ff ff 1 1 1 1 4 1 1 01 10 1 0 10 44 5 3 . ( . )( . ) ( .8 10 1 0 10 15 494 5 3 fl fl fi ff ff ) . m J 3 (c) To calculate final temperature Tf consider, T T V V f i f i fi ff ffl ffi fl fi fi fi 1 0 40 300 15 886 613 ( K) ( ) . K C The pressure involved in this process is about 45 atm. This is an adiabatic compression. The temperature of the gas is increased without any transfer of heat. Similar heating is used in automobile (diesel) engines. The fuel used in the engine is heated rapidly to such a high temperature that it ignites without any spark plug. (d) (i) Pressure in isothermal process is given by pi Vi = pf Vf p p V V f i i f = = 15 atm (ii) There will be no change in the temperature because it is an isothermal process. (iii) Work done in isothermal process is given by W nRT V V n f i fi ff ffl ffi fl fi fi ln , . . ( . ) 0 405 0 0405 831 300 0 270 fi 2726 J The work done during adiabatic process is very much less than the work done during isothermal process. Can you explain this? What happens to this work which is apparently 'lost' ? 1. Why is the p-V curve for adiabatic process steeper than that for isothermal process? 2. Explain formation of clouds at high altitude.
95 Can you tell? Remember this Can you tell? 6. Cyclic Process: A thermodynamic process that returns a system to its initial state is a cyclic process. In this process, the initial and the final state is the same. Figure 4.19 shows the p-V diagram of a cyclic process. For a cyclic process, the total change in the internal energy of a system is zero. (∆U = 0). According to the first law of thermodynamics, we have, for a cyclic process, Q = W --- (4.21) 1 2 Fig. 4.19: p-V diagram of cyclic process. Working of all heat engines is a cyclic process. When the temperature of a system is increased or decreased in an adiabatic heating or cooling, is there any transfer of heat to the system or from the system? 1. How would you interpret the Eq. 4.21 for a cyclic process? 2. An engine works at 5000 RPM, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine? Example 4.8: Cyclic process: The total work done in the cyclic process shown in the Fig. 4.19 is -1000 J. (a) What does the negative sign mean? (b) What is the change in internal energy and the heat transferred during this process. (c) What will happen when the direction of the cycle is changed? Solution: (a) Work done in the process from A to B along path 2 is given by the area under this curve. In this process, volume is increasing therefore the work done is positive. Work done in the process from B to A along path 1 is given by the area under this curve. In this process volume is decreasing therefore the work done is negative. The total work done during the complete cycle, from A to B along path 2 and form B to A along path 1 is the area enclosed by the closed loop. This is the difference between the area under the curve 1 and that under the curve 2. Since the area under curve 1 is negative and larger than the area under the curve 2, the area of the loop is also negative. That means the work is done by the system is negative. (b) This is a cyclic process which means the initial and the final state of the system is the same. For a cyclic process ∆U = 0, so Q = W = - 1000 J. That is, 1000 joules of heat must be rejected by the system. (c) If the direction of the cycle is changed the work done will be positive. The system will do work. Form this example we conclude that: The total work done in a cyclic process is positive if the process is goes around the cycle in a clockwise direction. The total work done in a cyclic process is negative if the process goes around the cycle in a counterclockwise direction. 7. Free Expansion: These expansions are adiabatic expansions and there is no exchange of heat between a system and its environment. Also, there is no work done on the system or by the system. Q = W = 0, and according to the first law of thermodynamics, ∆U = 0. For example, when a balloon is ruptured suddenly, or a tyre
96 is suddenly punctured, the air inside rushes out rapidly but there is no displacement of a piston or any other surface. Free expansion is different than other thermodynamic processes we have discussed so far because it is an uncontrolled change. It is an instantaneous change and the system is not in thermodynamic equilibrium. A free expansion cannot be plotted on a p-V diagram. Only its initial and the final state can be plotted. A heat engine receives heat from a source called reservoir and converts some of it into work. Remember that all the heat absorbed is not converted into work by a heat engine. Some heat is lost in the form of exhaust. A typical heat engine has the following elements: (1) A working substance: It is called the system. It can be an ideal gas for an ideal heat engine (to be discussed later). For a practical heat engine, the working substance can be a mixture of fuel vapour and air in a gasoline (petrol) or diesel engine, or steam in a steam engine. It is the working substance that absorbs heat and does work. (2) Hot and cold reservoir: The working substance interacts with the reservoirs. The hot reservoir is the source of heat. It is at a relatively high temperature and is capable of providing large amount of heat at constant higher temperature, TH. It is also called as the source. The cold reservoir absorbs large amount of heat from the working substance at constant lower temperature, TC. It is also called as the sink. (3) Cylinder: Generally, the working substance is enclosed in a cylinder with a moving, frictionless, and massless piston. The working substance does some work by displacing the piston in the cylinder. This displacement is transferred to the environment using some arrangement such as a crank shaft which transfers mechanical energy to the wheels of a vehicle. Heat engines are of two basic types. They differ in the way the working substance absorbs heat. In an external combustion engine, the working substance is heated externally as in case of a steam engine. In case of the internal combustion engine, the working substance is heated internally similar to an automobile engine using gasoline or diesel. Any heat engine works in following three basic steps. Example 4.9: A cyclic process ABCA Solution: Straight line form A to B on the T V V T p V 4.8 Heat Engines: As mentioned earlier, thermodynamics is related to study of different processes involving conversion of heat and work into each other. In this section, we will study the practical machines that convert some heat into work. These are heat engines. 4.8.1 Heat Engine: Heat engines are devices that transform heat partly into work or mechanical energy. Heat engines work by using cyclic processes and involve thermodynamic changes. Automobile engines are familiar examples of heat engines. is shown in V-T diagram (Fig (a)). It is performed with a constant mass of ideal gas. How will this transform to a p-V digram? V-T diagram means V ∝ T, i.e., 'p' is constant. V is constant and temperature decreases. Along B - C i.e., p should also decrease. Temperature is constant along CA. That means it is an isothermal process. The p-V curve would look as shown in the Fig (b). (Fig (a) (Fig (b)
97 Do you know? 1. The working substance absorbs heat from a hot reservoir at higher temperature. 2. Part of the heat absorbed by the working substance is converted into work. 3. The remaining heat is transferred to a cold reservoir at lower temperature. Heat engines are classified according to the working substance used and the way these steps are actually implemented during its operation. Heat engines are diagrammatically represented by an energy flow diagram schematically shown in Fig. 4.20. Energy exchange takes place during various stages of working of a heat engine. the pipeline indicated by W is proportional to the part of the heat converted into mechanical work. One single execution of the steps mentioned above is one operating ‘cycle’ of the engine. Several such cycles are repeated when a heat engine operates. The quantities QH and QC represent the amount of heat absorbed (positive) and rejected (negative) respectivety during one cycle of operation. Let QH be the heat absorbed by the working substance at the source, and QC be the heat rejected by it at the sink. In a heat engine, QH is positive and QC is negative. Also, let W be the work done by the working substance. In the Fig. 4.20, the circle represents the engine. The ‘heat pipelines’ shown in the diagram represent the heat absorbed, rejected, and converted into work. The width of the heat ‘pipeline’ indicated by QH, is proportional to the amount of heat absorbed at the source. Width of the branch indicated by QC is proportional to the magnitude | QC | of the amount of heat rejected at the sink. Width of the branch of The number of repetitions of the operating cycles of an automobile engine is indicated by its RPM or Revolutions Per Minute. The net heat Q absorbed per operating cycle is, Q Qfi ff fi fi H C Q Q fi ffl H c Q --- (4.22) The net work done in one operating cycle, by the working substance, is given by using the first law of thermodynamics. W Qfi fifi fi Q Q H c ff --- (4.23) Ideally, we would expect a heat engine to convert all the heat absorbed, QH , in to work. Practically, this is not possible. There is always some heat lost, i.e., QC ≠ 0. The thermal efficiency η of the heat engine is defined as, fi ff W QH --- (4.24) Thus, the thermal efficiency, or simply, the efficiency of a heat engine is the ratio of the work done by the working substance and the amount of heat absorbed by it. It is the ratio of the output, in the form of the work done W by the engine, and the input, in the form of the heat supplied QH. In simple words, efficiency of a heat engine is the fraction of the heat absorbed that is converted into work, Eq. (4.24). In terms of the energy flow diagram Fig. 4.20, the ‘pipeline’ representing the work is as wide as possible and the pipeline representing the exhaust is as narrow as possible for the most efficient heat engine. Fig. 4.20: Schematic energy flow diagram of a heat engine.
98 There is a fundamental limit on the efficiency of a heat engine set by the second law of thermodynamics, which we will discuss later. Using Eq. (4.22) and Eq. (4.23) we can write the efficiency of a heat engine as, fi ff ff fi fi ffl ff ffi W Q Q Q Q H Q C H c H 1 1 --- (4.25) Equation (4.25) gives the thermal efficiency of a heat engine. It is a ratio of the quantities which represent energy. Therefore, it has no units but, we must express W, QH, and QC in the same units. 4.8.2 The Heat Engine Cycle and the p-V Diagram: As discussed previously, the working of a heat engine is a well defined sequence of operations. It is a cyclic thermodynamic process. We know that a thermodynamic process can be represented by a p-V diagram. We will now discuss the p-V diagram of a heat engine. Keep in mind that this is a p-V diagram of a general heat engine. There are different ways of operating a heat engine. We will discuss some such heat engines in the following sections. A heat engine uses energy absorbed in the form of heat to do work and then rejects the heat which cannot be used to do work. Heat is absorbed in one part of the cycle, work is done in another part, and the unused heat is rejected in yet other part of the cycle. The p-V diagram of a typical heat engine is shown in Fig. (4.21). The operating cycle begins at the point A in the cycle. The working substance, the gas in this case, absorbs heat at constant volume and no work is done by the gas or on the gas. The pressure is increased till the point B is reached. The temperature of the gas also increases and its internal energy increases. The gas starts expanding by pushing the piston away and its volume changes from the point B to the point C. Because the gas expands, its pressure is reduced. The gas does work in this part of the cycle. When the point C is reached, the excess heat, the heat that is not utilized in doing work by the gas, is rejected. The gas cools down and its internal energy decreases. This process is again at constant volume. The pressure of the gas is reduced and point D on the p-V diagram is reached. The gas is now compressed. Its volume decreases and its pressure increases. The change continues till the point A is reached. The cycle is complete and the system is ready for the next cycle. Thus, the p-V diagram is a visual tool for the study of heat engines. The working substance of a heat engine is usually a gaseous mixture. Study of the p-V diagram helps us understand the behavior of the three state variables of a gas throughout the operational cycle. The operation of a heat engine is a cyclic process therefore, its p-V diagram is a closed loop. The area of the loop represents the work done during one complete cycle. Since work is done by the gas, or on the gas, only when its volume changes, the p-V diagram provides a visual interpretation of the work done during one complete cycle. Similarly, the internal energy of the gas depends upon its temperature. Hence, the p-V diagram along with the temperatures calculated from the ideal gas law determines Fig. 4.21: p-V diagram of a typical heat engine. the changes in the internal energy of the gas.
99 We can calculate the amount of heat added or rejected from the first law of thermodynamics. Thus, a p-V diagram helps us analyze the performance of any heat engine which uses a gas as its working substance. 4.9 Refrigerators and Heat Pumps: So far, we have discussed a heat engine which takes heat from a source at higher temperature and rejects it to a sink at lower temperature. The input provided to the working substance (a gas or a mixture of gasoline and air) in a heat engine is in the form of heat which is converted into mechanical work as output. Figure 4.20 shows this in the form of an energy flow diagram. Refrigerators and heat pumps are heat engines that work in backward direction. They convert mechanical work into heat. 4.9.1 Heat Flow from a Colder Region to a Hotter Region: According to the second law of thermodynamics (to be discussed in the next article), heat cannot flow from a region of lower temperature to a region of higher temperature on its own. We can force heat to flow from a region of lower temperature to a region of higher temperature by doing work on the system (or, on the working substance of a heat engine). Refrigerators or air-conditioners and heat pumps are examples of heat engines which cause heat to be transferred from a cold region to a hot region. Usually, this is achieved with the aid of phase change of a fluid, called the refrigerant. The refrigerant is forced to evaporate and then condense by successively decreasing and increasing its pressure. It can, therefore, ‘pump’ energy from a region at lower temperature to a region of higher temperature. It extracts the heat of vaporization of the refrigerant from the cold region and rejects it to the hotter region outside the refrigerator. This results in cooling down the cold region further. Figure 4.22 shows the concept of transferring heat from a cold region to a hot region in a schematic way. Heat from the cold region is carried to the hot region by the refrigerant. It extracts heat from a cold region due to forced evaporation. The heat of evaporation of the refrigerant thus absorbed is rejected by compressing and condensing it into liquid at a higher temperature. All this process is carried out in a mechanism involving a compressor and closed tubing such as seen at the back of a house hold refrigerator. Fig. 4.22: Schematic diagram of transferring heat from a cold region to a hot region. 4.9.2 Refrigerator: Refrigeration is a process of cooling a space or substance of a system and/or to maintain its temperature below its ambient temperature. In simple words, refrigeration is artificial cooling. Fig. 4.23 (a): Schematics of a refrigerator. A refrigerator extracts heat from a cold region (inside the chamber, or the compartments) and delivers it to the surrounding (the atmosphere) thus, further cooling the cold region. That’s the reason why if you place your hand behind a working
100 refrigerator, you can feel the warm air. But the interior of the refrigerator is cold. An air conditioner also works on similar principles. Figure 4.23 (a) shows the schematics of the mechanism used in a typical refrigerator. It consists of a compressor, an expansion valve, and a closed tube which carries the refrigerant. Part of the tube, called the cooling coil, is in the region which is to be cooed at lower temperature and lower pressure. The other part which is exposed to the surrounding (generally, the atmosphere) is at a higher temperature and higher pressure. A fluid such as (fluorinated hydrocarbons) is used as refrigerant. Normally, the cold and the hot part of the coil contain the refrigerant as a mixture of liquid and vapour phase in equilibrium. up as heat is transferred to it from the contents of the fridge. This takes place at constant pressure, so it's an isobaric expansion. Step 3: The gas is transferred to a compressor, which does most of the work in this process. The gas is compressed adiabatically, heating it and turning it back to a liquid. Step 4: The hot liquid passes through coils on the outside of the fridge, and heat is transferred to the atmosphere. This is an isobaric compression process. The compressor is driven by an external energy source and it does the work |W| on the working substance during each cycle. 4.9.3 Performance of a Refrigerator: Consider the energy flow diagram of a refrigerator Fig. 4.23 (b). It shows the relation between the work and heat involved in transferring heat from a low temperature region to a high temperature region. This is a cyclic process in which the working substance, the refrigerant in this case, is taken back to the initial state. For a refrigerator, the heat absorbed by the working substance is QC and the heat rejected by it is QH. A refrigerator absorbs heat at lower temperature and rejects it at higher temperature, therefore, we have, QC > 0, QH < 0, and W < 0. Hence, we write, |W| and |QH| = - QH. In this case, we apply the first law of thermodynamics to the cyclic process. For a cyclic process, the internal energy of the system in the initial state and the final state is the same, therefore, from Eq. (4.21), we have, Q Q H C fi ffW , or Q Q H C fi fi fi fffi fi W ffl 0 fiffQ Q H C fi fi ffl fi–W For a refrigerator, QH < 0, and W< 0, therefore, Q Q H c fi ff W ---(4.26) From the Fig.4.23 (b), we realize that the heat |QH| rejected by the working substance at the hot reservoir is always greater than the heat QC received by it at the cold reservoir. Note that the Eq. (4.26), derived for a refrigerator Figure 4.23 (b) shows the energy flow diagram of a refrigerator. As you can see, the heat extracted from a cold reservoir is supplemented by the mechanical work done (on the refrigerant) by the compressor and the total energy is rejected at the hot reservoir. The refrigerant goes through the following steps in one complete cycle of refrigeration. Step 1: The fluid passes through a nozzle and expands into a low-pressure area. Similar to the way carbon dioxide comes out of a fire extinguisher and cools down, the fluid turns into a gas and cools down. This is essentially an adiabatic expansion. Step 2: The cool gas is in thermal contact with the inner compartment of the fridge. It heats Fig. 4.23 (b): Energy flow diagram of a refrigerator.
