19 45 Physics Term 3 STPM Chapter 19 Oscillations 27. (a) Refer to page 2 (b) (i) No (ii) Yes, from definition (iii) No (iv) Yes. T = 2π ω , ω = constant (v) No. Acceleration is positive when x is negative. (vi) Yes. F = ma, a always towards O. (vii) No. Displacement x = A sin ωt. A = constant 28. (a) • Spring obeys Hooke’s law. • Amplitude e, equilibrium extension. (b) T1 = 2π m k 2 x = force constant of one spring T2 = 2π m 2k T1 = 2T2 (c) T = 2π m k = 2π l g g = 4π2 l T2 = 4π2 (gradient of graph) g = 9.70 m s–2 29. (a) (i) Torque = F × d (d = distance of force F from axis of rotation) (ii) Torque = I d2 θ dt 2 (b) (i) I = m1 l 2 + 9m2 l 2 I = Σ(mr2 ) = (m1 + 9m2 )l 2 (ii) Torque = m1 gl sin θ + m2 g(3l sin θ) d2 θ dt 2 = – (m1 + 3m2 )gθ (m1 + 9m2 )l sin θ = θ, small T = 2π (m1 + 9m2 )l (m1 + 3m2 )g (iii) f = 1 T = 0.547 Hz (iv) Frequency decreases because I increases. 30. (a)(b) Refer to page 2 (c) (i) vmax = ωA ω = 2πf = 1.89 m s–1 Kmax = maximum potential energy in spring = 1 2 kx2 = 2.70 J (ii) Amplitude, A′ = 1 2 A K’ max = 1 2 Kmax = 4.24 cm Frequency, f′ = 2 f = 7.07 Hz 31. (a) Refer to page 29 (b) (i) mx.. = –kx – bx. (ii) T = 1.28 s T = 2π m k (iii) 4.71 × 10–2 J Energy loss = 1 2 k (A2 – a2 ) a = Ae– bt 2m , t = 10T 32. (a) Refer to page 31 (b) f = 1 2π g l , l = Length of test tube below liquid surface when in equilibrium. (c) 7.07 × 10–2 m f = v λ = 1 2π g l mg = (πd2 )l 4 ρg 33. (a) Energy lost from system (b) Resonance (c) (i) ω = 2πf = 78.6 rad s–1 (ii) T = 1 f = 0.080 s (d) (i) f Amplitude 0 (ii) f Amplitude 0 (e) Microwave oven ; magnetic resonance machine, radio receiver; musical instruments.
46 20 Physics Term 3 STPM Chapter 20 Wave Motion Chapter 20 Wave Motion 46 20 Physics Term 3 STPM CHAPTER Bilingual Keywords Antinode: Antinod Compression: Mampatan Disturbance: Gangguan Lagging: Menyusul Leading: Menganjur Longitudinal: Membujur Medium: Medium Node: Nod Phasor: Fasor Plane-polarised: Terkutub satah Progressive: Maju Rarefaction: Renggangan Refl ection : Pantulan Standing: Pegun Superposition: Superposisi Transverse: Melintang Wave pulse: Denyutan gelombang Wave: Gelombang Wavefront: Muka gelombang Stationary Wave Principle of Superposition Electromagnetic Waves Wave Intensity Progressive Wave Wave Motion WAVE MOTION 20 Concept Map
47 20 Physics Term 3 STPM Chapter 20 Wave Motion INTRODUCTION 1. Waves such as water waves and sound waves require a material medium for its propagation. These are examples of mechanical waves. 2. On the other hand, electromagnetic waves such as light can also propagate through a vacuum. 3. A wave is a mean of transfer of energy through vibration. A O –A A B C D E F G A H l v l Long stretched string Figure 20.1 : Wave produced in a long stretched string 4. Figure 20.1 shows one end of a long stretched string attached to a mass hanging from a vertical helical spring. 5. When the mass is pulled down a distance A from the equilibrium position and then released, it performs simple harmonic motion. 6. The end A of the string vibrates vertically and energy of the oscillating mass is transferred along the string in a form of wave. 7. Each particle of the string oscillates in the same manner as A with the same amplitude, A and frequency, f. 8. The amplitude, A of a wave is the maximum distance from the equilibrium position for particles in the medium. 9. The frequency, f of a wave is the number of complete oscillations per second for particles in the medium. 10. The period, T is the time for a complete oscillation. Period T = 1 f 11. The wavelength, λ is the distance between successive particles that oscillate in phase, example the distance between AE, BF or DH. 12. The velocity, v of a wave is the distance travelled per second by a wave pulse. 13. In a time equal to the period T, the distance travelled by a wave pulse is equal to the wavelength λ. Hence, wave velocity, v = distance travelled time taken = λ T ( 1 T = f ) v = f λ
48 20 Physics Term 3 STPM Chapter 20 Wave Motion 20.1 Progressive Waves Students should be able to: • interpret and use the progressive wave equation y = A sin (ωt − kx) or y = A cos (ωt − kx) • sketch and interpret the displacement-time graph and the displacement-distance graph • use the formula φ = 2πx λ • derive and use the relationship v = fλ Learning Outcomes Transverse Waves 1. The wave set up along the string shown in Figure 20.1 is an example of transverse wave. 2. A transverse wave is a wave which propagates in a form of oscillation that is perpendicular to the direction of propagation. 3. In transverse mechanical waves, the particles in the medium oscillate in a direction perpendicular to the direction of motion of the wave. 4. Examples of transverse waves are water waves and electromagnetic waves such as light. Longitudinal Waves 1. In longitudinal waves, the direction of oscillation is parallel to the direction of propagation of the wave. 2. Figure 20.2 shows the longitudinal wave set up in a stretched slinky spring when one end of the spring is pushed and pulled repeatedly. Compression λ λ Rarefaction Figure 20.2 : Longitudinal wave in a slinky spring 3. As the wave moves through the spring, compressions and rarefactions are set up along the spring. The wavelength λ = distance between two successive compressions = distance between two successive rarefactions λ λ Displacement Distance A –A 0 Figure 20.3 : Displacement-distance graph 4. Figure 20.3 shows the displacement of particles along the spring at a particular instant. The amplitude of the wave is A, and its wavelength is λ. 2009/P2/Q3, 2010/P1/Q13, 2010/P2/Q6, 2012/P1/Q12, 2013/P3/Q3,Q18 2015/P3/Q3, 2016/P3/Q3 VIDEO Waves
49 20 Physics Term 3 STPM Chapter 20 Wave Motion 5. Figure 20.4 shows the displacement-time graph of a particular particle in the spring as the wave travels through the spring. The period of the wave is T. Displacement Time A –A 0 T 2T Figure 20.4 : Displacement-time graph Example 1 The graph represents the simple harmonic motion of a particle in the medium where a wave travels at 5.0 m s–1. What is (a) the frequency of oscillation? (b) amplitude of oscillation? (c) the wavelength? Solution: (a) From the graph, period T = 20 ms = 20 × 10–3 s Frequency, f = 1 T = 1 20 × 10–3 = 50 Hz (b) Amplitude = Maximum displacement = 2 μm (c) Using v = f λ Wavelength, λ = v f = 5.0 50 = 0.1 m Progressive Waves 1. The wave produced in a stretched string when one end of it is moved up and down, and the longitudinal wave travelling along a slinky spring are examples of progressive waves. 2. In a progressive wave, the wave profi le moves in the direction of the propagation of the wave, and energy is transferred outwards from the wave source. 3. Figure 20.5 shows the position of the wave profi le at time = 0, and at time = t. The distance moved by the wave profi le in time = t is vt. Time / ms –2 0 +2 10 20 30 40 vt v Wave profile at time = t Wave profile at time = 0 Figure 20.5 : Progressive wave
50 20 Physics Term 3 STPM Chapter 20 Wave Motion 4. A progressive wave may be represented by an equation known as the progressive wave equation. The progressive wave equation represents the variation of the displacement y of a particle in the medium with time t. 5. Figure 20.6 shows a progressive wave of wavelength λ travelling in a medium from left to right with a speed v. 6. Suppose that the displacement y of the particle at O is represented by y = A sin ωt where A = amplitude of oscillation, ω = 2πf, f = frequency of the wave 7. The particle P at a distance x from O, vibrates in the same manner as the particle at O but lags behind by a phase difference of φ = x λ (2π) (λ ≡ 2π rad) = 2πx λ Hence, the displacement y of the particle at P, at a distance x from O is represented by y = A sin (ωt – φ) y = A sin (ωt – 2πx λ ) .............................. a = A sin (2πft – 2πx λ ) (ω = 2πf) = A sin ( 2πvt λ – 2πx λ ) (f = v λ ) y = A sin 2π λ (vt – x) .............................. b v λ = 1 T Hence, y = A sin 2π ( t T – x λ ) .............................. c 8. Equations a, b and c above are the different forms of the progressive waves equation. The phase difference –2πx λ shows that the wave is travelling from left to right, resulting in the oscillation of the particle at P lagging behind the oscillation of the particle at O. 9. For a progressive wave moving from right to left, the particle at P would oscillate first before the particle at O. Hence, the oscillation at P leads the oscillation at O by a phase difference of φ = –2πx λ . The progressive wave equations are y = A sin (ωt + 2πx λ ) y = A sin 2π λ (vt + x) y = A sin 2π ( t T + x λ ) -A A 0 P x v Distance, x Displacement, y λ Figure 20.6 Exam Tips Just remember one equation for progressive waves. y = A sin (ωt – 2πx λ ) and ω = 2πf v = f λ