101 Remember this Do you know? Remember this Do you know? and the Eq. (4.23), derived for a heat engine, are the same. They are valid for a heat engine and also for a refrigerator. The ratio Q W c indicates the performance of a refrigerator and is called the coefficient of performance (CoP), K, or quality factor, or Q-value of a refrigerator. Larger is the ratio, better is the refrigerator. That means a refrigerator has the best performance when the heat extracted by the refrigerant at the cold reservoir is maximum by doing minimum work in one operating cycle. From Eq. (4.26), W Q fi ff c H Q fi ff ff ffl K Q W Q Q Q c c c H --- (4.27) All the quantities on the right side of Eq. (4.27) represent energy and are measured in the same energy units. The coefficient of performance, K of a refrigerator is, therefore, a dimensionless number. For a typical household refrigerator, K ≈ 5. an air conditioner is defined by K Q W c = . It is important to consider the rate of heat removed H and the power P required for removing the heat. We define the rate of heat removed as the heat current H Q t c = , where, t is the time in which heat Qc is removed. Therefore, the coefficient of performance of an air conditioner can be calculated as, K Q W Ht Pt H P c = = = --- (4.28) Typical values of K are 2.5 to 3.0 for room air conditioners. Capacity of a refrigerator is expressed in litre. It is the volume available inside a refrigerator. Capacity of an air conditioner is expressed in tonne. Do you know why? Before refrigerator and AC was invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today. 4.9.4 Air conditioner: Working of an air conditioner and a refrigerator is exactly similar. It differs from a refrigerator only in the volume of the chamber/ room it cools down. For an air conditioner, the evaporator coils are inside the room that is to be cooled and the condenser is outside the room. The air cooled by the evaporator coils inside the room is circulated by a fan placed inside the air conditioning unit. The performance of Refrigerator transfers heat from inside a closed space to its external environment so that inside space is cooled to temperature below the ambient temperature. 4.9.5 Heat Pump: Heat pump is a device which works similar to a refrigerator. It is used to heat a building or a similar larger structure by cooling the air outside it. A heat pump works like a refrigerator operating inside out. In this case, the evaporator coils are outside and absorb heat from the cold air from outside. The condenser coils are inside the building. They release the absorbed heat to the air inside the thus, warming the building. Heat flow from a hot object to a cold object is spontaneous whereas, work is always required for the transfer of heat from a colder object to a hotter object.
102 4.10 Second Law of Thermodynamics: 4.10.1 Limitations of the First Law of Thermodynamics: The First law of thermodynamics tells us that heat can be converted into work and work can also be converted into heat. It is merely a quantitative statement of the equivalence of heat and work. It has the following limitations. (a) It does not tell us whether any particular process can actually occur. According to the first law of thermodynamics, heat may, on its own, flow from an object at higher temperature to one at lower temperature and it can also, on its own, flow from an object at lower temperature to one at higher temperature. We know that practically, heat cannot flow from an object at lower temperature to another at higher temperature. The First law of thermodynamics does not predict this practical observation. (b) According to the First law, we could convert all (100%) of the heat available to us into work. Similarly, all the work could be converted into heat. Again, we know that practically this is not possible. Thus, the First law of thermodynamics does not prevent us from converting heat entirely into work or work entirely into heat. These limitations lead to the formulation of another law of thermodynamics called the Second law of thermodynamics. We will discuss this at a later stage in this chapter. We have seen earlier in section 4.7.3 that an irreversible process defines the preferred direction of an irreversible process. It is also found that it is impossible to build a heat engine that has 100% efficiency Eq. (4.25). That is, it is not possible to build a heat engine that can completely convert heat into work. Similarly, for a refrigerator it is impossible to remove heat without doing any work on a system. That is, the coefficient of performance, Eq. (4.28) of a refrigerator can never be infinite. These practical observations form the basis of a very important principle of thermodynamics, the Second law of thermodynamics. The Second law of thermodynamics is a general principle which puts constraints upon the direction of heat transfer and the efficiencies that a heat engine can achieve. Fig. 4.24 (a): Energy of an object at two different hights. Fig. 4.24 (b): Limitations on efficiency of a heat engine. Consider an object A at certain height of h above the ground and another object B of the same mass a height of h/2 as shown in Fig. 4.24 (a). We know that potential energy of the object B is half that of the object A. That means we can extract only half the energy from the object B. Similarly, if a heat engine as shown in Fig. 4.24 (b) operates between the temperatures of 800 K and 400 K, i.e., if it receives heat at 800 K and rejects it at 400 K, its maximum efficiency can be 50%. 4.10.2 The second law of thermodynamics, statement: We now know that heat can be converted into work by using a heat engine. However, our practical experience says that entire heat supplied to the working substance can never be converted into mechanical work. Second law of thermodynamics helps us to understand this. According to the second law of thermodynamics, “It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W. Some amount of A B
103 heat QC must be exhausted to a cold reservoir. This prohibits the possibility of a perfect heat engine”. Sometimes it is also called as the ‘Engine Law’ or the ‘Engine Statement’ of the Second law of thermodynamics. This means that energy will not flow spontaneously from an object at low temperature to an object at a higher temperature. This rules out the possibility of a perfect refrigerator. The statements about refrigerators are also applicable to air conditioners and heat pumps, which work on the same principles. This is the ‘Second form’ or Clausius statement of the Second law of thermodynamics. Sometimes it is also called as the ‘Refrigerator Law’ or the ‘Refrigerator Statement’ of the Second law of thermodynamics. Fig. 4.25 (a): Second law of thermodynamics. Fig. 4.25 (b): Energy flow diagram of Engine statement. Figure 4.25 is a diagrammatic representation of the application of the Second law of thermodynamics to a heat engine. As you can see from the diagram, Fig. 4.25 (a), all heat engines lose some heat to the environment of a perfect heat engine. All the heat QH, extracted can not be used to do work. Figure 4.25 (b) shows the energy flow diagram for such a situation. This form of statement of the Second law of thermodynamics is called as the KelvinPlanck statement or the ‘First form’ of the Second law of thermodynamics. We have seen how the efficiency of a heat engine is restricted by the second law of thermodynamics. Heat engine is one form of ‘heat – work conversion’. Let us see what happens in case of a refrigerator, the other form of ‘work – heat conversion’. “It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow”. Fig. 4.26 (a): Energy flow diagram of a practical refrigerator. Fig. 4.26 (b): Energy flow of prefect refrigerator. Figure 4.26 is a diagrammatic representation of the application of the Second law of thermodynamics to a refrigerator. As you can see form the energy flow diagram, Fig. 4.26 (a), a practical refrigerator requires work W to be done to extract heat QH from a cold reservoir and reject it to a hot reservoir. The statement means that spontaneous flow of heat from an object at cold temperature is not possible Fig. 4.26 (b).
104 Remember this 4.11 Carnot Cycle and Carnot Engine: In section 4.7.3, we have discussed the concept of a reversible and an irreversible process at length. Now we will discuss why reversibility is such a basic and important concept in thermodynamics. 4.11.1 Significance of Reversibility in Thermodynamics: We know that a reversible process is a ‘bidirectional’ process, i.e., it follows exactly the same steps in either direction. This requires the process to take place in infinitesimally small steps. Also, the difference between the state variables in the two infinitesimally close states should be very small. This would be possible if the system is in thermodynamic equilibrium with its environment throughout the change. An irreversible process, on the contrary, is a unidirectional process. It can take place in only in one direction. Any irreversible process is not in thermal equilibrium with its environment. 4.11.2 Maximum Efficiency of a Heat Engine and Carnot’s Cycle: We know that conversion of work to heat (refrigerator, section 4.9.2) is an irreversible process. A heat engine would convert maximum heat into work if all irreversible processes could be avoided. In that case, the efficiency of the heat engine can be maximum. Sadi Carnot, French engineer and scientist, proposed a hypothetical ideal engine in 1824 which has the maximum efficiency. In a carnot engine, there are basically two processes: (i) Exchange of heat (steps A to B and C to D in the Fig. 4.21). For this to be reversible, the heat exchange must be isothermal. This is possible if the working substance is at the temperature TH of the source while absorbing heat. The working substance should be at the temperature of the cold reservoir TC, while rejecting the heat. (ii) Work done (steps B to C and D to A). For work done to be reversible, the process should be adiabatic. Thus, the cycle includes two isothermal and two adiabatic processes for maximum efficiency. The corresponding p-V diagram will then be as shown in the Fig. 4.27. When we say that energy will not flow spontaneously from a cold object to a hot object, we are referring to the net transfer of energy. Energy can transfer from a cold object to a hot object either by transfer of energetic particles or electromagnetic radiation. However, in any spontaneous process, the net transfer of energy will be from the hot object to the cold object. Work is required to transfer net energy from a cold object to a hot object. By using the expression for work done during an adiabatic and an isothermal process Eq. (4.7) and (4.20), we can derive an expression for the efficiency of a Carnot cycle/engine as, fi ff ff fi fi ffl ff ffl W Q Q Q T H T c H C H 1 1 --- (4.29) Thus, while designing a heat engine for maximum efficiency, the source temperature TH should be as high as possible and the sink temperature TC should be as low as possible. Fig. 4.27: Carnot cycle AB: Isothermal expansion, BC: Adiabatic expansion, CD: Isothermal compression, DA Adiabatic compression.