51 20 Physics Term 3 STPM Chapter 20 Wave Motion Exam Tips ( ) ↑ ↑ (b) y = A sin 2πft + direction of wave propagation is negative (– x) direction ( ) (a) y = A sin 2πft – direction of wave propagation is positive (x) direction 2πx λ 2πx λ Example 2 A plane progressive wave is represented by the equation y = 0.2 sin (100 πt – 10πx 13 ) where y is in cm, t in second and x, the distance from the point O in metre. (a) Find (i) the frequency, (ii) the wavelength, (iii) the speed of the wave. (b) What is the phase difference between a point 0.25 m from O and another point 0.75 m from O? (c) Write the wave equation of a progressive wave having twice the amplitude, twice the frequency and moving in the opposite direction in the same medium. Solution: Comparing the equation y = 0.2 sin (100 πt – 10π 13 x ) with the wave equation y = A sin (ωt – 2πx λ ) (a) (i) ω = 2πf = 100π f = 50 Hz (ii) 2π λ = 10π 13 λ = 2 × 13 10 = 2.6 m (iii) Speed of wave, v = f λ = 50 × 2.6 = 130 m s–1 (b) Distance between the two points = (0.75 – 0.25) = 0.50 m One wavelength, λ corresponds to a phase difference of 2π radians Phase difference which corresponds to 0.50 m = 0.50 × 2π 2.6 = 1.21 rad (c) Amplitude, A = 2 × 0.2 = 0.4 cm Frequency, f = 2 × 50 = 100 Hz ω = 2πf = 200 π
52 20 Physics Term 3 STPM Chapter 20 Wave Motion Wavelength, λ = v f = 130 100 = 13 10 m 2πx λ = 2π 13 10 x = 20π 13 x Hence, progressive wave equation y = A sin (ωt + 2πx λ ) = 0.4 sin (200 πt + 20 π 13 x ) Example 3 A sinusoidal longitudinal wave travels along a slinky spring when a vibrator attached to one end of the spring vibrates with frequency of 25 Hz. The distance between two successive compressions is 24 cm. (a) Calculate the speed of the wave. (b) Deduce the wave equation, if the maximum displacement of a coil of the spring from the equilibrium position is 0.30 m and the wave travels in the negative (–x) direction. Assume that at time t = 0, the displacement of the spring at x = 0 is zero. Solution: (a) Frequency, f = 25 Hz Wavelength, λ = 24 cm Speed of wave, v = f λ = 25 × 24 = 600 cm s–1 (b) Amplitude, A = 0.30 m = 30 cm Wave equation, y = A sin (2πft + 2πx λ ) = 30 sin (2π × 25 t + 2π 24 x) = 30 sin (50π + πx 12 ) where x in cm, y in cm, and t in s. Quick Check 1 1. The diagram below shows an instantaneous position of a string as a transverse wave travels along it from left to right. 1 2 3 Which one of the following correctly shows the directions of the velocities of the points 1, 2, and 3 on the string? 1 2 3 A → ← → B ← → → C ↑ ↓ ↑ D ↑ ↓ ↓
53 20 Physics Term 3 STPM Chapter 20 Wave Motion 2. Figure (a) shows a stretched slinky spring in equilibrium. Figure (b) shows a snapshot of the slinky spring carrying a longitudinal wave moving from left to right. X Y Z Figure (b) Figure (a) Which of the following statements is correct? A The distance XY = wavelength B Y and Z are in phase C The displacement of X at that instant is zero D The instantaneous speed of Z is zero 3. The figure below is the displacement-time graph of a wave motion. Displacement 0 Time The graph shows the displacement of A particles along the medium at a particular instant B anyone particle in the medium at various instants C the particle at the source at the start of the wave motion D any one particle along the medium during a complete cycle 4. Figure (a) and (b) are the displacement-time graph and displacement-distance graph of a transverse wave. Displacement / cm 1 –1 0 Time / ms 20 40 60 80 Figure (a) Displacement / cm 1 –1 0 Distance / mm 2 4 6 8 Figure (b) What is the speed of the wave? A 0.05 m s–1 C 0.16 m s–1 B 0.10 m s–1 D 0.20 m s–1 5. A wave motion is represented by the equation y = A sin (ωt – kx) The graph below shows how the displacement y at a fixed point varies with time t. P R S Q y 0 A Time, t Which one of the labelled points shows the displacement equal to the position x = π 2k at time t = 0? A P C R B Q D S 6. A progressive wave is represented by y = A sin 2π ( t T + x λ ) The phase angle at a point, distance x from the wave source is given by A x λ C 2πx λ B – x λ D – 2πx λ 7. The figure below shows two sinusoidal curves R and S. _ R S y 0 A θ / rad. _ _ _ _ _ _ Which of the following pairs of equations represents the curves?
54 20 Physics Term 3 STPM Chapter 20 Wave Motion A yR = A cos θ; yS = A sin (θ – 5 8 π) B yR = A cos θ; yS = A cos (θ – 5 8 π) C yR = A cos θ; yS = A cos (θ + 5 8 π) D yR = A sin θ; yS = A sin (θ – 5 8 π) 8. Two sinusoidal waves M and N of the same frequency are shown in the figure. 0 1.0 M N y t / s What is the frequency and the phase relationship between M and N? Frequency/Hz Phase lead of N over M/rad A 0.4 – π 2 B 1.0 + π 4 C 2.5 + π 2 D 2.5 – π 2 9. A progressive wave is represented by the equation y = 5 sin (50 t + x) where x and y are in cm, and t in second. Find (a) the wave amplitude, (b) the wave frequency, (c) the wavelength, (d) the wave velocity. 10. The equation for a wave motion is represented by y = 6 sin 2π (4 t – 0.2 x) where x and y are in metre and t in second. Calculate (a) amplitude, (b) wavelength, (c) speed of the wave. 11. The equation of a transverse wave travelling along a string is y = 6.0 sin (0.20x + 5.0 t) where x and y are in cm and t in second. Calculate (a) amplitude, (b) wavelength, (c) frequency, (d) speed, (e) direction of propagation, (f) the maximum transverse speed of a particle on the string. 12. Write an equation representing a wave moving in the negative x-axis direction, having an amplitude of 0.50 m, frequency of 500 Hz, and a speed of 300 m s–1. 13. The graph shows a transverse wave travelling along a string in the x-direction at a particular instant. P, Q, R and S are points on the string. 0 x y P Q R S v Compare the instantaneous (i) velocity and (ii) acceleration of the points P, Q, R and S. State the direction of each. 14. (a) A wave travels along a stretched string at a speed of 5.0 cm s-1 in the +Ox direction. The variation of displacement with time t of a particle O is given by y = 2.0 sin 4πt. (i) Deduce the expression for the displacement y with time t of a particle which is at a distance x from O. (ii) What is the displacement y of the particle at x = 1.5 cm at time t = 3.0 s? (b) The equation of a transverse wave travelling along a string is y = 6.0 sin π(0.20x + 5.0t) where x and y are in cm and t in seconds. Calculate (i) wavelength, (ii) frequency, (iii) period, (iv) maximum transverse speed a particle of the string. (c) On the same axes, sketch the displacement-distance graph for (i) t = 0, and (ii) t = 0.20 s. Show the direction of wave propagation, and mark suitable values on both axes.
55 20 Physics Term 3 STPM Chapter 20 Wave Motion 20.2 Waves Intensity Students should be able to: • defi ne intensity and use the relationship I ∝ A2 • describe the variation of intensity with distance of a point source in space Learning Outcome 1. Energy is transferred outward from the source by progressive waves. 2. The intensity, I of a wave is the rate at which energy is transferred by the wave to a unit area perpendicular to the direction of propagation of the wave. 3. For a sinusoidal wave to move in a medium, particles of the medium should perform simple harmonic motion of total energy 1 2 mω2 A2 where ω = angular frequency and A = amplitude of oscillation. 4. Hence, the intensity I of a wave is directly proportional to the square of the amplitude. Intensity I ∝ (Amplitude)2 5. Figure 20.7 shows a point wave source emitting wave in all directions. If P is the energy emitted per second, then the intensity at a distance r from the source S is I = P 4πr2 (area of sphere of radius r is 4πr2 ) I ∝ 1 r2 Quick Check 2 1. Two progressive waves are represented by the equations y1 = 2 sin (5t – 3x) and y2 = 3 sin (5t – 3x) where y1 and y2 are in cm and t in seconds. Compare the intensity of y1 to y2 . 2. A point source emits waves in all directions. The intensity at a distance 2.0 m from the source is 10 W m-2. What is the intensity at a distance of (a) 1.0 m, and (b) 3.0 m from the source? 20.3 Principle of Superposition Students should be able to: • state the principle of superposition Learning Outcome 1. This principle states that the resultant displacement produced by two waves at a point is the vector sum of the individual displacements of the two waves. r I S Figure 20.7 2008/P1/Q11, 2016/P3/Q6, 2017/P3/Q3 VIDEO Superposition of Waves
56 20 Physics Term 3 STPM Chapter 20 Wave Motion 2. Figure 20.8 shows the application of the principle for two wave pulses on a string of the same amplitude A but travelling in opposite directions. (a) Wave pulses A and B approaching each other (b) During superposition, resultant displacement = A + A = 2A (c) After superposition, each wave continues moving in the same direction 3. When the resultant displacement is greater than the individual displacement, constructive superposition is said to have occurred. 4. Figure 20.9 shows destructive superposition of two waves having the same amplitude with displacements in opposite directions. Wave pulse B Wave pulse B Wave pulse A Resultant displacement = 0 Wave pulse A Figure 20.9 : Destructive superposition 5. Suppose that two progressive waves are represented by the equations y1 = A1 sin (ωt – 2πx λ ) and y2 = A2 sin (ωt – 2πx λ ) Using the principle of superposition, the resultant displacement y = y1 + y2 = A1 sin (ωt – 2πx λ ) + A2 sin (ωt – 2πx λ ) = (A1 + A2 ) sin (ωt – 2πx λ ) 0 0 0 A1 y1 –A1 Time A2 y2 –A2 Time (A1 + A2) –(A1 + A2) y = y1 + y2 Time Figure 20.10 : Superposition of waves Figure 20.8 : Superposition of waves A A 2A Wave pulse A Resultant displacement Wave pulse A Wave pulse B Wave pulse B (a) (b) (c)