105 Remember this Use your brain power depends on only the temperature difference of the hot and the cold reservoir. When the temperature difference is very small, the coefficient is very large. In this case, a large quantity of heat can be removed from the lower temperature to the higher temperature by doing very small amount of work. The coefficient of performance is very small when the temperature difference is large. That means a small quantity of heat will be removed even when a large amount of work is done. Suggest a practical way to increase the efficiency of a heat engine. Always Remember for a Carnot Engine: 1. Carnot engine is a hypothetical concept. 2. Every process must be either isothermal or adiabatic. 3. The system must maintain thermodynamic equilibrium throughout the cycle so that it is reversible. 4. The efficiency of a Carnot engine can never be 100% unless TC = 0. We know that this is not possible practically. That means even an ideal heat engine, the Carnot engine, cannot have 100% efficiency. 4.11.2 Carnot Refrigerator: We know that a refrigerator is nothing but a heat engine operated in the reverse direction. Because each step in the Carnot cycle is reversible, the entire Carnot cycle is reversible. If we operate the Carnot engine in the reverse direction, we get the Carnot refrigerator. Using the Eq. (4.28), we can write the coefficient of performance of a Carnot refrigerator as, K Q Q Q c c H fi ff = Q Q Q Q c H c H 1− --- (4.30) Using Q Q T T c H C H = , in Eq. (4.30) we have, the coefficient of performance of a Carnot refrigerator as, K T T T C H C fi ff --- (4.31) Equation (4.31) gives the coefficient of performance of an ideal refrigerator or, the Carnot refrigerator. It says that the coefficient of performance of a Carnot refrigerator also Example 4.10: Carnot engine: A Carnot engine receives 2.0 kJ of heat from a reservoir at 500 K, does some work, and rejects some heat to a reservoir at 350 K. (a) How much work does it do? (b) how much heat is rejected. (c) what is its efficiency? Solution: The heat QC rejected by the engine is given by Q Q T T C H C H fiff fiff fi ff J K K J ( ) 2000 350 500 1400 From the First law, the work W done by the engine is, W = QH + QC = 2000 J + (-1400 J) = 600 J Efficiency of the Carnot engine is, fi ffffl ffffl 1 1 ff ff 350 500 0 30 30 T T C H K K . % Is there any simple way to calculate efficiency? 4.11.3 The Second Law of Thermodynamics and the Carnot Cycle: "The Carnot engine is the most efficient heat engine. Also, all Carnot engines operating between the same two temperatures have the same efficiency, irrespective of the nature of the working substance". We have made two very important statements here. 1. Carnot engine is the most efficient heat engine, and
106 Internet my friend 2. Efficiency of a Carnot engine is independent of its working substance. We can also show that “A Carnot refrigerator has a greater coefficient of performance among all the refrigerators working between the same two temperatures.” 4.12 Sterling Cycle: Fig. 4.28: Sterling cycle, p-V diagram. This is a closed thermodynamic cycle. The Sterling engine is based on this cycle shown in Fig. 4.28. The working substance used in a Sterling engine is air, helium, hydrogen, nitrogen etc. All the processes in the Sterling cycle are reversible processes. When the gas is heated, the Sterling engine produces useful work. When work is done on the gas, it works as a refrigerator. This is reverse working of a Sterling cycle. The reversed Sterling cycle is extensively used in the field of cryogenics to produce extremely low temperatures or to liquefy air or gases mentioned above. The ideal Sterling cycle has two isothermal processes AB and CD. Two isobaric processes BC and DA connect the two isothermal processes. Heat is absorbed at constant temperature TH and rejected at constant temperature TC. The four processes in a Sterling cycle are described briefly in the following. • Isothermal expansion (AB): The gas is heated by supplying heat QH at constant temperature TH. Useful work is done by the gas in this part of the cycle. • Isochoric process (BC): Part of the heat absorbed (QH) by the gas in the previous part of the cycle is released by the gas to the refrigerator. This heat (Q) is used in the next part of the cycle. The gas cools down to temperature TC. • Isothermal compression (CD): The heat generated in this part of the cycle (QC) is rejected to the coolant (sink). The temperature of the gas is maintained at TC during this process. • Isobaric heat absorption (DA): The compressed gas absorbs heat (Q) during this process. Its temperature is increased to TH. The cycle repeats when the process reaches the point A. 1. https://opentextbc.ca/ physicstestbook2/chapter/the-firstlaw-of-thermodynamics/ 2. https://opentextbc.ca/ physicstestbook2/chapter/ introduction-to-the-second-law-ofthermodynamics-heat-engines-andtheir-efficiency/ 3. https://opentextbc.ca/ physicstestbook2/chapter/the-firstlaw-of-thermodynamics-and-somesimple-processes/ 4. https://courses.lumenlearning. com/boundless-physics/chapter/ introduction-8/ 5. http://heatengine-sundervallii. blogspot.com/2010/10/everydayexamples-of-heat-engine.html 6. http://hyperphysics.phy-astr.gsu.edu/ hbase/heacon.html#heacon 7. http://hyperphysics.phy-astr.gsu.edu/ hbase/heacon.html
107 1. Choose the correct option. i) A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in energy of the system? (A) 10J (B) 4J (C) -10J (D) - 4J ii) Which of the following is an example of the first law of thermodynamics? (A) The specific heat of an object explains how easily it changes temperatures. (B)While melting, an ice cube remains at the same temperature. (C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time. (D) After falling down the hill, a ball's kinetic energy plus heat energy equals the initial potential energy. iii) Efficiency of a Carnot engine is large when (A) TH is large (B) TC is low (C) TH - TC is large (D) TH - TC is small iv) The second law of thermodynamics deals with transfer of: (A) work done (B) energy (C) momentum (D) heat v) During refrigeration cycle, heat is rejected by the refrigerant in the : (A) condenser (B) cold chamber (C) evaporator (D) hot chamber 2. Answer in brief. i) A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed. (a) Has there been a transfer of heat? (b) Has work been done? ii) Give an example of some familiar process in which no heat is added to or removed form a system, but the temperature of the system changes. iii) Give an example of some familiar process in which heat is added to an object, without changing its temperature. iv) What sets the limits on efficiency of a heat engine? v) Why should a Carnot cycle have two isothermal two adiabatic processes? 3. i) A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assume that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics? ii) A resistor held in running water carries electric current. Treat the resistor as the system (a) Does heat flow into the resistor? (b) Is there a flow of heat into the water? (c) Is any work done? (d) Assuming the state of resistance to remain unchanged, apply the first low of thermodynamics to this process. iii) A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system, (a) Has heat been transferred ? (b) Has work been done? (c) What is the sign of ∆U ? iv) Draw a p-V diagram and explain the concept of positive and negative work. Give one example each. v) A solar cooker and a pressure cooker both are used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them. Exercises
108 V T p V p V 4. A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. [Ans: 106.65 J] 5. A system releases 125 kJ of heat while 104 kJ of work is done on the system. Calculate the change in internal energy. [Ans: ∆U = 21 kJ] 6. Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir. [Ans: 23ºC] 7. A Carnot refrigerator operates between 250ºK and 300ºK. Calculate its coefficient of performance. [Ans: 5] 8. An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it? [Ans: Zero] 9. An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure? [Ans: 5.6] 10. A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles. [Ans: (a)] [Ans: (b)] 12. A system is taken to its final state from initial state in hypothetical paths as shown figure calculate the work done in each case. [Ans: 7.85 × 104 J] 11. The figure shows the V-T diagram for one cycle of a hypothetical heat engine which uses the ideal gas. Draw (a) the p-V diagram and p-T diagram of the system. p T p V [Ans: AB = 2.4 × 106 J, CD = 8 × 105 J, BC and DA zero, because constant volume change]
109 Can you tell? Can you recall? 5.1 Introduction: Oscillation is a very common and interesting phenomenon in the world of Physics. In our daily life we come across various examples of oscillatory motion, like rocking of a cradle, swinging of a swing, motion of the pendulum of a clock, the vibrations of a guitar or violin string, up and down motion of the needle of a sewing machine, the motion of the prongs of a vibrating tuning fork, oscillations of a spring, etc. In these cases, the motion repeated after a certain interval of time is a periodic motion. Here the motion of an object is mostly to and fro or up and down. Oscillatory motion is a periodic motion. In this chapter, we shall see that the displacement, velocity and acceleration for this motion can be represented by sine and cosine functions. These functions are known as harmonic functions. Therefore, an oscillatory motion obeying such functions is called harmonic motion. After studying this chapter, you will be able to understand the use of appropriate terminology to describe oscillations, simple harmonic motion (S.H.M.), graphical representations of S.H.M., energy changes during S.H.M., damping of oscillations, resonance, etc. 5.2 Explanation of Periodic Motion: Any motion which repeats itself after a definite interval of time is called periodic motion. A body performing periodic motion goes on repeating the same set of movements. The time taken for one such set of movements is called its period or periodic time. At the end of each set of movements, the state of the body is the same as that at the beginning. Some examples of periodic motion are the motion of the moon around the earth and the motion of other planets around the sun, the motion of electrons around the nucleus, etc. As seen in Chapter 1, the uniform circular motion of any object is thus a periodic motion. Another type of periodic motion in which a particle repeatedly moves to and fro along the same path is the oscillatory or vibratory motion. Every oscillatory motion is periodic but every periodic motion need not be oscillatory. Circular motion is periodic but it is not oscillatory. The simplest form of oscillatory periodic motion is the simple harmonic motion in which every particle of the oscillating body moves to and fro, about its mean position, along a certain fixed path. If the path is a straight line, the motion is called linear simple harmonic motion and if the path is an arc of a circle, it is called angular simple harmonic motion. The smallest interval of time after which the to and fro motion is repeated is called its period (T) and the number of oscillations completed per unit time is called the frequency (n) of the periodic motion. 5. Oscillations 1. What do you mean by linear motion and angular motion? 2. Can you give some practical examples of oscillations in our daily life? 3. What do you know about restoring force? 4. All musical instruments make use of oscillations, can you identify, where? 5. Why does a ball floating on water bobs up and down, if pushed down and released? Is the motion of a leaf of a tree blowing in the wind periodic? 5.3 Linear Simple Harmonic Motion (S.H.M.): Place a rectangular block on a smooth frictionless horizontal surface. Attach one end
110 (a) Remember this of a spring to a rigid wall and the other end to the block as shown in Fig. 5.1. Pull the block of mass m towards the right and release it. The block will begin its to and fro motion on either side of its equilibrium position. This motion is linear simple harmonic motion. f k fi ff x --- (5.1) where, k is a constant that depends upon the elastic properties of the spring. It is called the force constant. The negative sign indicates that the force and displacement are oppositely directed. If the block is displaced towards left from its equilibrium position, the force exerted by the spring on the block is directed towards the right and its magnitude is proportional to the displacement from the mean position. (Fig. 5.1(c)) Thus, f = - kx can be used as the equation of motion of the block. Now if the block is released from the rightmost position, the restoring force exerted by the spring accelerates it towards its equilibrium position. The acceleration (a) of the block is given by, a f m k m fi fi ff x ffl ffi fl --- (5.2) where, m is mass of the block. This shows that the acceleration is also proportional to the displacement and its direction is opposite to that of the displacement, i.e., the force and acceleration are both directed towards the mean or equilibrium position. As the block moves towards the mean position, its speed starts increasing due to its acceleration, but its displacement from the mean position goes on decreasing. When the block returns to its mean position, the displacement and hence force and acceleration are zero. The speed of the block at the mean position becomes maximum and hence its kinetic energy attains its maximum value. Thus, the block does not stop at the mean position, but continues to move beyond the mean position towards the left. During this process, the spring is compressed and it exerts a restoring force on the block towards right. Once again, the force and displacement are oppositely directed. This opposing force retards the motion of the block, so that the Fig. 5.1 (a), (b) and (c): Spring mass oscillator. Fig. 5.1(b) shows the equilibrium position in which the spring exerts no force on the block. If the block is displaced towards the right from its equilibrium position, the force exerted by the spring on the block is directed towards the left [Fig. 5.1(a)]. On account of its elastic properties, the spring tends to regain its original shape and size and therefore it exerts a restoring force on the block. This is responsible to bring it back to the original position. This force is proportional to the displacement but its direction is opposite to that of the displacement. If x is the displacement, the restoring force f is given by, For such a motion, as a convention, we shall always measure the displacement from the mean position. Also, as the entire motion is along a single straight line, we need not use vector notation (only ± signs will be enough).
111 Activity Remember this Use your brain power speed goes on reducing and finally it becomes zero. This position is shown in Fig. 5.1(c). In this position the displacement from the mean position and restoring force are maximum. This force now accelerates the block towards the right, towards the equilibrium position. The process goes on repeating that causes the block to oscillate on either side of its equilibrium (mean) position. Such oscillatory motion along a straight path is called linear simple harmonic motion (S.H.M.). Linear S.H.M. is defined as the linear periodic motion of a body, in which force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position. If there is friction between a block and the resting surface, how will it govern the motion of the block? A complete oscillation is when the object goes from one extreme to other and back to the initial position. The conditions required for simple harmonic motion are: 1. Oscillation of the particle is about a fixed point. 2. The net force or acceleration is always directed towards the fixed point. 3. The particle comes back to the fixed point due to restoring force. Harmonic oscillation is that oscillation which can be expressed in terms of a single harmonic function, such as x a fi fisinfft or x a fi ficosfft Non-harmonic oscillation is that oscillation which cannot be expressed in terms of single harmonic function. It may be a combination of two or more harmonic oscillations such as x = a sinωt + b sin2ωt , etc. Some experiments described below can be performed in the classroom to demonstrate S.H.M. Try to write their equations. 5.4 Differential Equation of S.H.M.: In a linear S.H.M., the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position. As seen in Eq. (5.1), f = - kx where k is force constant and x is displacement from the mean position. According to Newton’s second law of motion, f = ma ∴ ma = - kx --- (5.3) The velocity of the particle is, vff= dx dt (a) A hydrometer is immersed in a glass jar filled with water. In the equilibrium position it floats vertically in water. If it is slightly depressed and released, it bobs up and down performing linear S.H.M. (b) A U-tube is filled with a sufficiently long column of mercury. Initially when both the arms of U tube are exposed to atmosphere, the level of mercury in both the arms is the same. Now, if the level of mercury in one of the arms is depressed slightly and released, the level of mercury in each arm starts moving up and down about the equilibrium position, performing linear S.H.M.