57 20 Physics Term 3 STPM Chapter 20 Wave Motion Phasor Diagram 1. The resultant amplitude from the superposition of two waves can be obtained by drawing a phasor diagram. 2. A phasor is a rotating vector used to represent a sinusoidal quantity, such as the displacement of a particle performing simple harmonic motion. 3. Figure 20.11 is the phasor diagram representing the displacement of a particle y1 = A1 sin ωt The line OP is the phasor. Its length represent the amplitude A. The phasor is rotated about O with uniform angular velocity ω. 4. At time = t, the angular displacement is ωt. The projection of OP onto the y-axis is y1 = A1 sin ωt 5. Figure 20.12 shows the use of phasor diagram to find the resultant amplitude of two progressive waves y1 = A1 sin ωt and y2 = A2 (sin ωt + φ) where φ is the phase difference. y O P y1 A1 x ωt ωt y O Q y2 A2 x ωt + φ ωt φ y O Q P x φ ωt (i) Phasor for y1 = A1 sin ωt (iii) Phasor for (y1 + y2 ) OP = A1 PQ = A2 OQ = amplitude of y1 + y2 (ii) Phasor for y1 = A2 (sin ωt + φ) Figure 20.12 : Resultant of two phasors If y1 = A1 sin wt and y2 = A2 sin (wt + φ) then the resultant amplitude A of the superposition of y1 and y2 can be obtained by drawing the vector diagram (or phasor diagram) Figure (a) or Figure (b) below. A2' A A1 B C O A φ Figure (b) A A2' A1 C A O φ Figure (a) Exam Tips Info Physics Laser uses the principle of superposition to produce a beam of high intensity y O P y1 A1 x ωt ωt Figure 20.11 : Phasor diagram
58 20 Physics Term 3 STPM Chapter 20 Wave Motion Example 4 (a) On axes immediately above each other, sketch two sinusoidal waves each of amplitude A, and with phase difference such that, when superposed, the waves would produce (i) maximum constructive superposition (ii) maximum destructive superposition What is the value of the phase difference between the two waves in (a)? (b) The phase difference between two sinusoidal waves is π 3 rad, but the waves have the same amplitude A. By means of a phasor diagram, (i) fi nd the amplitude of the resultant wave in term of A. (ii) calculate the ratio of the intensity of the resultant wave compared to the total intensity of the two component waves. Comment on your answer in relation to the principle of conservation of energy. Solution: (a) (i) (ii) y = y1 + y2 y1 y2 0 Distance A –A 0 Distance A –A 0 Distance 2A –2A y = y1 + y2 y1 y2 0 Distance A –A 0 Distance A –A 0 Distance (no wave) 2A –2A Maximum constructive superposition Maximum destructive superposition Phase difference = 0 Phase difference = π radian
59 20 Physics Term 3 STPM Chapter 20 Wave Motion (b) π 3 radian = 60° In triangle PQR, use the cosine rule, PR2 = PQ2 + QR2 – 2 (PQ)(QR) cos 120° = A2 + A2 – 2A2 cos 120° = 3A2 ∴ PR = 3 A (c) Intensity I ∝(Amplitude)2 Intensity of resultant wave Total intensity of the two waves = ( 3 A)2 A2 + A2 = 3 2 The results do not contradict the principle of conversation of energy because the intensity of the resultant wave is greater than the total intensity of the two waves where there is constructive superposition, and intensity Ic ∝ ( 3 A)2 = 3A2 When destructive superposition occurs, the resultant amplitude is A as shown in the phasor diagram. Intensity ∝ A2 Average intensity ∝ 3A2 + A2 2 ∝ 2A2 Hence, Average intensity Total intensity = 2A2 2A2 = 1 Quick Check 3 1. S1 and S2 are two sources of waves which have the same wavelength and speed, but there is a phase difference of π radian between the two sources. A point P is equidistance from S1 and S2 . The wave froms S1 has an amplitude of 2A, and the wave from S2 is of amplitude A. Which of the graphs below shows correctly the resultant oscillation at P? A 0 Time Displacement A –A C 0 Time Displacement A –A –3A 3A B 0 Time Displacement A D 0 Time Displacement –3A 3A A y = y1 + y2 P A Q R 60° 120° y2 y1 (y1 – y2) A A A y2 y1 60°
60 20 Physics Term 3 STPM Chapter 20 Wave Motion 2. The energy transferred per second by a progressive wave is directly proportional to the square of its amplitude. If two progressive waves which are in phase and have different amplitudes are superposed, the energy per second transferred by the resultant wave is directly proportional to A the sum of the amplitudes C the square of the sum of the amplitudes B the difference of the amplitudes D the square of the mean of the amplitudes 3. A point source emits waves informly in all directions. At a distance of 5.0 m from the source, the amplitude of the wave is A. What is the wave amplitude at a distance of 3.0 m from the source? A 3 5 A C 9 25A B 5 3 A D 25 9 A 20.4 Standing Waves Students should be able to: • use the principle of superposition to explain the formation of standing waves • derive and interpret the standing wave equation • distinguish between progressive and standing waves Learning Outcomes 1. Energy is transferred by progressive waves. The profi le of a progressive wave moves in the direction of wave propagation. 2. A wave whose profi le does not move and no energy is transferred from one end of the medium to the other end is known as a standing wave or stationary wave. 3. Standing waves are produced in most musical instruments such as the guitar, drum and trumpet. 4. A Standing wave is produced by the superposition of two progressive waves of the same frequency, same amplitude but travelling in opposite directions. Load Pulley Vibrator String P A A A A N N N N N A Figure 20.13 : Stationary waves 5. Figure 20.13 shows the stationary wave set up in a stretched string when one end of the string attached to a vibrator is set oscillating and the distance of the vibrator from the pulley is adjusted until resonance occurs. 6. The wave from the source at P travels along the string to the end N and is refl ected. Superposition of the wave from P to N, with the refl ected wave travelling in the opposition direction, NP produces the standing wave as shown in Figure 20.13. 2009/P1/Q13, 2012/P2/Q10, 2013/P3/Q4, 2017/P3/Q4
61 20 Physics Term 3 STPM Chapter 20 Wave Motion 7. In Figure 20.14, you will notice that there are points marked N, which are not displaced at all. These points are known as nodes. 8. Between successive nodes, are points marked A, which oscillate with the maximum amplitude. These points are known as antinodes. (1) + = N N N N NNNN (2) (3) NNNN NNNN (1) (2) (3) + = (a) At time = t (c) At time = (t + T 2 ) NNNN NNNN + = (1) (2) (3) NNNN NNNN (1) (2) (3) + = (b) At time = (t + T 4 ) (d) At time = (t + 3T 4 ) Figure 20.14 9. Figure 20.15 shows graphically how a standing wave is produced by superposition of two progressive waves of the same frequency, amplitude and speed but travelling in opposite directions. λ N N N N (a) (c) (b) and (d) Figure 20.15
62 20 Physics Term 3 STPM Chapter 20 Wave Motion 10. Figure 20.15 shows the resultant wave at (a) time = t, (b) time = t + T 4 , (c) time = t + T 2 and (d) time = t + 3T 4 where T = period of wave. 11. Figure 20.15 shows that the distance between successive nodes is λ 2 . The distance between successive antinodes is also λ 2 . 12. The production of a standing wave from the superposition of two progressive waves of the same frequency, amplitude and speed but travelling in opposite directions can be deduced mathematically. Suppose the two progressive waves are represented by y1 = A sin (ωt – 2πx λ ) y2 = A sin (ωt + 2πx λ ) Using the principle of superposition of wave, the resultant wave is y = y1 + y2 = A sin (ωt – 2πx λ ) + A sin (ωt + 2πx λ ) = 2 A sin ωt cos 2πx λ = (2 A cos 2πx λ ) sin ωt = Ax sin ωt where Ax = 2A cos 2πx λ is the amplitude of the resultant wave. The value of Ax depends on the value of x. From Ax = 2A cos 2πx λ Ax = 0 or nodes occur when cos 2πx λ = 0 2πx λ = π 2 or 2πx λ = 3 2 π or 2πx λ = 5π λ ,... x = λ 2 or x = 3 4 λ or x = 5 4 λ Distance between nodes = ( 3 4 λ – λ 4 ) or ( 5 4 λ – 3 4 λ) = λ 2
63 20 Physics Term 3 STPM Chapter 20 Wave Motion 13. At the antinodes, A is maximum, and it occurs when )cos 2πx λ * = 1 2πx λ = 0 or 2πx λ = π, or 2πx λ = 2π ,... x = 0 or x = λ 2 , or x = λ Distance between antinode = ( λ 2 – 0) or (λ – λ 2 ) = λ 2 14. From the above discussion, the differences between progressive wave and standing wave are shown in the table below. Progressive wave Standing wave 1. Wave profi le travels in the direction of wave propagation. 1. Wave profi le is stationary. 2. Energy is transferred along the direction of wave propagation. 2. No energy is transferred along the wave. 3. No nodes. 3. No displacement at the nodes. 4. Neighbouring particles in the medium have the same amplitude of oscillation. 4. Particles between a node and an antinode have different amplitudes. 5. Neighbouring particles do not oscillate in phase. 5. Particles between two successive nodes oscillate in phase. Quick Check 4 1. Two progressive waves of the same frequency of 5.0 Hz, amplitude of 2.0 m and wavelength of 0.50 m are travelling in opposite directions along a straight line. Superposition of the two progressive waves produces a stationary wave with the maximum displacement y at x = 0 at time t = 0. Which is the equation of the stationary wave? A y = 2.0 cos 2πx sin 10πt C y = 4.0 cos 4πx sin 10πt B y = 2.0 cos 10πx sin 2πt D y = 4.0 cos 10πx sin 4πt 2. Which one of the following comparison between the characteristics of progressive wave and standing wave is correct? Progressive wave Standing wave A No medium required. Requires a medium. B Distance between two particles oscillating in phase is equal to the wavelength. Distance between a node and an antinode is half a wavelength. C The energy at any point is always kinetic energy. Energy at any point changes from kinetic energy to potential energy and vice versa. D The amplitude is the same at all points. Neighbouring points have different amplitudes.