112 Can you tell? and its acceleration, a d dt = v = d x dt 2 2 Substituting it in Eq. (5.3), we get m d x dt kx 2 2 fi ff fi ff fififififi ffl d x dt k m x 2 2 0 --- (5.4) Substituting k m fiff2 , where ω is the angular frequency, d x dt x 2 2 2 fi ff ffl 0 --- (5.5) Eq. (5.5) is the differential equation of linear S.H.M. ∴ d x dt x 2 2 2 fi ffffl --- (5.6) But a d x dt = 2 2 is the acceleration of the particle performing S.H.M. ∴ a x fi ffffl2 --- (5.7) This is the expression for acceleration in terms of displacement x. From Eq. (5.6), we have d x dt x 2 2 2 fi ffffl ∴ d dt dx dt x fi ff ffl ffi fl 2 fi ff ffl d dt x v ffi2 fi ff ffl d dx dx dt x vff ffi2 fi ff v ffl dv dx ffi x 2 fi ff v vff ff d x fflffi dx 2 Integrating both the sides, we get fi fi v vff ff d x ff fflffi dx 2 fi ff ffl ffi v 2 2 2 2 2 fl x C , --- (5.8) where C is the constant of integration. Let A be the maximum displacement (amplitude) of the particle in S.H.M. When the particle is at the extreme position, velocity (v) is zero. Thus, at x A fi fi fi ff ,fififififi v 0 fi Substituting in Eq. (5.8), we get 0 2 2 2 fi ff ffl ffi A C fi ff C ffl ffi A 2 2 2 --- (5.9) Using C in Eq. (5.8), we get v 2 2 2 2 2 2 2 2 fi ff ffl ffi ffi x A fi ff v ffi fl ffl 2 2 2 2 A x fi ff v ffl ffi fl A x 2 2 --- (5.10) This is the expression for the velocity of a particle performing linear S.H.M. in terms of displacement x . Substituting v = dx dt in Eq. (5.10), we get Why is the symbol ω and also the term angular frequency used for a linear motion? 5.5 Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.: We can obtain expressions for the acceleration, velocity and displacement of a particle performing S.H.M. by solving the differential equation of S.H.M. in terms of displacement x and time t. From Eq. (5.5), we have d x dt x 2 2 2 fi ff ffl 0 Example 5.1 A body of mass 0.2 kg performs linear S.H.M. It experiences a restoring force of 0.2 N when its displacement from the mean position is 4 cm. Determine (i) force constant (ii) period of S.H.M. and (iii) acceleration of the body when its displacement from the mean position is 1 cm. Solution: (i) Force constant, k = f / x = (0.2)/ 0.04 = 5 N/m (ii) Period T fi fi fi 2ff ffl/ = 2 2 0 2 5 fi fi m k ff . ffiffiffiffiffiffi= 0.4π s (iii) Acceleration a x k m fi fi x . fi ff fiff fiff ffl fi . ff ff ffi2 2 5 0 2 0 04 1 m s
113 dx dt fi ff ffl A x 2 2 fi ff ffl dx A x dt 2 2 ffi Integrating both the sides, we get fi fi ff ffl dx A x dt 2 2 ffi sinfi ff ffl ffi fl 1 x A t --- (5.11) Here φ is the constant of integration. To know φ , we need to know the value of x at any instance of time t, most convenient being t = 0. fi ff x Afisinffi fl t ffl --- (5.12) This is the general expression for the displacement (x) of a particle performing linear S.H.M. at time t. Let us find expressions for displacement for two particular cases. Case (i) If the particle starts S.H.M. from the mean position, x = 0 at t = 0 Using Eq. (5.11), we get fi ff ffl ffi fl ffl sin ffi 1 0 x A orffi Substituting in Eq. (5.12), we get x A fi fffi sinffl ffi flt --- (5.13) This is the expression for displacement at any instant if the particle starts S.H.M. from the mean position. Positive sign to be chosen if it starts towards positive and negative sign for starting towards negative. Case (ii) If the particle starts S.H.M. from the extreme position, x A fi ff fifi fififiat 0 t fi fi ff ffl ffi fl ff sin 1 2 3 2 x A or Substituting in Eq. (5.12), we get x A fi fft x A t ffl ffi fl fi ff ffl ffi fl sin s in 2 3 2 or ∴ x A fi fffi cosfiffl ffi flt --- (5.14) This is the expression for displacement at any instant, if the particle starts S.H.M. from the extreme position. Positive sign for starting from positive extreme position and negative sign for starting from the negative extreme position. In the cases (i) and (ii) above, we have used the phrase, “if the particle starts S.H.M…...” More specifically, it is not the particle that starts its S.H.M., but we (the observer) start counting the time t form that instant. The particle is already performing its motion. We start recording the time as per our convenience. In other words, t = 0 (or initial condition) is always subjective to the observer. Expressions of displacement (x), velocity (v) and acceleration (a) at time t: From Eq. (5.12), x A fi ff fisinffl ffi fl t fi ff v ff ffl ffi fl dx dt A t cos fi ff a ff ffl ffi fl d dt A t v 2 sin Example 5.2: A particle performs linear S.H.M. of period 4 seconds and amplitude 4 cm. Find the time taken by it to travel a distance of 1 cm from the positive extreme position. Solution: x A fi ff fisinffl ffi fl t Since particle performs S.H.M. from positive extreme position, φ = π 2 and from data fifififi fi x A fi ff1 3 fi cm fi ff ffl ffi fl 3 4 2 2 ffi ffi ffi ffi sin ffi ffi T t ∴ 3 4 2 4 fi cos ff t = fi fi cos π 2 t ∴ fi fi 2 41 4 41 4 180 0 t ff ff ffl ffi fl . . c tfi fi =0 4. 6fis Or, . ffl ffl ffl ffl ffl . ffls 180 2 t t fi ff 41 4 0 fi 46 ffl ffi fl Example 5.3: A particle performing linear S.H.M. with period 6 second is at the positive extreme position at t = 0. The particle is found to be at a distance of 3 cm from this position at time t = 7s, before reaching the mean position. Find the
114 Can you tell? amplitude of S.H.M. Solution: x A fi ff sinffi( ) ffl ffi t Since particle starts (t = 0) from positive extreme position, φ = π/2 and x A fi ff 3 fi ff ffl ffi fl x Afi fi sin t 2 fi ff ffl ffi fl A A T 3 t 2 2 ffl fflsinffl fi ff ffl ffi fl A A 3 2 6 7 2 sin fi ff ffl ffi fl A A 3 7 3 2 sin fi ff ffl ffi fl ffl ffl A A 3 3 2 3 1 2 sin cos fi ff fi fi 2 6 A Affl fi ff A 6ffcm Example 5.4: The speeds of a particle performing linear S.H.M. are 8 cm/s and 6 cm/s at respective displacements of 6 cm and 8 cm. Find its period and amplitude. Solution: vfffi ff fl ffl ffi A x 2 2 ∴ 8 6 6 8 4 3 36 64 2 2 2 2 2 2 fi ffl ffi ff ffl ffi ff fi ffl ffi ff ffl ffi ff fl fl fl fl A A A A or ∴ A = 10fflcm v1 2 1 2 ffi ffi fi ff fl ffl ffi A x fi ff8 ffi fl ffl fi 2 10 6 2 2 T 8 2 fi 8ffl ff T ∴ fififiT fi fi = fi . 6 28s Extreme values of displacement (x), velocity (v) and acceleration (a): 1) Displacement: The general expression for displacement x in S.H.M. is x t fi ff Afisinffl ffi fl At the mean position, ffi fl fi ff t ffl = 0 or π ∴ xmin = 0. Thus, at the mean position, the displacement of the particle performing S.H.M. is minimum (i.e. zero). At the extreme position, fi ff ffl fl t ffi 2 or 3 2 π fi ff x t Afisinffi fl ffl fi ff x fflfi fi A sin ffi 2 fifi ∴ x A max fi ff Thus, at the extreme position the displacement of the particle performing S.H.M. is maximum. 2) Velocity: According to Eq. (5.10) the magnitude of velocity of the particle performing S.H.M. is v fi ffffi A x ffl2 2 At the mean position, x A max fi fi fi fi ff ffl0 fi fi v ff ffi fl . Thus, the velocity of the particle in S.H.M. is maximum at the mean position. At the extreme position, x A fi ff fi ff fivmin fi fi0.fi Thus, the velocity of the particle in S.H.M. is minimum at the extreme positions. 3) Acceleration: The magnitude of the acceleration of the particle in S.H.M is ω2 x At the mean position xfi fi =0 , so that the acceleration is minimum. ∴ amin fi 0 . At the extreme positions x A fi ff , so that the acceleration is maximum amax = ω2 A 1. State at which point during an oscillation the oscillator has zero velocity but positive acceleration? 2. During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa? 3. During which part of the oscillation the two are along the same direction? Example 5.5: The maximum velocity of a particle performing S.H.M. is 6.28 cm/s. If the length of its path is 8 cm, calculate its period. Solution: fi . v fi fi max fi fi 6 28 2 ficm s cm s ff and fi fi A=fi fi 4 cm vmax A A T fi fi ff 2ffl ∴ 2 4 2 fi fi ff T ∴ T = 4ffs
115 This result shows that the particle is at the same position after a time 2fi ff . That means, the particle completes one oscillation in time 2fi ff . It can be shown that t T fi fi 2ff ffl is the minimum time after which it repeats. Hence its period T is given by T fi 2ff ffl From Eq.(5.4) and Eq.(5.5) fi2 ff ff k m force per unit displacement mass ff ff = acceleration per unit displacement fi ff T acceleration per unit displacement 2ffl ff ff ff ff Also, T = 2π m k --- (5.15) 5.6.3 Frequency of S.H.M.: The number of oscillations performed by a particle performing S.H.M. per unit time is called the frequency of S.H.M. In time T , the particle performs one oscillation. Hence in unit time it performs 1 T oscillations. Hence, frequency n of S.H.M. is given by n T k m fi fi fi 1 2 1 2 ff ffl ffl fflffl --- (5.16) Combination of springs: A number of springs of different spring constants can be combined in series (Figure A) or in parallel (Figure B) or both. Series combination (Figure A): In this case, all the springs are connected one after the other forming a single chain. Consider an arrangement of two such springs of spring constants k1 and k2 . If the springs are massless, each will have the same stretching force as f. For vertical arrangement, it will be the weight mg. If e1 and e2 are the respective extensions, we can write, f k e k e e f k e f k fi fi fi fi 1 1 2 2ff fi1 fi fi fi 1 2 2 and The total extension is e e e f k k fi ff fi ff ffl ffi fl 1 2 1 2 1 1 . 5.6: Amplitude(A), Period(T) and Frequency (n) of S.H.M. : 5.6.1 Amplitude of S.H.M.: Fig. 5.2 S.H.M. of a particle. Consider a particle P performing S.H.M. along the straight line MN (Fig. 5.2). The centre O of MN is the mean position of the particle. The displacement of the particle as given by Eq. (5.12) is x A fi ff fisinffl ffi fl t The particle will have its maximum displacement when fisinffi fl fi ff t ffl 1, i.e., when x A fi ff . This distance A is called the amplitude of S.H.M. The maximum displacement of a particle performing S.H.M. from its mean position is called the amplitude of S.H.M. 5.6.2 Period of S.H.M.: The time taken by the particle performing S.H.M. to complete one oscillation is called the period of S.H.M. Displacement of the particle at time t is given by x A fi ff fisinffl ffi fl t After a time t t fi ff ffl ffi fl 2 the displacement will be x A fi fft ffl ffi fl ff fisin 2 fi ff x Afisinffi fl t ffl ffl 2 fi ff x Afisinffi fl t ffl Example 5.6: The maximum speed of a particle performing linear S.H.M is 0.08 m/s. If its maximum acceleration is 0.32 m/ s2 , calculate its (i) period and (ii) amplitude. Solution: (i) fiamax vmax A A T fi fi fi ff ffl ffl ffl ffi 2 2 0 32 0 08 2. . fi ff T fifi fifififififi fi ff T s fi . 1 57 (ii) vmax A A T fi fi ff 2ffl fi ff A 2ffcm
116 5.7 Reference Circle Method: Figure 5.3 shows a rod rotating along a vertical circle in the x-y plane. If the rod is illuminated parallel to x-axis from either side by a linear source parallel to the rod, as shown in the Fig. 5.3, the shadow (projection) of the rod will be produced on the y-axis. The tip of this shadow can be seen to be oscillating about the origin, along the y-axis. If ks is the effective spring constant (as if there is a single spring that gives the same total extension for the same force), we can write, e f k f k k k k k fi fi ff ffl ffi fl fi ff s s 1 1 1 1 1 1 2 1 2 For a number of such (massless) springs, in series, 1 1 1 1 1 2 k k k k s i i fi ff fffflffl fi ffi fl For only two massless springs of spring constant k each, in series, k k k k k s fi ff fi 1 2 1 2 Product Sum For n such identical massless springs, in series, k k n s = Fig. B Let f k e f k e 1 1 fi fi ,ffi ,ffi 2 2 ffff be the individual restoring forces. If k p is the effective spring constant, a single spring of this spring constant will be stretched by the same extension e, by the same stretching force f. fi ff f k pe f ff ffl1 2 f k fflffiff ffl 1 2 e k e fflffi fi ff k k p i 1 2 ffl ffl k k ffi ff fl For m such identical massless springs of spring constant k each, in parallel, k mk p = Fig 5.3: Projection of a rotating rod. We shall now prove that motion of the tip of the projection is an S.H.M. if the corresponding motion of the tip of the rod is a U.C.M. For this, we should take the projections of displacement, velocity, etc. on any reference diameter and confirm that we get the corresponding quantities for a linear S.H.M. Figure 5.4 shows the anticlockwise uniform circular motion of a particle P, with centre at the origin O. Its angular positions are decided with the reference OX. It means, if the particle is at E, the angular position is zero, at Parallel combination (Figure B): In such a combination, all the springs are connected between same two points, one of them is the support and at the other end, the stretching force f is applied at a suitable point. Irrespective of their spring constants, each spring will now have the same extension e. The springs now share the force such that in the equilibrium position, the total restoring force is equal and opposite to the stretching force f. Fig. A
117 F it is 90° = π 2 c , at G it is 180°= πc , and so on. If it comes to E again, it will be 360° = 2πc (and not zero). Let r = OP be the position vector of this particle. Projection of velocity: Instantaneous velocity of the particle P in the circular motion is the tangential velocity of magnitude rω as shown in the Fig. 5.5. Its projection on the reference diameter will be vy fi fi r r fl cos c fl fl os fl ffl ffi t ff . This is the expression for the velocity of a particle performing a linear S.H.M. Projection of acceleration: Instantaneous acceleration of the particle P in circular motion is the radial or centripetal acceleration of magnitude rω2 , directed towards O. Its projection on the reference diameter will be a r r t y y fiff fiff ffi fl ffl fi ff 2 2 2 sin sin . Again, this is the corresponding acceleration for the linear S.H.M. From this analogy it is clear that projection of any quantity for a uniform circular motion gives us the corresponding quantity of linear S.H.M. This analogy can be verified for any diameter as the reference diameter. Thus, the projection of a U.C.M. on any diameter is an S.H.M. 5.8 Phase in S.H.M.: Phase in S.H.M. (or for any motion) is basically the state of oscillation. In order to know the state of oscillation in S.H.M., we need to know the displacement (position), the direction of velocity and the oscillation number (during which oscillation) at that instant of time. Knowing only the displacement is not enough, because at a given position there are two possible directions of velocity (except the extreme positions), and it repeats for successive oscillations. Knowing only velocity is not enough because there are two different positions for the same velocity (except the mean position). Even after this, both these repeat for the successive oscillations. Hence, to know the phase, we need a quantity that is continuously changing with time. It is clear that all the quantities of linear S.H.M. (x, v, a etc) are the projections taken Fig 5.4: S.H.M. as projection of a U.C.M. At t = 0, let the particle be at P0 with reference angle φ . During time t, it has angular displacement ωt . Thus, the reference angle at time t is fi ff ffi fl t ffl . Let us choose the diameter FH along y-axis as the reference diameter and label OM as the projection of r = OP on this. Projection of displacement: At time t, we get the projection or the position vector OM = OPsin s fi ff ffi ffi y r in t flffl . This is the equation of linear S.H.M. of amplitude r. The term ω can thus be understood as the angular velocity of the reference circular motion. For linear S.H.M. we may call it the angular frequency as it decides the periodicity of the S.H.M. In the next section, you will come to know that the phase angle fi ff ffi fl t ffl of the circular motion can be used to be the phase of the corresponding S.H.M. Fig 5.5: Projection of velocity.
118 on a diameter, of the respective quantities for the reference circular motion. The angular displacement fi ff ffi fl t ffl can thus be used as the phase of S.H.M. as it varies continuously with time. In this case, it will be called as the phase angle. Special cases: (i) Phase θ = 0 indicates that the particle is at the mean position, moving to the positive, during the beginning of the first oscillation. Phase angle fi ff ffl 360 2 0 florfl c is the beginning of the second oscillation, and so on for the successive oscillations. (ii) Phase fi ff ffl 1800 ffi ffi or c indicates that during its first oscillation, the particle is at the mean position and moving to the negative. Similar state in the second oscillation will have phase fi ff ffl ffi fl 360 180 2fl ffiff 0 or c , and so on for the successive oscillations. (iii) Phase fi ff ffl ffi fl 90 2 0 ffiorffi c indicates that the particle is at the positive extreme position during first oscillation. For the second oscillation it will be fi ff ff ffl ffi fl ffi 360 90 2 2 0 or c , and so on for the successive oscillations. (iv) Phase fi ff ffl ffi fl 270 3 2 0 florfl c indicates that the particle is at the negative extreme position during the first oscillation. For the second oscillation it will be fi ff ff ffl ffi fl ffi 360 270 2 3 2 0 or c , and so on for the successive oscillations. Example 5.7: Describe the state of oscillation if the phase angle is 11100 . Solution: 1110 3 360 30 0 0 0 fi ff ffl 3 3600 × plus something indicates 4th oscillation. Now, A A sin 30 2 0 = Thus, phase angle 11100 indicates that during its 4th oscillation, the particle is at +A/2 and moving to the positive extreme. Example 5.8: While completing its third oscillation during linear S.H.M., a particle is at − 3 2 A , heading to the mean position. Determine the phase angle. Solution: A A c c sin ffi fi fi ff ff ff ff 1 1 3 2 3 2 3 ffl ffi fl ffl ffi orffi From negative side, the particle is heading to the mean position. Thus, the phase angle is in the fourth quadrant for that oscillation. fi ff ffl ffi fl 1 2 3 c As it is the third oscillation, phase fi ff fi fi ff ff ff ff ff ff ffl ffi fl ffl fl ffl ffl 2 2 4 2 3 6 3 17 3 1 c 5.9. Graphical Representation of S.H.M.: (a) Particle executing S.H.M., starting from mean position, towards positive: As the particle starts from the mean position Fig (5.6), towards positive, φ = 0 fi ff displacementx Asinfflt Velocity v = A t ω ω ficos Acceleration fi fi a A fi ff ffl ffl sin t 2 (t) 0 T/4 T/2 3T/4 T 5T /4 (θ ) 0 π 2 π 3 2 π 2π 5 2 π (x) 0 A 0 - A 0 A (v) Aω 0 - Aω 0 Aω 0 (a) 0 -Aω2 0 Aω2 0 -Aω2 Conclusions from the graphs: • Displacement, velocity and acceleration of S.H.M. are periodic functions of time. • Displacement time curve and acceleration time curves are sine curves and velocity time curve is a cosine curve. • There is phase difference of π/2 radian between displacement and velocity. • There is phase difference of π/2 radian between velocity and acceleration.