64 20 Physics Term 3 STPM Chapter 20 Wave Motion 3. Superposition of two waves of the same frequency and amplitude but travel in opposite directions produces A a progressive wave of twice the amplitude C a standing wave of the same frequency B a progressive wave of twice the frequency D beats of the same frequency 20.5 Electromagnetic waves Students should be able to: • state that electromagnetic waves are made up of electrical vibrations E = E0 sin (ωt − kx) and magnetic vibrations B = B0 sin (ωt − kx); • state the characteristics of electromagnetic waves; • compare electromagnetic waves with mechanical waves; • state the formula c = 1 ε0 fi0 and explain its signifi cance; • state the orders of the magnitude of wavelengths and frequencies for different types of electromagnetic waves. Learning Outcomes 1. To understand the natural of electromagnetic waves, let us consider the transmission of radio waves from a dipole aerial. (a) (b) (c) Vertical electric field Dipole aerial + – Magnetic field Dipole aerial a.c. ~ supply c in the direction of the vector product E fi B E = E0 sin ( ωt – kx ) c B = B0 sin ( ωt – kx ) (d) Plane-polarized electromagnetic wave Figure 20.16 2. Figure 20.16 (a) shows the vertical electric fi eld between the poles of a dipole aerial when a battery is connected across the aerial. 3. Figure 20.16 (b) shows the magnetic fi eld around the wire carrying the current. The plane of the magnetic fi eld is perpendicular to the plane of the electric fi eld. 4. When a sinusoidal alternating voltage is connected to the dipole, (a) the electric fi eld across the dipole varies sinusoidally with time, and may be represented by the equation E = E0 sin ωt (b) the magnetic fi eld due to the alternating current may be represented by the equation B = B0 sin ωt 2014/P3/Q3, Q16
65 20 Physics Term 3 STPM Chapter 20 Wave Motion 5. The energy is transmitted from the dipoles in the form of waves, known as electromagnetic waves. 6. The electromagnetic wave is propagated in the form of vibrations of electric field and magnetic field which are mutually perpendicular to each other as shown in Figure 20.16 (d) and the direction of propagation. 7. The variation with time t for the electric field E at a distance x from the dipole is given by E = E0 sin (ωt – kx) where ω = 2πf, f = frequency of wave E0 = amplitude of electric field 8. Similarly, at a distance x from the dipole, the magnetic field is given by B = B0 sin (ωt – kx) where B0 = amplitude of magnetic field 9. Both the electric field E and magnetic field B are in phase. 10. The electromagnetic field shown in Figure 20.16(d) is plane polarized because the vibration of the electric field is limited to one plane, the vertical plane, and the vibration of the magnetic field in a plane perpendicular to that of the electric field. 11. Most electromagnetic waves, such as light waves from a lamp are not plane-polarized. The vibrations of the electric field, and the magnetic field are in an infinite number of planes as shown in Figure 20.17. Relationship between ε0, μ0 and c 1. In 1862 James Clerk Maxwell predicted the existence of a type of waves, which we know now as electromagnetic waves, based on his theoretical research on electromagnetic induction and the magnetic effect of a current. 2. According to Maxwell, all electromagnet waves travel in vacuum with the same speed given by c = ABBBB 1 ε0μ0 where ε0 = the permittivity of free space = 8.85 × 10–12 F m–1 and μ0 = permeability of free space = 4π × 10–7 H m–1 3. Substituting the values of ε0 and μ0 into the equation c = ABBBB 1 ε0μ0 = 1 (8.85 × 10–12) (4π × 10–7 ABBBBBBBBBBBBBBBBBB) = 3.0 × 108 m s–1 which equals the speed of light 4. The significant of this equation c = ABBBB 1 ε0μ0 is that it shows that light is a type of electromagnetic wave. The speed of light, 3.0 × 108 m s–1 had been determined before Maxwell made known his equation. Plane of vibration Figure 20.17 Info Physics Microwave oven uses the principle of resonance to generate heat in food. Frequency of microwave used in the oven is 2.5 GHz which causes water molecules in food to resonant.
66 20 Physics Term 3 STPM Chapter 20 Wave Motion Comparison between Electromagnetic Waves and Mechanical Waves 2009/P1/Q38, 2009/P1/Q39, 2009/P2/Q12(a), 2010/P1/Q37, 2010/P1/Q38, 1. Mechanical waves are produced by vibrations of mechanical systems. Examples of mechanical waves are waves along a stretched string, water waves and sound waves. 2. Table 20.1 shows the comparison of electromagnet waves with sound waves, an example of mechanical waves. Table 20.1 Properties Electromagnetic waves Sound waves (mechanical wave) 1. Able to transfer energy Yes Yes 2. Medium of propagation • Can travel in vacuum • Gas, liquids, and solids • A medium is not necessary for its propagation • Require a material medium such as gas, liquid or solid 3. Speed • All electromagnetic waves travel with the same speed in vacuum, 3 × 108 m s–1 • Slower in other medium • Cannot travel in a vacuum. • Slower speed e.g. speed of sound in air 330 m s–1 • Sound travelsfasterin liquids, and solids 4. Types of wave • All electromagnetic waves are transverse waves • Sound waves are longitudinal waves • There are mechanical waves which are transverse waves, e.g. water waves 5. Wave properties: Reflection, refraction, diffraction, interference, plane-polarization Yes Yes, except longitudinal waves such as sound waves cannot be plane-polarized Electromagnetic Spectrum Table 20.2 Electromagnetic Spectrum Types Wavelength (m) Sources Detector Radio waves > 1 Electrical oscillations in LC circuit Tuning circuit Microwaves 10–4 – 1 Electrical oscillations in LC circuit Tuning circuit Infrared 10–6 – 10–4 Hot objects Bolometer, thermopile Light 4 × 10–7, – 7 × 10–7 Sun, lamp Eye Ultraviolet 10–9 – 10–7 Sparks, discharge tube, sun Photo cell X-rays 10–11 – 10–9 Collisions of fast electrons with heavy metals Ionizing chamber G-M tube γ-rays <10–10 Radioactive decays G-M tube
67 20 Physics Term 3 STPM Chapter 20 Wave Motion 1. Table 20.2 shows the electromagnetic spectrum. Although the different types of electromagnetic waves travel with the same speed in vacuum, they have difference wavelengths and frequencies. 2. There is no clear cut boundaries between different types of electromagnetic waves. There is overlapping of wavelengths between two neighbouring types of electromagnetic waves. 3. The different types of electromagnetic waves exhibit wave properties such refl ection, refraction, diffraction, interference and plane-polarization. 4. The relationship between frequency v, wavelength λ and speed c is c = vλ 5. According to the Quantum theory, energy of electromagnetic waves is quantized, i.e. comes in discrete packets, or quanta. A packet of electromagnet radiation energy is known as photon. The energy of a photon, E = hff where h = Planck’s constant, 6.63 × 10–34 J s ff = frequency of the radiation 6. Since E = hff, a photon of X-ray has more energy than a photon of red light, because the frequency of the X-ray is very much higher than the frequency of red light. 7. Defi nition From E = hff Planck’s constant, h = E v = Energy of a photon of electromagnetic wave Frequency of the wave Example 5 A radio station emits radio waves of frequency 100 MHz at a rate of 200 kW. (a) What is the energy of a photon of radio wave? (b) Calculate the number of photons emitted per second. Solution: (a) Frequency ff = 100 MHz = 1.00 × 108 Hz Energy of a photon = hff = (6.63 × 10–34) (1.00 × 108 ) = 6.63 × 10–26 J (b) Power = (Number of photon s–1) (Energy of photon) Number of photons s–1 = 200 × 103 6.63 × 10–26 = 3.02 × 1030 Example 6 A beacon emits millisecond fl ashes of red light of frequency 4.3 × 1014 Hz uniformly in all directions at one millisecond interval. The light power during each fl ash is 10 W. An observer has a telescope with an aperture area of 2.5 × 10–3 m2 . His eye-brain system perceives a fl ash of light when 50 or more photons are incident on a small area of his retina in less than 0.1 s.
68 20 Physics Term 3 STPM Chapter 20 Wave Motion Estimate (a) the number of photons emitted by the beacon in one millisecond, (b) the maximum range at which the beacon might be detected by the observer using the telescope. Solution: (a) Power of a fl ash, P = 10 W Energy in 1.0 ms = Pt Energy of 1 photon = hff Number of photon in 1.0 ms = Pt hv = 10 × 10–3 (6.63 × 10–34) (4.3 × 1014) = 3.5 × 1016 (b) Suppose that the telescope is at a maximum distance r from the beacon. In a fl ash, (3.5 × 1016) photons are emitted in all directions. Number of photons received by the aperture of telescope = (3.5 × 1016) × A surface area of sphere of area A in a fl ash Hence, (3.5 × 1016) (2.5 × 10–3) 4 πr2 = 50 r2 = (3.5 × 1016) (2.5 × 10–3) 200 π r = 3.7 × 105 m Beacon r A = 2.5 × 10–3 m2 Quick Check 5 1. Which of the following arranges the electromagnetic waves in order of increasing frequency? A gamma-rays, ultraviolet, radio waves B light, infrared, ultraviolet C microwaves, ultraviolet, X-rays D radio waves, light, infrared 2. Which of the following describes correctly the changes in the characteristics of electromagnetic waves from infrared radiations to X-rays? Frequency Wavelength Speed in in vacuum vacuum A Decreases Increases Constant B Decreases Decreases Increases C Increases Increases Decreases D Increases Decreases Constant 3. A photon is A a unit energy B a positively charged particle C a quantum of electromagnetic radiation D a rate of energy transferred
69 20 Physics Term 3 STPM Chapter 20 Wave Motion C increases because its wavelength increases D remains constant because its frequency remains unchanged 8. A beam of blue light of wavelength 400 nm and n photons per second when incident on a thermopile is able to maintain an increase in temperature of ∆θ. How many photons per second in a beam of red light of wavelength 700 nm is required to maintain the same increase in temperature? A 3 7 n C 3 4 n B 4 7 n D 7 4 n Important Formulae 1. v = fλ 2. Intensity of wave ∝ (amplitude)2 3. Progressive wave equation: y = A sin (2πft – 2π λ x) 4. Standing wave equation: y = (2A cos 2π λ x) sin ωt 5. Speed of light, c = ABBBB 1 ε0 μ0 6. Energy of photon, E = hv, c = vλ 4. Which of the following has the largest energy content? A 102 photons of γ-rays of wavelength 1 pm. B 104 photons of X-rays of wavelength 2 nm. C 105 photons of infrared radiations of wavelength 5 μm. D 106 photons of yellow light of wavelength 600 nm. 5. Which one of the following arranges the following photons in order of increasing magnitude? I. a photon of microwave II. a photon of infrared radiation III. a photon of ultraviolet radiation A I, II, III C II, III, I B I, III, II D III, II, I 6. A body emits infrared radiations of wavelength λ at a constant rate P. If h is Planck’s constant, and c is the speed of light, what is the number of photons emitted per second? A Pc hλ C hc Pλ B Pλ hc D λc Ph 7. When a photon of light travels from air to glass, its energy A increases because its wavelength decreases B decreases because its speed decreases STPM PRACTICE 20 1. Two wave pulses P and Q travel at 3 cm s–1 in opposite directions. The positions of the wave pulses at time t = 0 is shown in the fi gure below. y / cm x / cm 3 0 2 4 P Q 6 8 10 12 -3 Which of the following shows the resultant wave pulse at time t = 1 s? A x / cm 3 0 2 4 6 8 10 12 -3 y / cm
70 20 Physics Term 3 STPM Chapter 20 Wave Motion B x / cm 3 0 2 4 6 8 10 12 -3 y / cm C x / cm 3 0 2 4 6 8 10 12 -3 y / cm D x / cm 3 0 2 4 6 8 10 12 -3 y / cm 2. The displacement-time graph of a wave is shown below. -3.0 3.0 0 2.0 4.0 t / s y / cm At a particular instant, the displacement is 2.0 cm. What is the displacement after 1.0 s? A –1.0 cm C 1.0 cm B – 2.0 cm D 2.0 cm 3. A progressive wave has a frequency of 60 Hz is reflected from a wall. A standing wave is produced by the superposition of an incident wave and reflected wave. The distance between successive antinodes of the standing wave is 12.0 cm. What is the speed of the progressive wave? A 7.2 m s–1 C 21.6 m s–1 B 14.4 m s–1 D 28.8 m s–1 4. Electromagnetic wave consists of electrical vibrations E and magnetic vibration B which are A in antiphase B parallel to each other C in the direction of propagation D mutually perpendicular to each other 5. The electrical vibration of an electromagnetic wave is y = E0 sin (2πf – kx). Which equation represents the magnetic vibration of the electromagnetic wave? A x = B0 sin (2πf – kx) B y = B0 sin (2πf + kx) C z = B0 sin (2πf – kx) D y = B0 sin (2πf + kx) 6. Which of the waves is a longitudinal wave? A Light wave B Wave along a vibrating guitar string C Wave in a flute D Microwave 7. What are the wavelengths of ultraviolet and infrared radiations? Ultraviolet Infrared A 10 nm 1000 pm B 100 nm 100 μm C 10 pm 100 pm D 100 nm 1000 nm 8. Which of the statement is not true of electromagnetic waves? A Electromagnetic waves are transverse waves B The vibrations of the electric and magnetic fields are in phase C The wave is deflected by an electric field D Speed of electromagnetic waves, c = 1 ε0 µ0 9. A progressive wave is represented by the equation y = 2.0 sin (50πt + πx 4 ) where x and y are in metre and t is in second. What is the wavelength λ and speed v of the wave? A λ = 4.0 m, v = 50 m s–1 B λ = 8.0 m, v = 100 m s–1 C λ = 8.0 m, v = 200 m s–1 D λ = 50 m, v = 200 m s–1
71 20 Physics Term 3 STPM Chapter 20 Wave Motion 10. A progressive wave P in a medium is represented by the equation y = 6.0 sin (200 πt – 0.5πx) Another similar progressive wave Q in the medium has the same amplitude but has a wavelength twice of P and travels in the opposite direction. Which of the following is the equation representing Q? A y = 6.0 sin (100πt + πx 4 ) B y = 6.0 sin (400πt + πx 4 ) C y = 6.0 sin (200πt + πx) D y = 6.0 sin (400πt – πx) 11. The graph below shows the variation of displacement y with time t for a particle P in a progressive wave of wavelength λ. A –A Which of the following is the displacementtime graph of a particle Q at a distance of λ 2 from P? A A –A B A –A C A –A D A –A 12. Which statement is true about a standing wave? A Points at the nodes have the greatest frequency B Neighbouring points have the same amplitude. C Points between two nodes vibrate in phase. D Standing waves are only transverse waves. 13. A progressive wave is given by y = 10 cos (5x + 25t) where x and y are in metres and t in seconds. What are the wavelength, frequency and direction of propagation of the wave? Wavelength Frequency Direction of propagation A 0.80 m 3.98 Hz Negative-x B 0.80 m 12.5 Hz Positive-x C 1.26 m 12.5 Hz Positive-x D 1.26 m 3.98 Hz Negative-x 14. Two progressive waves P and Q which moving in the +x-direction are shown in the figure. The intensity of each of the waves is I0 . Superposition of waves occurs. y 0 x A 0.5A P Q What is the phase difference φ between the waves, and what is the intensity I and of the resultant wave? A φ = π 6 rad and I = 2.0I0 B φ = π 2 rad and I = 2.0I0 C φ = π 6 rad and I = 3.7I0 D φ = π 2 rad and I = 3.7I0 15. The equation representing a progressive wave is y = 0.25 sin(5t – 1.5x) where x and y are in metres and t in seconds. What is the speed of a particle at x = 2.0 m at time t = 5.0 s? A 0.85 m s–1 C 1.16 m s–1 B 1.00 m s–1 D 1.25 m s–1
72 20 Physics Term 3 STPM Chapter 20 Wave Motion 16. A pulse of transverse wave travels along a rope. In Figure (a) the end P of the rope is fixed. In Figure (b) the end P of the rope is tied to a light ring which is free to move. Draw the reflected wave pulse in Figures (a) and (b). P P 17. (i) Microwaves that used for television transmission are plane-polarised. State the characteristic of microwaves that enables polarisation of microwaves. (ii) An electron moves in simple harmonic motion along the y-axis as shown in the figure below. An electromagnetic wave is produced and propagates along the positive z-direction. x z y e- State the directions of vibration of the electric field E, and magnetic field B of the electromagnetic wave. 18. A progressive wave is represented by the equation y = 5.0 sin (50t – 2.5x) where x and y are in metres and t in seconds. (a) What is the direction of the wave propagation? (b) What is the velocity of the wave? (c) Determine the velocity of a particle at the point x = 2.0 m at time t = 0.60 s. 19. (a) A particle charged +q moves along the x-direction in a region of electric field E and magnetic field B that vary sinusoidally with distance x from O. E B q v x (i) State the condition for the particle to continue moving along the x-axis. (ii) The amplitude of the magnetic field B = 0.24 T. The charge particle is able to move along the x-axis with a velocity of 4.0 × 105 m s-1. Determine the amplitude of the electric field E. (b) The speed of an electromagnetic wave of frequency 6.0 × 109 Hz in a material is 2.0 × 108 m s-1. (i) What is its wavelength in air. Hence identify the electromagnetic wave. (ii) Find the refractive index of the material. 20. (a) State the difference between transverse wave and longitudinal wave. (b) What is meant by a plane-polarised wave? (c) Explain whether it is possible to planepolarise (i) sound waves (ii) light waves 21. Transverse waves of frequency 200 Hz move along a stretched string with a speed of 5.0 m s–1. Each particle of the string oscillates through a distance of 6.0 cm. (a) Write an equation to represent the progressive wave along the string. (b) What is its wavelength? 22. The above figure shows a progressive wave at time zero moving in the negative x-direction. The wave frequency is 50 Hz. (a) Find (i) amplitude, (ii) wavelength, and (iii)speed of the wave.
73 20 Physics Term 3 STPM Chapter 20 Wave Motion (b) Write the equation representing the wave. 23. (a) Estimate the maximum and minimum value of a photon of visible light. (b) State two similarities and two differences between light waves and sound waves. (c) Yellow light has wavelength 5.62 × 10–7 m. Calculate (i) its frequency, (ii) the energy of a photon, (iii) the rate of emission of photons from a light power of 60 W. 24. (a) Give an expression for E, the energy of a photon, in terms of f, its frequency, and h, the Planck’s constant. (b) Give a typical wavelength of visible light and for the wavelength, calculate the energy of a photon. 25. (a) Explain the difference between transverse wave and longitudinal wave. Give an example of each wave. (b) A progressive wave is represented by the equation y = 0.25 sin (6 000 t – 20x) Calculate (i) the wave frequency, (ii) the wavelength, and (iii) the speed of the wave (c) Write an equation to represent a similar wave of twice the amplitude and frequency and travelling with the same speed but in the opposite direction. 26. (a) Explain the meaning of intensity of a wave. What is the relation between intensity and amplitude? (b) O P x Figure (a) x / cm y / mm P 1 2 3 4 5 6 7 8 9 10 11 8.0 4.0 1.0 0 Figure (b) Figure (a) shows circular waves from a vibrator at O in a ripple tank. Figure (b) shows the instantaneous displacement of water particles along the line of propagation Ox of the ripples. (i) What is the wavelength? (ii) Explain why the amplitude of the wave decreases as the distance x increases. (iii) Use the data shown in Figure (b) to calculate the ratio Intensity at P Intensity at O . (c) The wave in a ripple tank is represented by the equation y = 0.60 sin (20t – 4x) where x and y are in cm and t in second. (i) Calculate the speed of the wave. (ii) Write an equation to represent a wave on t he ripple tank which has half the amplitude and twice the frequency but travels with the same speed.