119 • There is phase difference of π radian between displacement and acceleration. • Shapes of all the curves get repeated after 2π radian or after a time T. 5.10 Composition of two S.H.M.s having same period and along the same path: Consider a particle subjected simultaneously to two S.H.M.s having the same period and along same path (let it be along the x-axis), but of different amplitudes and initial phases. The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant. Equations of displacement of the two S.H.M.s along same straight line (x-axis) are x1 = A1 sin (ωt + φ1 ) and x2 = A2 sin (ωt + φ2 ) The resultant displacement (x) at any instant (t) is given by x = x1 + x2 x = A1 sin (ωt + φ1 ) + A2 sin (ωt + φ2 ) ∴ x = A1 sin ωt cos φ1 + A1 cos ωt sinφ1 + A2 sin ωt cos φ2 + A2 cos ωt sin φ2 A1 , A2 ,φ1 and φ2 are constants and ωt is variable. Thus, collecting the constants together, x = (A1 cosφ1 + A2 cos φ2 ) sin ωt + (A1 sin φ1 + A2 sin φ2 ) cos ωt As A1 , A2 , φ1 and φ2 are constants, we can combine them in terms of another convenient constants R and δ as Fig. 5.6: (a) Variation of displacement with time, (b) Variation of velocity with time, (c) Variation of acceleration with time. Fig. 5.7: (a) Variation of displacement with time, (b) Variation of velocity with time, (c) Variation of acceleration with time. (a) (b) (c) (b) Particle performing S.H.M., starting from the positive extreme position. As the particle starts from the positive extreme position Fig. (5.7), fi ffffl 2 fi ff displacement, s x A inffi fl t A ffl / 2 ff cost Velocity, v = dx dt d A t dt fi A t ff ffl fi ffi ff ffl cos fi sin fl fl fl Acceleration, fi fi a cos d dt d A t dt fi fi A t ffl ffi ff ffl ffi fi ff ffl ffi v fl fl fl fl sin 2 (t) 0 T/4 T/2 3T/4 T 5T /4 (θ ) π 2 π 3 2 π 2π 5 2 π 3 π (x) A 0 -A 0 A 0 (v) 0 -Aω 0 Aω 0 -Aω (a) -Aω2 0 Aω2 0 -Aω2 0 (a) (b) (c)
120 Activity R cos δ = A1 cos φ1 + A2 cosφ2 --- (5.17) and R sin δ = A1 sinφ1 + A2 sinφ2 --- (5.18) ∴ x = R (sin ωt cos δ + cos ωt sin δ) ∴ x = R sin (ωt + δ) This is the equation of an S.H.M. of the same angular frequency (hence, the same period) but of amplitude R and initial phase δ. It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring a long the same path is also an S.H.M. Resultant amplitude, R R fi ff ffl sin c fl fl ffi ff ffl R os 2 2 Substituting from Eq. (5.17) and Eq. (5.18), we get R2 = A1 2 + A2 2 + 2A1 A2 cos(φ1 - φ2 ) fi ff R A1 ffl ffl A A A fl ffi 2 2 2 1 2 1 2 ffl ffl 2 fflffl cos - (5.19) Initial phase (δ) of the resultant motion: Dividing Eq. (5.18) by Eq. (5.17), we get R R A A A A sin cos sin sin cos cos fi fi ff ff ff ff ffl ffi ffi 1 1 2 2 1 1 2 2 ∴ tan δ = A A A A 1 1 2 2 1 1 2 2 sin sin cos cos fi fi fi fi ff ff ∴ fi ff ff ff ff ffl ffi ffi fl tan sin sin cos cos 1 1 1 2 2 1 1 2 2 A A A A --- (5.20) Special cases: (i) If the two S.H.M.s are in phase, (φ1 - φ2 ) = 0° , ∴ cos (φ1 - φ2 ) = 1. fi ff R A1 ffl ffl A A A A ff ffifl ffl A 2 2 2 1 2 1 2 2 fflffl . Further, if A1 = A2 = A, we get R = 2A (ii) If the two S.H.M.s are 90° out of phase, (φ1 - φ2 ) = 90° ∴cos (φ1 - φ2 ) = 0. ∴ R A fi ff 1 A 2 2 2 Further, if A1 = A2 = A, we get, R = 2 A (iii) If the two S.H.M.s are 180° out of phase, (φ1 - φ2 ) = 180° ∴cos (φ1 - φ2 ) = -1 ∴ R A fi ff 1 A A ffl A 2 2 2 1 2 2 ∴ R A fi ff1 2 A Further, if A1 = A2 = A, we get R = 0 Tie a string horizontally tight between two vertical supports. To this string, tie three pendula, two of them (A and B) of equal lengths. Third one (C) need not have the same length, but not very different. Oscillate the pendula A and B in a plane perpendicular to the horizontal string. It will be observed that pendulum C also starts oscillating in the same plane, with the same period as those of A and B. With this system and procedure, we are imposing two S.H.M.s of the same period. The resultant energy transfers through the strings into the third pendulum C and it starts oscillating. Special cases (i), (ii) and (iii) above can be verified by making suitable changes. 5.11: Energy of a Particle Performing S.H.M.: While performing an S.H.M., the particle possesses speed (hence kinetic energy) at all the positions except at the extreme positions. In spite of the presence of a restoring force (except at the mean position), the particle occupies various positions. This is an indication that work is done and the system has potential energy (elastic - in the case of a spring, gravitational - for a pendulum, magnetic - for a magnet, etc.). Total energy of the particle performing an S.H.M. is thus the sum of its kinetic and potential energies. Consider a particle of mass m, performing a linear S.H.M. along the path MN about the mean position O. At a given instant, let the particle be at P, at a distance x from O. Fig. 5.8: Energy in an S.H.M. Velocity of the particle in S.H.M. is given as v fi ff A x fi ffl A t ffi fl 2 2 cos ,
121 Can you tell? where x is the displacement of the particle performing S.H.M. and A is the amplitude of S.H.M. Thus, the kinetic energy, E m A x k A x k fi ff ffl ffi fi ff ffl ffi 1 2 1 2 2 2 2 2 2 fl --- (5.21) This is the kinetic energy at displacement x. At time t, it is E m mA t kA t k fi fi ffl ffi ff fi ff ffl ffi 1 2 1 2 1 2 2 2 2 2 2 2 v fl fl fl cos cos --- (5.22) Thus, with time, it varies as cos 2 θ . The restoring force acting on the particle at point P is given by f = - kx where k is the force constant. Suppose that the particle is displaced further by an infinitesimal displacement dx against the restoring force f. The external work done (dW) during this displacement is dW fi ff f d ffl ffi x k fi ff x d ffl ffi ff x fi kxdx The total work done on the particle to displace it from O to P is given by W dW kx dx kx x x fi fi fi ff ff 0 0 1 2 2 ffi ffi This should be the potential energy (P.E.) E p of the particle at displacement x. fi ff E kx m ff x p 1 2 1 2 2 2 2 ffl --- (5.23) At time t, it is E kx kA t mA t p fi fi ffl ffi ff fi ff ffl ffi 1 2 1 2 1 2 2 2 2 2 2 2 sin cos fl fl fl Thus, with time, it varies as sin2 θ . The total energy of the particle is the sum of its kinetic energy and potential energy. fi ff E Ek p ffl E Using Eq. (5.21) and Eq. (5.23), we get E m fi ff ffl ffi A x fl m x 1 2 1 2 2 2 2 2 2 E m fi fi A kA m fi ff ffl max 1 2 1 2 1 2 2 2 2 2 ffi ffl v ---(5.24) This expression gives the total energy of the particle at point P. As m, ω and A are constant, the total energy of the particle at any point P is constant (independent of x and t). In other words, the energy is conserved in S.H.M. If n is the frequency in S.H.M., fi ff ffl 2 n . Using this in Eq. (5.24), we get E m n A n A m m A T fi ff ffl fi fi 1 2 2 2 2 2 2 2 2 2 2 2 2 ffi ffi ffi --- (5.25) Thus, the total energy in S.H.M. is directly proportional to (a) the mass of the particle (b) the square of the amplitude (c) the square of the frequency (d) the force constant, and inversely proportional to square of the period. To start a pendulum swinging, usually you pull it slightly to one side and release. • What kind of energy is transferred to the mass in doing this? • Describe the energy changes that occur when the mass is released. • Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases? Special cases: (i) At the mean position, x = 0 and velocity is maximum. Hence E E fi ff ffl k fi m A max 1 2 2 2 ffi and potential energy fi ff E p ffl min 0 (ii) At the extreme positions, the velocity of the particle is zero and x A fi ff Hence E E fi ff ffl p fi m A max 1 2 2 2 ffi and kinetic energy fi ff Ek ffl min 0 As the particle oscillates, the energy changes between kinetic and potential. At the mean position, the energy is entirely kinetic; while at the extreme positions, it is entirely potential. At other positions the energy is partly kinetic and partly potential. However, the total energy is always conserved. (iii) Iffi . K E.fi=fi . P E.fi,fi 1 2 2 2 2 m A fi ffl ffi ff x fl fifi 1 2 2 2 m x ω fi ff ffl x A 2
122 Thus at x A fi ff 2 , the K.E. = P.E. = E 2 for a particle performing linear S.H.M. (iv) At x A fi ff 2 , P E. . fi fi ff ffl ffi fl fi 1 2 1 4 1 2 4 2 2 kx kA E fi ff K E. .ffl Pffl ffi .3 E Thus, at x A fi ff 2 , the energy is 25% potential and 75% kinetic. The variation of K.E. and P.E. with displacement in S.H.M. is shown in Fig. (5.9) The distance between the point of suspension and centre of gravity of the bob (point of oscillation) is called the length of the pendulum. Let m be the mass of the bob and T' be the tension in the string. The pendulum remains in equilibrium in the position OA, with the centre of gravity of the bob, vertically below the point of suspension O. If now the pendulum is displaced through a small angle θ (called angular amplitude) and released, it begins to oscillate on either side of the mean (equilibrium) position in a single vertical plane. We shall now show that the bob performs S.H.M. about the mean position for small angular amplitude θ . Fig. 5.9: Energy in S.H.M. Example 5.9: The total energy of a particle of mass 200 g, performing S.H.M. is 10-2 J. Find its maximum velocity and period if the amplitude is 7 cm. Solution: E m A E m v fi ff fi ffl ffi max ff 1 2 1 2 2 2 2 fl fflfflfflffl fflffl ∴ vmax E m = 2 fi ff ffl ff ffi vmax 2 10 0 2 0 3162 2 . . m s/ v v max max A T A T A fi fi ffl ff fi fi 2 2 ffi ffi 1 3. 9s 5.12 Simple Pendulum: An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support. A practical simple pendulum is a small heavy (dense) sphere (called bob) suspended by a light and inextensible string from a rigid support. Fig.5.10: Simple pendulum. In the displaced position (extreme position), two forces are acting on the bob. (i) Force T' due to tension in the string, directed along the string, towards the support and (ii) Weight mg, in the vertically downward direction. At the extreme positions, there should not be any net force along the string. The component of mg can only balance the force due to tension. Thus, weight mg is resolved into two components; (i) The component mg cosθ along the string, which is balanced by the tension T ' and (ii) The component mg sinθ perpendicular to the string is the restoring force acting on mass m tending to return it to the equilibrium position. Rigid support
123 ∴Restoring force, F = - mg sinθ --- (5.26) As θ is very small (θ <10°), we can write sinθ ≅ θ c fi ff F mfi–fi gffl From the Fig. 5.10, the small angle fi , ff ff x L ∴ F mg x L fi ff --- (5.27) As m, g and L are constant, F ∝- x Thus, for small displacement, the restoring force is directly proportional to the displacement and is oppositely directed. Hence the bob of a simple pendulum performs linear S.H.M. for small amplitudes. From Eq. (5.15), the period T of oscillation of a pendulum from can be given as, fi 2ff ffl = 2π acceleration pff ff er unit dff ff isplacement Using Eq. (5.27), F mg x L fi ff ∴ ma mg x L fi ff ∴ a g x L fi ff fi ff fifififi fififi ffl ff ffi fl a x g L g L infimagnitude Substituting in the expression for T, we get, T L g fi 2ff --- (5.28) The Eq. (5.28) gives the expression for the time period of a simple pendulum. However, while deriving the expression the following assumptions are made. (i) The amplitude of oscillations is very small (at least 20 times smaller than the length). (ii) The length of the string is large and (iii) During the oscillations, the bob moves along a single vertical plane. Frequency of oscillation n of the simple pendulum is n T g L fi fi 1 1 2ff --- (5.29) From the Eq. (5.28), we can conclude the following for a simple pendulum. (a) The period of a simple pendulum is directly proportional to the square root of its length. (b) The period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity. (c) The period of a simple pendulum does not depend on its mass. (d) The period of a simple pendulum does not depend on its amplitude (for small amplitude). These conclusions are also called the 'laws of simple pendulum'. 5.12.1 Second’s Pendulum: A simple pendulum whose period is two seconds is called second’s pendulum. Period T L g fi 2ff fi ff Forffiaffisecond s pendulum ffi ' , 2 2ffl L g s where Ls is the length of second’s pendulum, having period T = 2s. fi ff fi fi L g s ffl2 --- (5.30) Using this relation, we can find the length of a second’s pendulum at a place, if we know the acceleration due to gravity at that place. Experimentally, if Ls is known, it can be used to determine acceleration due to gravity g at that place. Example 5.10: The period of oscillations of a simple pendulum increases by 10%, when its length is increased by 21 cm. Find its initial length and initial period. Solution: fifififiT l g fi 2ff ∴ 100 110 1 2 = l l ∴ 10 11 0 21 1 1 fi ff l l . ∴ 1 21 0 21 1 1 1 1 . . l l fi ff ffl fi fl fl l m ∴ Period T l g fi fi 2 2 1 9 8 ff ff . =fi . fi2 006 s
124 Activity When you perform the experiment to determine the period of simple pendulum, it is recommended to keep the amplitude very small. But how small should it be? And why? To find this it would be better to measure the time period for different angular amplitudes. Let T L g 0 fi 2ff be the period for (ideally) very small angular amplitude and Tθ be the period at higher angular amplitude θ . Experimentally determined values of the ratio T T θ 0 are as shown in the table below. θ 20° 45° 50° 70° 90° T T θ 0 1.02 1.04 1.05 1.10 1.18 It shows that the error in the time period is about 2% at amplitude of 20°, 5% at amplitude of 50°, 10% at amplitude of 70° and 18% at amplitude of 90°. Thus, the recommended maximum angular amplitude is less than 20°. It also helps us in restricting the oscillations in a single vertical plane. Example 5.11: In summer season, a pendulum clock is regulated as a second’s pendulum and it keeps correct time. During winter, the length of the pendulum decreases by 1%. How much will the clock gain or lose in one day. (g = 9.8 m/s2 ) Solution: In summer, with period Ts = 2 s, the clock keeps correct time. Thus, in a day of 86400 seconds, the clock’s pendulum should perform 86400 2 = 43200 oscillations, to keep correct time. L L w s fi fi ffff1 0 %fl lessflthanflsummer .99 T L g fi 2ff fi ff fiffl fiffl fi ffl T L T T L L T T w s w s w w ff ffff . ff . 2 0 99 1 99 s With this period, the pendulum will now perform 86400 1 99 43417 . = oscillations per day. Thus, it will gain 43417 - 43200 = 217 oscillations, per day. Per oscillations the clock refers to 2 second. Thus, the time gained, per day = 217 × 2 = 434 second = 7 minutes, 14 second. Conical pendulum Simple pendulum 1 Trajectory and the plane of the motion of the bob is a horizontal circle Trajectory and the plane of motion of the bob is part of a vertical circle. 2 K.E. and gravitational P.E. are constant. K.E. and gravitational P.E. are interconverted and their sum is conserved. 3 Horizontal component of the force due to tension is the necessary centripetal force (governing force). Tangential component of the weight is the governing force for the energy conversions during the motion. 4 Period, T L g fi 2ff cosffl Period, T L g fi 2ff 5 String always makes a fixed angle with the horizontal and can never be horizontal. With large amplitude, the string can be horizontal at some instances. 6 During the discussion for both, we have ignored the stretching of the string and the energy spent for it. However, the string is always stretched otherwise it will never have tension (except at the extreme positions of the simple pendulum). Also, non-conservative forces like air resistance are neglected.
125 5.13: Angular S.H.M. and its Differential Equation: Figure 5.11 shows a metallic disc attached centrally to a thin wire (preferably nylon or metallic wire) hanging from a rigid support. If the disc is slightly twisted about the axis along the wire, and released, it performs rotational motion partly in clockwise and anticlockwise (or opposite) sense. Such oscillations are called angular oscillations or torsional oscillations. This motion is governed by the restoring torque in the wire, which is always opposite to the angular displacement. If its magnitude happens to be proportional to the corresponding angular displacement, we can call the motion to be angular S.H.M. Angular S.H.M. is defined as the oscillatory motion of a body in which the torque for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement. The time period T of angular S.H.M. is given by, T fi 2ff ffl fi 2ff angularaccelerationperunit angulardisplacement 5.13.1 Magnet Vibrating in Uniform Magnetic Field: If a bar magnet is freely suspended in the plane of a uniform magnetic field, it remains in equilibrium with its axis parallel to the direction of the field. If it is given a small angular displacement θ (about an axis passing through its centre, perpendicular to itself and to the field) and released, it performs angular oscillations Fig. (5.12). Let µ be the magnetic dipole moment and B the magnetic field. In the deflected position, a restoring torque acts on the magnet, that tends to bring it back to its equilibrium position. [Here we used the symbol µ for the magnetic dipole moment as the symbol m is used for mass]. The magnitude of this torque is fi ff ffl µ Bsin If θ is small, sinfi fi ff c fi ff ffl ffi µB For clockwise angular displacement θ , the restoring torque is in the anticlockwise direction. Fig. 5.11: Torsional (angular) oscillations. Thus, for the angular S.H.M. of a body, the restoring torque acting upon it, for angular displacement θ, is fi ff ffl ffi or fi ff ffl ffic --- (5.31) The constant of proportionality c is the restoring torque per unit angular displacement. If I is the moment of inertia of the body, the torque acting on the body is given by, fi ff ffl I Where α is the angular acceleration. Using this in Eq. (5.31) we get, I c fi ff ffl ffi fi ff I ffl d dt c 2 2 0 ffi ffl ffl ffi --- (5.32) This is the differential equation for angular S.H.M. From this equation, the angular acceleration α can be written as, fi ff ff ffl ffl ffi d dt c I 2 2 Since c and I are constants, the angular acceleration α is directly proportional to θ and its direction is opposite to that of the angular displacement. Hence, this oscillatory motion is called angular S.H.M. Fig. 5.12: Magnet vibrating in a uniform magnetic field.
126 ∴ τ fi fi I µ ffl ffi ff B where I is the moment of inertia of the bar magnet and α is its angular acceleration. fi ff fflffi fl µB I --- (5.33) Since µ, B and I are constants, Eq. (5.33) shows that angular acceleration is directly proportional to the angular displacement and directed opposite to the angular displacement. Hence the magnet performs angular S.H.M. The period of vibrations of the magnet is given by T fi fi 2 2 ff angularaccelerationperunit angulardisplacement ff ffl ffi fi ff T I µB 2ffl --- (5.34) 5.14 Damped Oscillations: Example 5.12: A bar magnet of mass 120 g, in the form of a rectangular parallelepiped, has dimensions l = 40 mm, b = 10 mm and h = 80 mm. With the dimension h vertical, the magnet performs angular oscillations in the plane of a magnetic field with period π s. If its magnetic moment is 3.4 A m2 , determine the influencing magnetic field. Solution: T I B fi 2ff ffl fi ff ffl ffl ffi 2 I B fi ff B 4I ffl For a bar magnet, moment of inertia I M l b fi ffl ff ffi fl 2 2 12 fi ff ffi ffl fl ff I 0 12 1600 100 12 10 1 7 10 6 5 2 . . Am fi ff ffl ffl ff ffl ffi ffi ffi B or 4 1 7 10 3 4 2 10 5 5 2 . . Wbm T Example 5.13: Two magnets with the same dimensions and mass, but of magnetic moments µ1 = 100 A m2 and µ2 = 50 A m2 are jointly suspended in the earth’s magnetic field so as to perform angular oscillations in a horizontal plane. When their like poles are joined together, the period of their angular S.H.M. is 5 s. Find the period of angular S.H.M. when their unlike poles are joined together. Solution: T I µB fi 2ff With like poles together, the effective magnetic moment is ff ffl µ µ 1 2 fi ∴ T I µ µ BH 1 1 2 fi 2 ffl ffi ff fl With unlike poles together, the effective magnetic moment is ff ffl µ µ 1 2 fi fi ff ffi fl ffl T I µ µ BH 2 1 2 2 ∴ T T µ µ µ µ 1 2 1 2 1 2 fi ffl ffi ff ffl ffi fl ∴ 5 1 3 T2 = fi ff T2 75 ff 8.665 s Fig. 5.13: A damped oscillator.
127 If the amplitude of oscillations of an oscillator is reduced by the application of an external force, the oscillator and its motion are said to be damped. Periodic oscillations of gradually decreasing amplitude are called damped harmonic oscillations and the oscillator is called a damped harmonic oscillator. For example, the motion of a simple pendulum, dies eventually as air exerts a viscous force on the pendulum and there may be some friction at the support. Figure 5.13 shows a block of mass m that can oscillate vertically on a spring. From the block, a rod extends to vane that is submerged on a liquid. As the vane moves up and down, the liquid exerts drag force on it, and thus on the complete oscillating system. The mechanical energy of the block-spring system decreases with time, as energy is transferred to thermal energy of the liquid and vane. The damping force (Fd ) depends on the nature of the surrounding medium and is directly proportional to the speed v of the vane and the block fi ff F b d fififi ffl v Where b is the damping constant and negative sign indicates that Fd opposes the velocity. For spring constant k, the force on the block from the spring is F kx s fi ff . Assuming that the gravitational force on the block is negligible compared to Fd and Fs , the total force acting on the mass at any time t is F F fi ffd s F fi ff ma F F d s ffl fi ff ma ffl ffl bv kx fi ff ma fi 0 bv ff ffl kx fi ff m ff ffl d x dt b dx dt kx 2 2 ffl ffl 0 --- (5.35) The solution of Eq. (5.35) describes the motion of the block under the influence of a damping force which is proportional to the speed. The solution is found to be of the form x Ae t bt m fi ff ffl ffi fl 2 ffcos ffff --- (5.36) Ae bt m fi ff ffl 2 ff is the amplitude of the damped harmonic oscillations. Fig. 5.14: Displacement against time graph. As shown in the displacement against time graph (Fig 5.14), the amplitude decreases with time exponentially. The term cosffl ffi fl fit ff shows that the motion is still an S.H.M. The angular frequency, fi ff ffl ffi fl k m b 2m 2 Period of oscillation, T = 2 2 2 2 fi ff fi ff ff ffl ffi fl k m b m The damping increases the period (slows down the motion) and decreases the amplitude. 5.15 Free Oscillations, Forced Oscillations and Resonance: Free Oscillations: If an object is allowed to oscillate or vibrate on its own, it does so with its natural frequency (or with one of its natural frequencies). For example, if the bob of a simple pendulum of length l is displaced and released, it will oscillate only with the frequency n g l fi 1 2ff which is called its natural frequency and the oscillations are free oscillations. However, by applying a periodic force, the same pendulum can be made to oscillate with different frequency. The oscillations then will be forced oscillations and the frequency is driver frequency or forced frequency.
128 Consider the arrangement shown in the Fig. 5.15. There are four pendula tied to a string. Pendula A and C are of the same length, pendulum B is shorter and pendulum D is longer. Pendulum A is having a solid rubber ball as its bob and will act as the driver pendulum or source pendulum. Other three pendula are having hollow rubber balls as their bobs and will act as the driven pendula. As the pendula A and C are of the same lengths, their natural frequencies are the same. Pendulum B has higher natural frequency as it is shorter and pendulum D is of lower natural frequency than that of A and C. absorbs maximum energy from the source. In such case, it is said to be in resonance with the source (pendulum A). For unequal natural frequencies on either side (higher or lower), the energy absorbed (hence, the amplitude) is less. If the activity is repeated for a set of pendula of different lengths and squares of their amplitudes are plotted against their natural frequencies, the plot will be similar to that shown in the Fig. 5.16. The peak occurs when the forced frequency matches with the natural frequency, i.e., at the resonant frequency. Fig 5.15: Forced oscillations. Pendulum A is now set into oscillations in a plane perpendicular to the string. In the course of time it will be observed that the other three pendula also start oscillating in parallel planes. This happens due to the transfer of vibrational energy through the string. Oscillations of A are free oscillations and those of B, C and D are forced oscillations of the same frequency as that of A. The natural frequency of pendulum C is the same as that of A, as it of the same length as that of A. It can also be seen that among the pendula B, C and D, the pendulum C oscillates with maximum amplitude and the other two with smaller amplitudes. As the energy depends upon the amplitude, it is clear that Pendulum C has absorbed maximum energy from the source pendulum A, while the other two absorbed less. It shows that the object C having the same natural frequency as that of the source Fig 5.16: Resonant frequency. In Chapter superposition of waves, you will see that most of the traditional musical instruments use the principle of resonance. In the topic AC circuits, the resonance in the L.C. circuits is discussed. Internet my friend 1. https://hyperphysics.phy-astr.gsu.edu/ hbase/shm.html 2. https://hyperphysics.phy-astr.gsu.edu/ hbase/pend.html 3. https://en.wikipedia.org/wiki/ simpleharmonicmotion 4. https://opentextbc.ca/physicstextbook 5. https://physics.info
129 1. Choose the correct option. i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is (A) 3 2 A (B) 2 3 A (C) A 2 (D) 1 2 A ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is (A) 36 J (B) 9 J (C) 27 J (D) 18 J iii) The length of second's pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface] (A) 1/6 m (B) 6 m (C) 1/36 m (D) 1 6 m iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio (A) 1:4 (B) 1:2 (C) 2:1 (D) 4:1 v) The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph? (A) The acceleration is maximum at time T. (B) The force is maximum at time 3T/4. (C) The velocity is zero at time T/2. (D) The kinetic energy is equal to total energy at time T/4. Exercises 2. Answer in brief. i) Define linear simple harmonic motion. ii) Using differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M. iii) Obtain the expression for the period of a simple pendulum performing S.H.M. iv) State the laws of simple pendulum. v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M. 3. Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M. 4. Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter. 5. Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position (b) the positive extreme position. Deduce your conclusions from the graph. 6. Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends. 7. Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends. 8. At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
130 9. In SI units, the differential equation of an S.H.M. is d x dt x 2 2 fi ff36 . Find its frequency and period. [Ans: 0.955 Hz, 1.05 s] 10. A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration 1 30 fi ff ffl ffi fl s after it has crossed the mean position. [Ans: 34.2 m/s2 ] 11. Potential energy of a particle performing linear S.H.M is 0.1 π2 x2 joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz] 12. The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s] 13. A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s] 14. A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 × 10-2 N] 15. Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2 . [Ans: Decreases by 0.0051 m] 16. At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm] 17. A particle performing linear S.H.M. of period 2π seconds about the mean position O is observed to have a speed of bfi fi 3 m / s , when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the time required by the particle, to travel a further distance b. [Ans: π/3 s] 18. The period of oscillation of a body of mass m1 suspended from a light spring is T. When a body of mass m2 is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m1 and m2 [Ans: 1/3] 19. The displacement of an oscillating particle is given by x a fi ff sin c ffl ffl t b os t where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A a fi ff b 2 2 ff ff 20. Two parallel S.H.M.s represented by x1 = 5sin (4π t + π/3) cm and x2 = 3sin (4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23'] 21. A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60o and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (π3 ≈ 31) [Ans: 0.04133 N m] 22. Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 × 10-5 Wb/m2 . The magnet has moment of inertia 3 × 10-6 kg/m2 and magnetic moment 3 A m2 . [Ans:38.19 osc/min.] 23. A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π2 ≈ 10 m s -2) [Ans: nmax = 1/s, E/m = 1.25 J/kg]
131 Can you recall? 6.1 Introduction: You may be familiar with different waves like water waves, sound waves, light waves, mechanical waves, electromagnetic waves etc. A mechanical wave is a disturbance produced in an elastic medium due to periodic vibrations of particles of the medium about their respective mean positions. In this process, energy and momentum are transferred from one particle to another. Thus, a wave carries or transfers energy from one point to another., but there is no transfer of matter or particles of the medium in which the wave is travelling. Another type of waves, known as electromagnetic waves, do not require material medium for their propagation; these are non-mechanical waves. We have studied sound waves (which are mechanical waves), their properties and various phenomena like echo, reverberation, Doppler effect related to these waves in earlier classes. In this Chapter, we will study mechanical waves, reflection of these waves, principle of superposition of waves, various phenomena like formation of stationary waves, beats, and their applications. 6.2 Progressive Wave: Have you seen ripples created on the surface of water when a stone is dropped in it? The water is displaced locally where the stone actually falls in water. The disturbance slowly spreads and distant particles get disturbed from their position of rest. The wave disturbs the particles for a short duration during its path. These particles oscillate about their position of rest for a short time. They are not bodily moved from their respective positions. This disturbance caused by the stone is actually a wave pulse. It is a disturbance caused locally for a short duration. A wave, in which the disturbance produced in the medium travels in a given direction continuously, without any damping and obstruction, from one particle to another, is a progressive wave or a travelling wave e.g., the sound wave, which is a pressure wave consisting of compressions and rarefactions travelling along the direction of propagation of the wave. 6.2.1 Properties of progressive waves: 1) Each particle in a medium executes the same type of vibration. Particles vibrate about their mean positions performing simple harmonic motion. 2) All vibrating particles of the medium have the same amplitude, period and frequency. 3) The phase, (i.e., state of vibration of a particle), changes from one particle to another. 4) No particle remains permanently at rest. Each particle comes to rest momentarily while at the extreme positions of vibration. 5) The particles attain maximum velocity when they pass through their mean positions. 6) During the propagation of wave, energy is transferred along the wave. There is no transfer of matter. 7) The wave propagates through the medium 6. Superposition of Waves 1. What is wave motion? 2. What is a wave pulse? 3. What are common properties of waves? 4. What happens when a wave propagates? 5. What are mechanical waves? 6. What are electromagnetic waves? 7. How are mechanical waves different from electromagnetic waves? 8. What are sound waves?