74 20 Physics Term 3 STPM Chapter 20 Wave Motion 1 1. D 2. C 3. B 4. B 5. B 6. D 7. B 8. D 9. (a) Amplitude = 5 cm (b) f = 7.96 Hz 2πf = 50 (c) λ = 6.28 cm 2π λ = 1 (d) v = 50.0 cm s–1 v = f λ 10. (a) Amplitude = 6.0 m (b) λ = 5.0 m (c) 20.0 m s–1 f = 4 Hz 11. (a) Amplitude = 6.0 cm (b) λ = 31.4 cm 2π λ = 0.2 (c) f = 0.796 Hz 2πf = 5.0 (d) 25.0 cm s–1 v = fλ (e) negative x-direction (f) 30 cm s–1 vmax = ωA 12. y = 0.50 sin (1 000 πt + 10πx 3 ) 13. (i) vP = vR = vS = 0, vQ is maximum and upwards. (ii) AQ = 0, AP = AS maximum downwards, AR is maximum and upwards. 14. (a) (i) φ = x 2.5(2π) y = 2.0 sin (4πt – φ) = 2.0 sin 4π(t – x 5 ) (ii) When x = 3.0 cm, t = 1.5 s y = 1.2 cm (b) (i) λ = 10.0 cm 2π λ = 0.20π (ii) f = 2.5 Hz 2πf = 5πt (iii) T = 1 f = 0.40 s (iv) vmax = ωA = 94 cm s-1 (c) t = 0.20 s = T 2 Distance travelled = λ 2 0 5 10 15 y / cm 6.0 –6.0 x / cm t = 0 t = 0.20 s 2 1. Amplitude, A1 = 2 cm, A2 = 3 cm. Intensity, I ∝ A2 . I1 /I 2 = (22 )/(3)2 = 4/9 2. For a point source, I ∝ 1 r2 (a) r = 1.0 m, I = (2.0)2 (1.0)2 (10 W m–2) = 40 W m–2 (b) r = 3.0 m, I = (2.0)2 (3.0)2 (10 W m–2) = 4.4 W m-2 3 1. A 2. C 3. B 4 1. C 2. D 3. C 5 1. C 2. D 3. C 4. A 5. A 6. B 7. D 8. D STPM Practice 20 1. A 2. B: Period T = 2.0 s, 1.0 s = 1 2 T, y = –2.0 cm 3. B: f = 60 Hz, λ = 2(12.0) cm = 24.0 cm v = fλ = (60.0)(24.0) cm s–1 = 14.4 m s–1 4. D 5. C: E and B vibrations in phase and travel in the x-direction. 6. C 7. D 8. C 9. C 10. A 11. C 12. C 13. D : Compare y = 10 cos (5x + 25t) with y = A cos ( ωt – 2π λ x) 2π λ = 5, λ = 1.26 m, ω = 2πf = 25, f = 3.98 Hz, + 5x implies negative-x direction. 14. C : y = y1 + y2 = (2A cos π 12 ) sin (ωt – kx + π 6 ) I ∝ (2A cos π 12 ) 2 15. D : v = dy dt = 0.25(5) cos (5t – 1.5x) When t = 5.0 s and at x = 2.0 m, v = 0.25(5) cos (5 × 5.0 – 1.5 × 2.0) m s–1 = 1.25 m s–1 ANSWERS
75 20 Physics Term 3 STPM Chapter 20 Wave Motion 16. (a) Phase change of 180o when the end P is fixed. (b) No phase change when the end P is free to move. 17. (i) Microwaves are transverse waves. (ii) Direction of propagation (or velocity v) of the EM wave is in the direction of the cross product (E × B) Since the electron oscillates along the yaxis, the electric field oscillation E is in the direction of the y-direction. The direction of oscillation of the magnetic field B is in the x-direction. 18. (a) Direction of wave propagation: positive x-axis. (b) Compare y = 5.0 sin (50t – 2.5x) with y = A sin (2πft – 2π λ x) 2πf = 50 and 2π λ = 2.5 Velocity, v = f λ = (50)/(2.5) m s–1 = 20 m s–1 (c) Velocity of particle = dy dt = (5.0)(50) cos (50t – 2.5x) When t = 0.6 s, x = 2.0 m, v = (250) cos (30 – 5) m s–1 = 248 m s–1 19. (a) (i) Condition: velocity of particle v = E B (ii) v = E0 B0 , E0 = 9.6 × 104 V m-1 (b) (i) l = c f = 0.050 m microwave (ii) n = c v = 1.5 20. (a) Refer to page 48 (b) Oscillations of wave in one plane only. (c) (i) No, sound waves cannot be plane polarised. (ii) Yes, light waves are transverse waves 21. (a) y = 3.0 sin (400 πt – 4π 5 x) x, y in cm: t in s. (b) λ = 0.025 m 22. (a) (i) 6.0 cm (ii) 20 cm (iii) 10 m s–1 (b) y = 6.0 sin (100 πt + πx 10 ) 23. (a) E max = 5 × 10–19 J (λ min = 4 × 10–7 m) E min = 3 × 10–19 J (λ max = 7 × 10–7 m) (b) Similarities: (Any two wave properties) Reflection, refraction, diffraction, interference Differences: • Electromagnetic waves/mechanical waves • Need no medium/Require a material medium • Speed 3 × 108 m s–1 in vacuum/330 m s–1 in air • Can be plane polarized/cannot (c) (i) f = 5. 34 × 1014 Hz f = —c λ (ii) 3.54 × 10–19 J E = hf (iii) 1.69 × 1020 s–1 24. (a) E = hf (b) λ = any value from 4 × 10–7 m to 7 × 10–7 m. Energy from 3 × 10–19 J to 5 × 10–19 J 25. (a) Refer to page 48 (b) (i) 95.5 Hz (ii) 0.314 m (iii) 30 m s–1 (c) y = 0.50 sin (12 000 t + 40x) 26. (a) Intensity ∝ (Amplitude)2 (b) (i) λ = 4.0 cm (ii) Circumference of circle increases. Same energy distributed over a greater length of the circumference. (iii) 1 64 (I ∝ A2 ) (c) (i) 5.0 cm s–1 (ii) y = 0.30 sin (40 t – 8x)
21 Physics Term 3 STPM Chapter 21 Sound Waves Chapter 21 Sound Waves 21 Physics Term 3 STPM CHAPTER SOUND WAVES 21 Concept Map Propagation of Sound waves Sources of Sound Doppler Effect f = ( v u0 v us )f Intensity Level of Sound = 10 log I I0 Beats f = f 1 – f 2 Stretched String f 0 = 1 2l T µ 2f 0 3f 0 Closed Pipe Open Pipe Sound Waves 76 Bilingual Keywords Air column: Turus udara Beats: Rentak Diaphragm: Diafragma Doppler effect: Kesan Doppler End correction: Pembetulan hujung Fundamental: Asas Open pipe: Paip terbuka Overtone: Nada lampau Resonant tube: Tiub resonans Sound intensity: Keamatan bunyi
77 21 Physics Term 3 STPM Chapter 21 Sound Waves 21.1 Propagation of Sound Waves Students should be able to: • explain the propagation of sound waves in air in terms of pressure variation and displacement; • interpret the equations for displacement y = y0 sin(ωt − kx) and pressure p = p0 sin (ωt − kx – π 2) Learning Outcomes Figure 21.1 1. Sound wave is a longitudinal wave which requires a medium for its propagation. Sound can travel through a gas, liquid or solid. INTRODUCTION 1. Sound waves are produced by vibrating objects. 2. The vibrating string of a guitar or a piano produces sound. The vibrating cone of a loudspeaker produces sound. Sound is produced by the vibrating membrane of a drum. 3. Energy from the vibrating source is transferred to the surroundings by sound waves. The frequency of the sound wave is the same as the frequency of vibration of the source. 4. The normal audible frequency range of a person is 20 Hz to 20 kHz. 5. The speed of sound in air is about 330 m s–1. 2010/P1/Q14, 2012/P1/Q13, 2015/P3/Q4, 2016/P3/Q5, 2017/P3/Q5
78 21 Physics Term 3 STPM Chapter 21 Sound Waves 2. Figure 21.1(a) shows the situation when sound wave from a source such as a loudspeaker travels through the air. Air molecules are displaced from the respective equilibrium positions and vibrate along the direction of wave propagation. 3. Figure 21.1(b) is the displacement-distance graph showing the instantaneous displacements of air molecules labelled 1 to 13. 4. Compressions and rarefactions are formed. The compressions and rarefactions move in the direction of wave propagation. 5. The distance between successive compressions or successive rarefactions is the wavelength, λ. 06. The variation of displacement y with time t for an air molecule at distance x from a reference point O can be represented by the progressive wave equation y = y0 sin (ωt – kx) where y0 = amplitude of wave ω = 2π f, f = wave frequency k = constant 7. Figure 21.1 (c) shows the variation of pressure change p with distance along the direction of wave propagation. The increase in pressure is maximum, p0 at a compression. The decrease in pressure is maximum at a rarefaction. The pressure change p lags behind the displacement by π 2 radians. 8. The variation of pressure change p with time t can be represented by the equation p = p0 sin (ωt – kx – π 2 ) where p0 = amplitude of pressure change. 9. The relation between speed of sound wave v, its frequency f, and its wavelength λ is v = fλ. 10. The speed of sound in a gas < speed of sound in liquid < speed of sound in solid. In air, speed of sound is about 300 m s–1, in water 1500 m s–1 and in steel 5000 m s–1. 11. In air the speed of sound increases when the temperature increases. Example 1 Sound of frequency 500 Hz is emitted from a loudspeaker along a straight line in the positive x-direction. The graph shows the instantaneous displacement of the air molecules from their respective equilibrium positions. (a) Among the points A, B, C and D, which point has maximum, and which point has minimum instantaneous pressure? (b) Calculate the time taken by the sound to travel from B to E. A A B B C C D D E E
79 21 Physics Term 3 STPM Chapter 21 Sound Waves Solution: (a) (i) The pressure at A is the maximum because the displacement of the air molecule to the left of A is positive, and the displacement of the air molecule to the right of A is negative. Hence, the air at A is compressed and the pressure is maximum. (ii) At C, the pressure is minimum, because the displacements of the air molecules on the left and right of C are as shown. The air at C is rarefi ed. Hence, pressure is minimum. (b) Period, T = 1 f = 1 500 = 2.0 × 10–3 s BE = 3 4 λ Time taken to travel distance λ is T. Time taken to travel 3 4 λ = 3 4 T = 3 4 × 2.0 × 10–3 s = 1.5 × 10–3 s 21.2 Sources of Sound Students should be able to: • use the standing wave equation to determine the positions of nodes and antinodes of a standing wave along a stretched string; • use the formula v = T µ to determine the frequencies of the sound produced by different modes of vibration of the standing waves along a stretched string; • describe, with appropriate diagrams, the different modes of vibration of standing waves in air columns, and calculate the frequencies of sound produced, including the determination of end correction. Learning Outcomes 1. Sound is produced by vibrating objects. Musical instruments use standing wave to produce sound. 2. When a guitar string is plucked, transverse standing wave set up in the string causes the air around it to vibrate and a sound is produced. Sound is produced in a fl ute and organ pipe by longitudinal waves set up in air column. 3. Sound is produced in a drum by standing wave set up in a vibrating membrane. Pull Standing wave N N A Continuous reflection (a) (b) Figure 21.2 A C 2017/P3/Q6
80 21 Physics Term 3 STPM Chapter 21 Sound Waves 4. Figure 21.2(a) shows a string stretched between two points is pulled at the mid-point. When the string is released, progressive transverse waves travel towards the ends of the wire. Continuous superposition of the waves reflected at the ends produces the transverse standing wave along the string. 5. The simplest mode, known as the fundamental mode of vibration of the standing wave is shown in Figure 21.2(b). The displacements at the fixed ends are perpetually zero and are known as nodes (N). The point where the displacement is maximum is known as antinode (A). 6. The separation between nodes, and between antinodes can also be obtained using the standing wave equation: y = (2A cos 2π λ x) sin ωt • At antinodes, amplitude of standing wave, (2A cos 2π λ x) = 2A cos 2π λ x = 0, π, 2π, … x = 0, λ 2 , λ, … Distance between successive antinodes, AA = ( λ 2 – 0) = λ 2 • At nodes, amplitude of standing wave, (2A cos 2π λ x) = 0 cos 2π λ x = p 2 , 3π 2 , 3π 5 ,… x = λ 4 , 3λ 4 , 5λ 4 ,… Distance between successive antinodes, AA = ( 3λ 4 – λ 4 ) = λ 2 7. Figure 21.3 shows four lowest modes of the standing wave along the string. 8. The speed of transverse progressive wave along a stretched string is v = T µ where T is the tension in the string and µ is the mass per unit length of the string. 9. For the fundamental mode, • Length of string l = distance between nodes, λ0 2 where λ0 is the wavelength of the transverse wave. Hence λ0 = 2l • Fundamental frequency, f 0 = v λ0 = 1 2l T µ 10. When the mid-point of the string is touched lightly, while the string is plucked at a point 1 4 from one end, the standing wave setup is as shown in figure 21.3. The note produced is known as the first overtone. Wavelength of first overtone, λ1 = l, length of string. λ = 2l 0 λ = l 1 λ = l 2 l N N N N N N A N A A A A A A A A A N N N N N N N 2_ 3 λ = l 3 1_ 2 (a) Fundamental frequency, (b) First overtone frequency, (c) Second overtone frequency, (d) Third overtone frequency, Figure 21.3 v 2l f 0 = v l f 1 = = 2f 0 3v 2l f 2 = = 3f 0 2v l f 3 = = 4f 0
81 21 Physics Term 3 STPM Chapter 21 Sound Waves Hence, frequency of fi rst overtone, f 1 = v l 1 = v l = 2f 0 11. For the second overtone (Figure 21.3(c)) l = 3 2 λ2 λ2 = 2 3 l and frequency, f 2 = v λ2 = 3v 2l = 3f 0 12. For the third overtone (Figure 21.3(d)) l = 2λ3 λ3 = l 2 Frequency, f 3 = 2v l =4f 0 13. The notes that can be emitted by a stretched string fi xed at both ends are f 0 , 2f 0 , 3f 0 , 4f 0 , … where f 0 = v 2l . That is all the harmonics are obtainable. Example 2 A wire stretched between two points 0.60 m apart is plucked near one end. What are the three lowest resonant frequencies if the speed of transverse progressive wave along the string is 600 m s–1? Solution: The lowest resonant frequency is the fundamental frequency f 0 . l = 0.60 m = λ0 2 λ0 = 1.20 m Fundamental frequency, f 0 = v λ0 = 600 1.20 = 500 Hz Hence, the three lowest resonant frequencies are f 0 , 2f 0 and 3f 0 : 500 Hz, 1 000 Hz and 1 500 Hz. Exam Tips Always draw a sketch to show the mode of vibration of the stretched string in order to relate its length l with λ.
82 21 Physics Term 3 STPM Chapter 21 Sound Waves Example 3 The lowest resonant frequency of a piano string of length 0.90 m is 200 Hz. What is the speed of transverse wave on the string? Solution: The lowest frequency is the fundamental frequency Hence, l = 0.90 m = λ0 2 λ0 = 1.80 m Speed of transverse wave = f λ0 = 200(1.80) = 360 m s–1 Example 4 The fi gure shows one end of a string attached to a vibrator which produces transverse motion. The other end of the string passes over a pulley, so that the length of the string between the vibrator and the pulley is 63.0 cm. A mass hangs from the free end of the string. When the frequency of the vibrator is varied, the wire vibrates with large amplitude at certain points along its length for certain frequencies of the vibrator. (a) Explain why this happens. (b) If the speed of transverse wave along the string is 150 m s–1, fi nd the three lowest frequencies in which the effect is observed. Sketch the mode of vibration of the string in each case. Solution: (a) When the end of the string attached to the vibrator vibrates, a transverse wave travels along the wire to the pulley where the wave is refl ected. Superposition of the incident wave and refl ected wave produces a standing wave along the stretched wire. For certain frequencies of the vibrator, the driving frequency is equal to the natural frequency of the stretched string, and resonance occurs. (b) The end of the string attached to the vibrator is an antinode, since it is constantly moved transversely. Hence, λ0 4 = 63.0 cm λ0 = 252 cm f 0 = v λ0 = 150 2.52 = 60 Hz 63.0 cm A N (i)
83 21 Physics Term 3 STPM Chapter 21 Sound Waves In (ii) 3 4 λ1 = 63.0 cm λ1 = 84.0 cm f1 = v λ1 = 150 0.84 = 179 Hz In (iii) 5 4 λ2 = 63.0 cm λ2 = 50.4 cm f2 = 150 0.504 = 298 Hz Quick Check 1 63.0 cm A N A (ii) (iii) 1. When sound wave of angular frequency ω travels through air, the displacement y at a distance x from the source varies with time t according to the expression y = A sin (ωt + kx). Which is the correct expression for the variation of pressure p with time t at a distance x from the source? A p = p0 sin (ωt + kx – π 2 ) B p = p0 sin (ωt + kx + π 2 ) C p = p0 sin (ωt + kx – π) D p = p0 sin (ωt + kx + π) 2. A string of length l fi xed at both ends and is plucked at its midpoint and emits its fundamental note of frequency f. When the string is plucked at a different point, the fi rst overtone frequency f′ is produced. Which of the following gives the frequency f’ of the fi rst overtone in terms of f and v the speed of transverse wave along the string? A f’ = 1 2 f, v = fl C f’ = 2f, v = 2fl B f’ = 1 2 f, v = 2fl D f’ = 2f, v = fl 3. A standing wave is set up on a stretched string XY as shown in the fi gure. Which point(s) vibrates exactly in phase with P? X Y P A 1 and 2 C 1 only B 2 and 3 D 3 only 4. The diagram below shows a string stretched between X and Y. The string is made to vibrate transversely and a standing wave is setup with X, P, Y on the string as nodes. R P S X Y The vibration of the two points R and S shown on the string have A the same amplitude and in antiphase B different amplitude and in phase C the same amplitude and in phase D different amplitude and in antiphase 5. The graph shows the displacement of air molecules along the direction of propagation of sound in the Ox direction. At which point the pressure of the air is maximum and minimum? Maximum pressure Minimum pressure A 1 2 B 4 3 C 2 4 D 3 1
84 21 Physics Term 3 STPM Chapter 21 Sound Waves 6. The frequency of the fundamental note produced by a stretched string of length 1.0 m is 256 Hz. When the string is shortened to 0.4 m at the same tension, the fundamental frequency is A 102 Hz C 416 Hz B 312 Hz D 640 Hz 7. A wire is stretched between two points distance L apart. Sketch diagrams to show the two lowest frequency modes of transverse vibration of the string, and write expression for their frequencies in terms of L and c, the speed of propagation of transverse waves on the string. 8. A stretched wire of length 1.20 m is fixed at both ends. A transverse wave of speed 300 m s–1 is propagated along the wire and forms a standing wave. In a certain mode of vibration, the distance between successive nodes is 0.40 m. (a) What is the frequency for this mode? (b) What lower resonant frequencies are possible? 9. The figure shows a standing wave on a string stretched between two points A and F which are separated by a distance L. A B C D E F (a) Describe the oscillations of the points B, C, D and E. Compare these oscillations in terms of their relative phases and amplitudes. (b) What is the wavelength in terms of L? Standing Wave in Air Column and Closed Pipe 2008/P1/Q13, 2008/P2/Q3, 2016/P3/Q18 1. If you blow softly across the opened end of a test tube, sound of a certain frequency is produced. When you blow harder, the sound produced has a higher pitch. 0 1 2 3 Distance Displacement (a) Closed pipe (b) Instantaneous (c) Displacement-distance graph Figure 21.4 2. Figure 21.4 shows situation when the air column in a closed pipe is disturbed by someone blowing across the opened end. 3. Figure 21.4(a) shows longitudinal wave travelling towards the closed end, and is then reflected. Superposition of the incident wave and the reflected wave produces standing longitudinal wave in the closed pipe. 4. Figure 21.4(b) shows the instantaneous displacement of air molecules along the axis of the pipe in the fundamental mode of vibration. Particle 3 at the opened end has the greatest displacement. The displacements of particle 2 and 1 become smaller and finally particle 0 at the closed end is not displaced at all.
85 21 Physics Term 3 STPM Chapter 21 Sound Waves 5. Figure 21.4(c) is the displacement-distance graph for the standing wave drawn inside the tube. This graph shows the fundamental mode of vibration. The sound emitted is known as the fundamental note. 6. The closed end is a node, N and the opened end is an antinode. Hence, the length of the opened pipe, l = λ0 4 where λ0 is the fundamental wavelength. Fundamental frequency, f 0 = v λ0 = v 4l where v = speed of sound in air. 7. When you blow harder across the opened end of the pipe, notes of higher frequencies are obtained. Figure 21.6 shows the modes of vibration for the notes obtained. (a) Fundamental mode of vibration (b) First overtone (c) Second overtone A N A A N N A A A N N N (a) Fundamental (b) First overtone (c) Second overtone (a) mode of (a) vibration Figure 21.6 8. For the first overtone, length of pipe, l = 3 4 λ1 λ1 = 4 3 l Frequency, f1 = v 4l 3 = 3f 0 9. For the second overtones, length of pipe, l = 5 4 λ2 λ2 = 4 5 l Frequency, f2 = v 4l 5 =5f 0 10. Hence, the notes available are in frequencies f 0 , 3f 0 , 5f 0 , 7f 0 , … that is all the odd harmonics. A N Figure 21.5
86 21 Physics Term 3 STPM Chapter 21 Sound Waves Example 5 The length of a closed pipe is 18.0 cm. If the speed of sound in air is 330 m s–1, fi nd the three lowest frequencies for the sound emitted when one blows across the opened end of the pipe. Solution: The three lowest modes of vibrations of stationary waves in the pipe are as shown on the right. For the fundamental note, l = λ0 4 λ0 = v f 0 = v 4f 0 f 0 = v 4l = 330 4(0.18) = 458 Hz The other two frequencies are 3f 0 = 1 374 Hz, and 5f 0 = 2 290 Hz. Quick Check 2 1. A pipe opened at one end and closed at the other is with length 0.6 m. If the speed of sound is 300 m s–1, calculate the two lowest resonant frequencies. 2. A well with vertical wall resonates with a note of 7.0 Hz and does not resonant with frequencies less than 7.0 Hz. Estimate the depth of the well if the speed of sound in air is 350 m s–1. 3. When a student blows softly across the opened end of a test tube, the fundamental frequency produced is f. If the student blows harder, what are the frequencies of the notes produced? 4. A loudspeaker which emits sound of frequency 2 500 Hz is placed facing a plane reflective wall inside a closed container containing a gas. A microphone connected to a cathode ray oscilloscope is used to detect nodes and antinodes along the line from the loudspeaker to the wall. The microphone is moved from one node to another node through 20 antinodes in a distance of 1.900 m. What is the speed of sound in the gas? 5. Two loudspeakers X and Y emit sound of frequency 1 100 Hz. A microphone is moved along the line joining X and Y with a speed of 3.0 m s–1. The intensity of sound detected rises and falls periodically. Explain why. If the speed of sound is 330 m s–1, calculate the frequency of the rise and fall of the intensity of sound detected by the microphone. End Correction 1. Air molecules at the opened end of a closed pipe is free to vibrate. Hence, when a standing wave is set up in the closed pipe, vibrations of the air molecules at the opened end extend outside the pipe by a small distance.