132 Can you tell? with a certain velocity. This velocity depends upon properties of the medium. 8) Progressive waves are of two types - transverse waves and longitudinal waves. 9) In a transverse wave, vibrations of particles are perpenduclar to the direction of propogation of wave and produce crests and troughs in their medium of travel. In longitudinal wave, vibrations of particles produce compressions and rarefactions along the direction of propagation of the wave. 10) Both, the transverse as well as the longitudinal, mechanical waves can propagate through solids but only longitudinal waves can propagate through fluids. You might recall that when a mechanical wave passes through an elastic medium, the displacement of any particle of the medium at a space point x at time t is given by the expression y xfi ff , ff t f ffl ffi fi ff x t v --- (6.1) where v is the speed at which the disturbance travels through the medium to the right (increasing x). The factor ff ffl x t fi v appears because the disturbance produced at the point x = 0 at time t reaches the point x = x′ on the right at time (t + x′/v) or we say that the disturbance of the particle at time t at position x = x′ actually originated on the left side at time (t - x′/v). Thus Eq. (6.1) represents a progressive wave travelling in the positive x-direction with a constant speed v. The function f depends on the motion of the source of disturbance. If the source of disturbance is performing simple harmonic motion, the wave is represented as a sine or cosine function of (x - vt) multiplied by a term which will make (x - vt) dimensionless. Generally we represent such a wave by the following equation y xfi ff , ff t Affl ffi sinfi ff kx flt --- (6.2) where A is the amplitude of the wave, k = 2π/λ is the wave number, λ and ω are the wavelength and the angular frequency of the wave and v =ω /k is the speed. The SI units of k, λ and ω are rad m-1, m and rad s-1 respectively. If T is the time period of oscillation, then n = 1/T = ω /(2π) is the frequency of oscillation measured in Hz (s-1). If the wave is travelling to the left i.e., along the negative x-direction, then the equation for the disturbance is y xfi ff , ff t A ffl ffi sin ff fi ff kx flt --- (6.3) Fig. 6.1: Reflection of a wave pulse sent as a crest from a rarer medium to a denser medium. Example 1 • Take a long light string AB. Attach one end of the string to a rigid support at B. (Here, for the wave pulse traveling on the string, the string is the rarer medium and the rigid support acts as a denser medium.) • By giving a jerk to the free end A of the string, a crest is generated in the string. • Observe what happens when this crest moves towards B? • Observe what happens when the crest reaches B? What is the minimum distance between any two particles of a medium which always have the same speed if a sine wave travels through the medium? 6.3 Reflection of Waves: When a progressive wave, travelling through a medium, reaches an interface separating two media, a certain part of the wave energy comes back in the same medium. The wave changes its direction of travel. This is called reflection of a wave from the interface. Reflection is the phenomenon in which the sound wave traveling from one medium to another comes back in the original medium with slightly different intensity and energy. To understand the reflection of waves, we will consider three examples below. 6.3.1 Reflection of a Transverse Wave:
133 • Take a long light string AB. Attach the end B of the string to a ring which can slide easily on a vertical metal rod without friction. (Here string is the denser medium while end B attached to the sliding ring is at the interface of a rarer medium as it can move freely.) • Give a jerk to free end A of the string. • Observe what happens when crest reaches the point B attached to the ring. • Try to find the reason of the observed movement. Example 3 • Observe the part of wave pulse reflected back on the heavy string. • Produce a wave pulse as a crest on the light string Q moving towards the junction point O. • Observe the part of wave pulse reflected on the light string. • What difference do you observe when the wave pulse gets reflected on the light string and when the wave pulse gets reflected on the heavy string? • Try to find reasons behind your observations. In example 1, when crest moves along the string towards B, it pulls the particles of string in upward direction. Similarly when the crest reaches B at rigid support, it tries to pull the point B upwards. But being a rigid support, B remains at rest and an equal and opposite reaction is produced on the string according to Newton’s third law of motion. The string is pulled downwards. Thus crest gets reflected as a trough (Fig. 6.1) or a trough gets reflected as a crest. Hence from example 1, we can conclude that when transverse wave is reflected from a rigid support, i.e., from a denser medium, a crest is reflected as a trough and a trough is reflected as a crest. You have learnt in Xth and XIth Std. that there is a phase difference of π radian between the particles at a crest and at a trough. Therefore we conclude that there is a phase change of π radian on reflection from the fixed end, i.e., from a denser medium. In example 2, we observe that when the crest reaches the point B, it pulls the ring upwards and causes the ring to move upward. The wave is seen to get reflected back as a crest and no phase change occurs on reflection from a rarer medium (Fig. 6.2). In example 3, we find that a crest travelling from the heavy string gets reflected as a crest from the lighter string, i.e., reflection at the surface when a wave is travelling from a denser medium to a rarer medium causes a crest to be reflected as a crest (Fig. 6.3 (a)). Fig.6.2: Reflection of a wave pulse sent as a crest from a denser medium to a rarer medium. Fig. 6.3: Reflection of a crest from (a) denser medium (in this case a heavy string) and (b) rarer medium (in this case a light string). • Perform the same activity repeatedly and observe carefully. Try to find the reasons of movements in above observations. Example 2 • Take a heavy string P and a light string Q and join them. Suppose they are joined at point O. (Heavy string acts as a denser medium and light string is the rarer medium.) • Produce a wave pulse as a crest on the heavy string P moving towards the junction O. (a) (b)
134 But in example 3 (Fig. 6.3 (b)), when a crest travels from the lighter string to the heavy string, the crest is reflected as a trough and vice versa. 6.3.2 Reflection of a Longitudinal Wave: Consider a longitudinal wave travelling from a rarer medium to a denser medium. In a longitudinal wave compression is a high pressure region while rarefaction is a low pressure region. When compression reaches the denser medium, it tries to push the particles of that medium. But the energy of particles in the rarer medium is not sufficient to compress the particles of denser medium. According to Newton’s third law of motion, an equal and opposite reaction comes into play. As a result, the particles of rarer medium get compressed. Thus, when the longitudinal wave travels from a rarer medium to a denser medium, a compression is reflected as a compression and a rarefaction is reflected as a rarefaction. There is no change of phase during this reflection (Fig. 6.4). 6.4 Superposition of Waves: Suppose you wish to listen to your favourite music. Is it always possible particularly when there are many other sounds from the surroundings disturbing you. How can the background sounds be blocked? Of course, the mobile lover generation uses headphones and enjoys listening to its favorite music. But you cannot avoid the background sound completely. Why? We know that sound waves are longitudinal waves propagating through an elastic medium. When two waves travelling through a medium cross each other, each wave travels in such a way as if there is no other wave. Each wave sets the particles of the medium into simple harmonic motion. Thus each particle of the medium is set into two simple harmonic motions due to the two waves. The total displacement of the particles, at any instant of time during travelling of these waves, is the vector sum of the two displacements. This happens according to the principle of superposition of waves, which states that, when two or more waves, travelling through a medium, pass through a common point, each wave produces its own displacement at that point, independent of the presence of the other wave. The resultant displacement at that point is equal to the vector sum of the displacements due to the individual wave at that point. As displacement is a vector, we must add the individual displacements by considering their directions. There is no change in the shape and nature of individual waves due to superposition of waves. This principle applies to all types of waves like sound waves, light Fig. 6.4: Reflection of a longitudinal wave from a denser medium. Fig. 6.5: Reflection of a longitudinal wave from a rarer medium. When longitudinal wave travels from a denser medium to a rarer medium (Fig. 6.5), a compression is reflected as a rarefaction. Here reversal of phase takes place, i.e., phase changes by π radians. When compression reaches a rarer medium from denser medium, it pushes the particles of rare medium. Due to this, particles of the rarer medium get compressed and move forward and a rarefaction is left behind. Thus a compression gets reflected as a rarefaction. Similarly a rarefaction gets reflected as a compression (Fig. 6.5).
135 waves, waves on a string etc. and we say that interference of waves has taken place. You might have seen singers using a special type of headphones during recording of songs. Those are active noise cancellation headphones, which is the best possible solution to avoid background sound. Active noise cancellation headphones consist of small microphones one on each earpiece. They detect the ambient noise that arrives at the ears. A special electronic circuit is built inside the earpiece to create sound waveforms exactly opposite to the arriving noise. This is called antisound. The antisound is added in the earphones so as to cancel the noise from outside. This is possible due to superposition of waves, as the displacements due to these two waves cancel each other. The phenomena of interference, beats, formation of stationary waves etc. are based on the principle of superposition of waves. Fig. 6.6: Superposition of two wave pulses of equal amplitude and same phase moving towards each other. Fig. 6.7: Superposition of two wave pulses of equal amplitude and opposite phases moving towards each other. propagation after superposition are shown in Figs. 6.6 (a) to 6.6 (f). Suppose two waves cross each other between t = 2 s and t = 4 s, as shown in Figs. 6.6 (c), (d) and (e). Here the two wave pulses superpose, the resultant displacement is equal to the sum of the displacements (full line) due to individual wave pulses (dashed lines). This is constructive interference. The displacement due to wave pulses after crossing at t = 5 s and t = 6 s are shown in Figs. 6.6 (f) and (g). After crossing each other, both the wave pulses continue to maintain their individual shapes. 6.4.2 Superposition of Two Wave Pulses of Equal Amplitude and Opposite Phases Moving towards Each Other : Let us consider superposition of two wave pulses in two different ways. 6.4.1 Superposition of Two Wave Pulses of Equal Amplitude and Same Phase Moving towards Each Other : The propagation of approaching wave pulses, their successive positions after every second, their superposition and their The propagation of approaching wave pulses, their successive positions after every second, their superposition and propagation after superposition are shown in Fig. 6.7 (a) to Fig. 6.7 (e). These wave pulses superimpose at t = 2 s and the resultant displacement (full line) is zero, due to individual displacements (dashed lines) differing in phase exactly by 180°. This is destructive interference. Displacement due to one wave pulse is cancelled by the displacement due to the other wave pulse when they cross each other (Fig. 6.7 (c)). After crossing each other, both
136 the wave pulses continue and maintain their individual shapes. 6.4.3 Amplitude of the Resultant Wave Produced due to Superposition of Two Waves: Consider two waves having the same frequency but different amplitudes A1 and A2 . Let these waves differ in phase by ϕ . The displacement of each wave at x = 0 is given as y A t 1 1 fi sinff y A t 2 2 fi ff sinffl ffi fl According to the principle of superposition of waves, the resultant displacement at x = 0 is y y fi ff y 1 2 or, y A fi ff 1 2 ffl s sinfl fl t A inffl ffi t ff y A fi fft A t A ff t 1 2 2 ffl ffl sin s ffl fflin ffl c cos ffi ffl os ffl sinffi y A fi ff ffl ffi A t ff A t 1 2 2 cos s fl sin infl cos If we write A A 1 2 fi ff fflcosffl ffi A cos --- (6.4) and A A 2sin s fi ff ffl in --- (6.5) we get y A fi ff cos ffl ffi sin s t A inffl ffi cos t ∴ fi fi y A fi ff sinffl ffi fl t --- (6.6) This is the equation of the resultant wave. It has the same frequency as that of the interfering waves. The resultant amplitude A is given by squaring and adding Eqs. (6.4) and (6.5). A A A A A 2 2 2 2 1 2 2 2 2 2 cos s fi fi ffl ffi in fl ffl cos s ff ff ffl in A A A A A A 2 1 2 1 2 2 2 2 2 2 2 fi ff 2 cos c ffl ffl ff ff os sin ffl ∴ A A fi ff 1 A A ff A 2 1 2 2 2 2ffl cosffl --- (6.7) Special cases: 1. When ϕ = 0, i.e., the waves are in phase, the resultant amplitude is A A fi ff 1 A A ff fi A A ff A 2 1 2 2 2 1 2 2 2 0 ffi ( cos ffi ) = A A 1 2 + The resultant amplitude is maximum when ϕ = 0. If the amplitudes of the waves are equal i.e., A1= A2 = A (say), then the resultant amplitude is 2A. 2. When ϕ = π, i.e., the waves are out of phase, the resultant amplitude is A A fi ff 1 A A cos A ff fi A A ffl 2 1 2 2 2 1 2 2 2ffl ffi (ffl ) = |A1 - A2 | The resultant amplitude is minimum when ϕ = π. If the amplitudes of the waves are equal i.e., A1= A2 = A (say), then the resultant amplitude is zero. Thus, the maximum amplitude is the sum of the two amplitudes when the phase difference between the two waves is zero and the minimum amplitude is the difference of the two amplitudes when the phase difference between the two waves is π. The intensities of the waves are proportional to the squares of their amplitudes. Hence, when ϕ = 0 I A max m fifi( ax ) fi fi( fi) A A 2 1 2 2 ff ffl --- (6.8) and when ϕ = π I A min m fififi( in ) fi(fiA A fi) 2 1 2 2 ff ffl --- (6.9) Therefore intensity is maximum when the two waves interfere in phase while intensity is minimum when the two waves interfere out of phase. You will learn more about superposition of waves in Chapter 7 on Wave Optics. Example 6.1: The displacements of two sinusoidal waves propagating through a string are given by the following equations y x t 1 fi ff 4 2 sinffl ffi 0 30 y x t 2 fi ff 4 2 sinffl ffi 5 40 where x and y are in centimeter and t is in second. a) Calculate the phase difference between these two waves at the points x = 5 cm and t = 2 s. b) When these two waves interfere, what are the maximum and minimum values of the intensity? Solution: Given y x t 1 fi ff 4 2 sinffl ffi 0 30
137 6.5 Stationary Waves: We have seen the superposition of two wave pulses, having same amplitudes and either same phase or opposite phases, and changes in the resultant amplitude pictorially in section 6.4. We have also derived the mathematical expression for the resultant displacement when two waves of same frequency superimpose as given by Eqs. (6.4) to (6.6). Now we are going to study an example of superposition of waves having the same amplitude and the same frequency travelling in opposite directions. 6.5.1 Formation of Stationary Waves: Imagine a string stretched between two fixed points. If the string is pulled at the middle and released, we get what is know as a stationary wave. Releasing of string produces two progressive waves travelling in opposite directions. These waves are reflected at the fixed ends. The waves produced in the string initially and their reflected waves combine to produce stationary waves as shown in Fig. 6.8 (a). and y x t 2 fi ff 4 2 sinffl ffi 5 40 a) To find phase difference when x = 5 cm and t = 2 s: y1 4 20 5 30 2 4 100 60 4 40 fi ff ffi fl ffl ff fi ffl ffi fl fi sin sin sin y2 4 25 5 40 2 4 125 80 4 45 fi ff ffi fl ffl ff fi ffl ffi fl fi sin sin sin ∴ Phase difference is 5 radian because ϕ = |45 – 40| = 5 radian. b) To find the maximum and minimum values of the intensity : Amplitudes of the two waves are A1 = 4 cm and A2 = 4 cm, ∴ I A max fi ff ffl ffi 1 2 A fi ff ffl ffi fi 2 2 4 4 64 when the phase difference is zero and I A min fi ff ffl ffi 1 2 A fi ff ffl ffi fi 2 2 4 4 0 when the phase difference is π. Example 6.2: A progressive wave travels on a stretched string. A particle on this string takes 4.0 ms to move from its mean position to one of its extreme positions. The distance between two consecutive points on the string which are at their mean positions (at a certain time instant) is 2.0 cm. Find the frequency, wavelength and speed of the wave. Solution : ∴ frequency n = 1/T = (1/16) × 103 s-1 = 62. 5 Hz As shown in Fig. (b), points A, B, and C correspond to mean positions, but the string is moving in one direction at point A and in the opposite direction at point B. Thus, out of the two consecutive particles at their mean positions, one will be moving upwards while the other will be moving downwards. The distance between them is 2.0 cm. Therefore distance between two consecutive particles moving in the same direction will be 2 × 2 cm = 4 cm. Thus the wavelength λ = 4 cm = 0.04 m Speed of wave v = n × λ = 62.5 × 0.04 = 2.5 m/s. (a) (b) A particles takes 4.0 × 10-3 s to travel from its mean position to extreme position. This is a quarter of the complete oscillation as shown in Fig. (a). Hence, the particle will take 4 × 4.0 × 10-3 s = 16 × 10-3 s to complete one oscillation. Fig. 6.8 (a): Formation of stationary waves on a string. The two sides arrows indicate the motion of the particles of the string.
138 6.5.2 Equation of Stationary Wave on a Stretched String: Consider two simple harmonic progressive waves of equal amplitudes (a) and wavelength (λ) propagating on a long uniform string in opposite directions (remember 2π/λ = k and 2πn = ω). The equation of wave travelling along the x-axis in the positive direction is y a nt x 1 fi ff 2 ffl ffi fl sin{ } --- (6.10) The equation of wave travelling along the x-axis in the negative direction is y a nt x 2 fi ff 2 ffl ffi fl sin --- (6.11) When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as y y fi ff y 1 2 y a nt x a nt x fi ff ffl ffi fl ffl ffi fl sin s 2 2 in By using, sin s C D in sin C D C D fi ff ffl fi ffi fl ffl ffi fl 2 2 2 cos ,we get y a nt x fi 2 2ff ffl 2 ffsin ffi ffi fl cos y = 2 a fi s cos in 2 2 fi ff fi x ffl ffi nt or, --- (6.12) Using 2 2 a x ffcos A fi ff ffl in Eq. (6.12), we get y A fi sinffi( ff nt)2 As ω = 2πn, we get, y A fi sinfft . This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves. It may be noted that the terms in position x and time t appear separately and not as a combination 2π (nt ± x/λ). Hence, the wave is not a progressive wave. x is present only in the expression for the amplitude. The amplitude of the resultant wave is given as A a x fi 2 2 ffcos ff ffl . It is a periodic function of x i.e., the amplitude is varying periodically in space. The amplitudes are different for different particles but each point on the string oscillates with the same frequency ω (same as that of the individual progressive wave). All the particles of the string pass through their mean positions simultaneously twice during each vibration. The string as a whole is vibrating with frequency ω with different amplitudes at different points. The wave is not moving either to the left or to the right. We therefore call such a wave a stationary wave or a standing wave. Particles move so fast that the visual effect is formation of loops. It is therefore customary to represent stationary waves as loops. In case of a string tied at both the ends, loops are seen when a stationary wave is formed because each progressive wave on a string is a traverse wave. When two identical waves travelling along the same path in opposite directions interfere with each other, resultant wave is called stationary wave. Condition for node: Nodes are the points of minimum displacement. This is possible if the amplitude is minimum (zero), i.e., 2 2 a 0 x fflcos fi ff ffl , or, cos 2fi ff x = 0, or, 2 2 3 2 5 2 fi ff x fi fi fi fflffi ,ffiffi ,ffiffi ,………. ∴ x = fi fi fi 4 3 4 5 4 ,flfl ,flfl ,ffffff i.e., x p fi ff ffl ffi 2 1 4 fl where p = 1, 2, 3, ……… The distance between two successive nodes is λ 2 . Condition for antinode: Antinodes are the points of maximum displacement, i.e., A a fi fi fi ff2 ∴ 2 2 a 2 x ffcos a fi ff ffl ffi or, ficos 2 1 fi ff x ffl ffi ∴ fifi , , , 2 0 2 3 fi ff fi fi fi x ffl ffi
139 or, x fi ff 0 2 3 2 ,flfl ,flfl ,flfl , .. ffl ffl ffl i.e., x p fi ff 2 where p = 0, 1, 2,3…. The distance between two successive antinodes is λ 2 . Nodes and antinodes are formed alternately. Therefore, the distance between a node and an adjacent antinode is λ 4 . When sin ωt = 1, at that instant of time, all the particles for which cos kx is positive have their maximum displacement in positive direction. At the same instant, all the particles for which cos kx is negative have their maximum displacement in negative direction. When sin ωt = 0, all the particles cross their mean positions, some of them moving in the positive direction and some in the negative direction. Longitudinal waves e.g. sound waves travelling in a tube /pipe of finite length are relected at the ends in the same way as transverse waves along a string are reflected at the ends. Interference between these waves travelling in opposite directions gives rise to standing waves as shown in Fig. 6.8 (b). We represent longitudinal stationary wave by a loop but the actual motion of the particles is along the length of the loop and not perpendiculat to it. through a medium along the same path in opposite directions. 2. If two identical transverse progressive waves superimpose or interfere, the resultant wave is a transverse stationary wave as shown in Fig. 6.8 (a). • When a transverse stationary wave is produced on a string, some points on the string are motionless. The points which do not move are called nodes. • There are some points on the string which oscillate with greatest amplitude (say A). They are called antinodes. • Points between the nodes and antinodes vibrate with values of amplitudes between 0 and A. 3. If two identical longitudinal progressive waves superimpose or interfere, the resultant wave is a longitudinal stationary wave. Figure 6.8 (b) shows a stationary sound wave produced in a pipe closed at one end. • The points, at which the amplitude of the particles of the medium is minimum (zero), are called nodes. • The points, at which the amplitude of the particles of the medium is maximum (say A), are called antinodes. • Points between the nodes and antinodes vibrate with values of amplitudes between 0 and A 4. The distance between two consecutive nodes is λ 2 and the distance between two consecutive antinodes is λ 2 . 5. Nodes and antinodes are produced alternately. The distance between a node and an adjacent antinode is λ 4 . 6. The amplitude of vibration varies periodically in space. All points vibrate with the same frequency. 7. Though all the particles (except those at the nodes) possess energy, there is no propagation of energy. The wave is localized and its velocity is zero. Therefore, we call it a stationary wave. Fig. 6.8 (b): Figure on the left shows standing waves in a conventional way while figure on the right shows the actual oscillations of material particles for a longitudinal stationary wave. Points A and N denote antinodes and nodes respectively. 6.5.3 Properties of Stationary Waves: 1. Stationary waves are produced due to superposition of two identical waves (either transverse or longitudinal waves) traveling
140 Do you know? 6. All particles between two consecutive nodes are moving in the same direction and are in phase while those in adjacent loops are moving in opposite directions and differ in phase by 180° in stationary waves but in a progressive wave, phases of adjacent particles are different. • What happens if a simple pendulum is pulled aside and released? • What happens when a guitar string is plucked? • Have you noticed vibrations in a drill machine or in a washing machine? How do they differ from vibrations in the above two cases? • A vibrating tuning fork of certain frequency is held in contact with table top and vibrations are noticed and then another vibrating tuning fork of different frequency is held on table top. Are the vibrations produced in the table top the same for both the tuning forks? Why? Example 6.3: Find the distance between two successive nodes in a stationary wave on a string vibrating with frequency 64 Hz. The velocity of progressive wave that resulted in the stationary wave is 48 m s-1 . Solution: Given: Speed of wave = v = 48 m s-1 Frequency n = 64 Hz We have v fi nff ∴ λ = = = v n ff . 48 64 0 75m We know that distance between successive nodes = fi 2 0 75 2 ff . = 0.375 m 8. All the particles between adjacent nodes (i.e., in one loop) vibrate in phase. There is no progressive change of phase from one particle to another particle. All the particles in the same loop are in the same phase of oscillation, which reverses for the adjacent loop. Musical instruments such as violin, tanpura, are based on the principle of formation of stationary waves or standing waves. 6.6 Free and Forced Vibrations: The frequency at which an object tends to vibrate when hit, plucked or somehow disturbed is known as its natural frequency. In these vibrations, object is not under the influence of any outside force. When a simple pendulum is pulled aside and released, it performs free vibrations with its natural frequency. Similarly when a string of guitar is plucked at some point it performs free vibrations with its natural frequency. In free vibration, the body at first is given an initial displacement and the force is then withdrawn. The body starts vibrating and continues the motion on its own. No external force acts on the body further to keep it in motion. Free vibration of a system means that the system vibrates at its natural frequency. In case of free vibrations, a body continuously 6.5.4 Comparison of Progressive Waves and Stationary Waves: 1. In a progressive wave, the disturbance travels form one region to the other with definite velocity. In stationary waves, disturbance remains in the region where it is produced, velocity of the wave is zero. 2. In progressive waves, amplitudes of all particles are same but in stationary waves, amplitudes of particles are different. 3. In a stationary wave, all the particles cross their mean positions simultaneously but in a progressive wave, this does not happen. 4. In progressive waves, all the particles are moving while in stationary waves particles at the position of nodes are always at rest. 5. Energy is transmitted from one region to another in progressive waves but in stationary waves there is no transfer of energy.
141 loses energy due to frictional resistance of surrounding medium. Therefore, the amplitude of vibrations goes on decreasing, the vibrations of the body eventually stop and the body comes to rest. The vibrations in a drill machine and in a washing machine are forced vibrations. Also the vibrations produced in the table top due to tuning forks of two different frequencies are different as they are forced vibrations due to two tuning forks of different frequencies. In forced vibrations, an external periodic force is applied on a body whose natural period is different from the period of the force. The body is made to vibrate with a frequency equal to that of the externally impressed force. The amplitude of forced vibrations depends upon the difference between the frequency of external periodic force and the natural frequency of the body. If this difference is small, the amplitude of forced vibrations is large and vice versa. If the frequencies exactly match, it is termed as resonance and the amplitude of vibration is maximum. An object vibrating with its natural frequency can cause another nearby object to vibrate. The second object absorbs the energy transmitted by the first object and starts vibrating if the natural frequencies of the two objects match. You have seen the example of two simple pendula supported from a string in the previous chapter. The second object is said to undergo forced vibrations. Strings or air columns can also undergo forced oscillations if the frequency of the external source of sound is close to the natural frequency of the system. Resonance is said to occur and we hear a louder sound. 6.7 Harmonics and Overtones: When a string or an air column is set into vibrations by some means, the waves are reflected from the ends and stationary waves can be formed. An important condition to form stationary waves depends on the boundary conditions that constrain the possible wavelengths or frequencies of vibration of the system. These are called the natural frequencies of normal modes of oscillations. The minimum of these frequencies is termed the fundamental frequency or the first harmonic. The corresponding mode of oscillations is called the fundmental mode or fundamental tone. The term overtone is used to represent higher frequencies. The first frequency higher than the fundamental frequency is called the first overtone, the next frequency higher is the second overtone and so on. The term 'harmonic' is used when the frequency of a particular overtone is an intergral multiple of the fundamental frequency. In strings and air columns, the frequencies of overtoners are integral multiples of the fundamental frequencies, hence they are termed as harmonics. But all harmonics may not be present in a given sound. The overtones are only those multiples of fundamental frequency which are actually present in a given sound. The harmonics may or may not be present in the sound so produced. To understand the concept of harmonics and overtones, let us study vibrations of air column. 6.7.1 End Correction: When an air column vibrates either in a pipe closed at one end or open at both ends, boundary conditions demand that there is always an antinode at the open end(s) (since the particles of the medium are comparatively free) and a node at the closed end (since there is hardly any freedom for the particles to move). The antinode is not formed exactly at the open end but it is slightly beyond the open end as air is more free to vibrate there in comparison to the air inside the pipe. Also as air particles in the plane of open end of the pipe are not free to move in all directions, reflection takes place at the plane at small distance outside the pipe.