87 21 Physics Term 3 STPM Chapter 21 Sound Waves 2. The position of the antinode A is at a small distance c from the opened end. This distance c is known as the end correction (Figure 21.7) 3. The value of the end correction c depends on the radius of the pipe and the wavelength of the sound. Nevertheless the value of c is approximately a constant. 4. Hence, for the fundamental note, (l + c) = λ0 4 where l = length of pipe. For the first overtone, (l + c) = 3 4 λ1 and for the second overtone, (l + c) = 5 4 λ2 Determination of the Speed of Sound (a) (b) (c) Figure 21.8 : Determination of the speed of sound 1. The speed of sound in air can be determined using the apparatus shown in Figure 21.8. A small loudspeaker which is connected to the output of an audio frequency generator is placed at the opened end of a resonant tube T. 2. Starting with the water level near the opened end, the length of the air column is increased slowly by lowering the reservoir until a sound of maximum intensity is heard. The resonant length l 1 of the air column is measured (Figure 21.8(b)). Then (l 1 + c) = λ 4 .............................. a where c = end correction. 3. The water level in the resonant tube is lowered further until resonance occurs again (Figure 21.8(c)). If l 2 = second resonant length of air column, then l2 + c = 3 4 λ .............................. b Equation b – a l2 – l 1 = λ 2 λ = v f = v 2f Speed of sound in air, v = 2f(l 2 – l 1 ), where f = frequency of audio frequency generator. A N Figure 21.7 : End Correction
88 21 Physics Term 3 STPM Chapter 21 Sound Waves 4. An alternative method is based on the equation l1 + c = λ 4 (λ = v f ) = v 4f The procedure to determine the fi rst resonant length l 1 is repeated using different values for the frequency of the audio frequency generator. 5. A graph of l 1 against 1 f is then plotted (Figure 21.8). From the equation, l 1 + c = v 4f Gradient of graph = v 4 Speed of sound in air, v = 4 (gradient of graph) where 1 f = 0, l 1 = –c Hence, the end correction c can be determined from the graph. Example 6 A student recorded the following readings for the length l of the air column in a tube which resonates with a tuning fork of the given frequency. Frequency/Hz Resonant length, l 825 30.5 cm 1 650 15.5 cm and 25.5 cm No resonance occurs between 15.5 cm and 25.5 cm for the tuning fork of frequency 1 650 Hz. Explain these readings and determine the speed of sound in air. Solution: Since no resonance occurs between 15.5 cm and 25.5 cm for f = 1 650 Hz, then λ 2 = (25.5 – 15.5) cm = 10.0 cm λ = 20.0 cm and speed of sound in air v = f λ = 1 650 × 0.20 = 330 m s–1 15.5 cm ≈ 3 4 λ and 25.5 cm ≈ 5 4 λ. The modes of vibration for these resonant lengths are shown as following. Figure 21.9
89 21 Physics Term 3 STPM Chapter 21 Sound Waves c = end correction First overtone Second overtone When f = 825 Hz, λ = v f = 330 825 = 0.40 m Resonant length, 30.5 cm ≈ 3 4 λ The note emitted is the fi rst overtone. Example 7 Fine sand The fi gure on the right shows the apparatus used to show resonance in an air column. Very fi ne sand is spread along the length of a horizontal tube fi tted with a horizontal piston. A small loudspeaker which emits sound of frequency 500 Hz is placed at the opened end and the length of the air column varied by adjusting the position of the piston. When resonance occurs, the sand in the tube forms the pattern as shown and the length of the air column is 0.78 m. (a) Explain the pattern of the fi ne sand. (b) If the speed of sound in air is 320 m s–1, calculate (i) wavelength of sound in the tube. (ii) distance between successive nodes. (iii) end correction of the tube. Solution: (a) When there is resonance, stationary wave of large amplitude is set up along the tube. At the antinodes, the sand is displaced away from the antinodes. At the nodes, the amplitude of vibration is zero and sand from the antinodes heap at the nodes.
90 21 Physics Term 3 STPM Chapter 21 Sound Waves (b) (i) Wavelength λ = v f = 320 500 = 0.64 m (ii) Distance between nodes = λ 2 = 0.32 m (iii) From the fi gure, 0.78 + c = 5 4 λ c = 5 4 × 0.64 – 0.78 = 0.02 m Quick Check 3 1. A vertical tube is fi lled with water. A small loudspeaker which emits sound of a certain frequency is placed above the opened end of the tube and water is slowly run off from the bottom. Various resonant lengths are obtained. The fi rst resonant length is 7.0 cm, and resonance also occurs when the length of the air column is 39 cm. Calculate two other possible resonant length. 2. A source of sound of frequency 250 Hz is used with a resonant tube closed at one end to measure the speed of sound in air. The resonant lengths are at 0.30 m and 0.96 m. Calculate (a) the speed of sound. (b) the end correction of the tube. 3. The shortest length of the air column in a tube closed at one end that resonates with a tuning fork of frequency 384 Hz is 206 mm. The second resonant length is 624 mm. Calculate the speed of sound in air. 4. The water level in a vertical glass tube of length 1.0 m can be adjusted. A tuning fork of frequency 660 Hz is held above the opened end of the tube. If the speed of sound in air is 330 m s–1, calculates the three shortest resonant lengths. 5. A tuning fork is held above the opened end of a tube which is closed at the other end. The length of the air column in the tube is adjusted until resonance occurs. (a) Describe the motion of air molecules along the length of the tube at resonance. (b) Sketch a diagram to show the mode of vibration. Explain what is represented by the curves you have drawn in the tube. 6. A vibrating tuning fork is held at the opened end of a tube which is closed at the other end. The length of the tube is adjusted until the shortest resonant length l is obtained. The experiment is repeated using tuning fork of different frequencies, and the corresponding resonant lengths l are measured as shown in the table. f/Hz 256 288 320 384 512 l/m 0.307 0.269 0.238 0.195 0.138 Calculate the speed of sound in air.
91 21 Physics Term 3 STPM Chapter 21 Sound Waves Standing Wave in Open Pipe 2009/P2/Q3(b), 2014/P3/Q4 1. An open pipe has both its ends opened. A stationary wave is produced in an open pipe when you blow across one end of the pipe. 2. For the fundamental mode of vibration, molecules of air at both the opened ends vibrate with the largest amplitude, forming antinodes. 3. The amplitude of vibration decreases as the distance from the ends increases. At the midpoint along the pipe, the amplitude is zero, a node is obtained (Figure 21.10(a)). 4. Figure 21.10(b) shows the instantaneous displacement y of air molecules along the length of the pipe for the fundamental mode of vibration. Length of pipe l = distance between antinodes = λ0 2 λ0 = v f 0 = v 2f 0 Fundamental frequency f 0 = v 2l A N A A A N A N A A N N N A A (a) Fundamental (b) First overtone (c) Second note overtone Figure 21.11 Exam Tips Fundamental frequency of opened pipe = twice fundamental frequency of closed pipe of the same length. 5. Figure 21.11 shows the modes of vibration for the fundamental note, first overtone, and second overtone. (a) For the first overtone, l = λ1 = v f 1 Frequency of first overtone, f1 = v l = 2f 0 (b) For the second overtone, l = 3 2 λ2 = 3 2 v f 2 Frequency of second overtone, f 2 = 3v 2l = 3f 0 Hence, frequencies of notes are f 0, 2f 0, 3f 0, 4f 0, … that is all the harmonics where f 0 = v 2l .
92 21 Physics Term 3 STPM Chapter 21 Sound Waves 6. If the end corrections, c is taken into consideration, (Figure 21.12), then for the fundamental note (l + 2c) = λ0 2 = v 2f 0 Fundamental frequency, f 0 = v 2(l + 2c) Example 8 A pipe of length 0.33 m is opened at one end and closed at the other. If the speed of sound in air is 330 m s–1, calculate (a) The frequencies of the fundamental note and the fi rst overtone. (b) The length of the pipe opened at both ends which has a fundamental frequency equal to the difference in the frequencies in (a). Solution: (a) 0.33 m = λ0 4 λ0 = 4 × 0.33 m Fundamental frequency, f0 = v λ0 = 330 4 × 0.33 = 250 Hz Frequency of fi rst overtone, f 1 = 3f 0 = 750 Hz (b) For the open pipe, fundamental frequency f’ 0 = f 1 – f 0 = (750 – 250) Hz = 500 Hz Length of open pipe, l = λ ′ 0 2 = v 2f ′ 0 = 330 2 × 500 = 0.33 m Figure 21.12
93 21 Physics Term 3 STPM Chapter 21 Sound Waves Example 9 A pipe opened at both ends, and a pipe opened at one end and closed at the other have the same length. Compare the fundamental resonant frequencies of the pipes. Solution For the closed pipe, l = λ0 4 = v 4f 0 f0 = v 4l For the open pipe, l = λ ′ 0 2 f′ 0 = v 2l = 2f 0 Quick Check 4 1. The standing wave set up in a resonant tube open at both ends always has A an odd number of nodes. B an even number of antinodes. C an even number of nodes and antinodes. D antinodes at the open ends. 2. A pipe opened at both ends has a length of 6.0 cm. If the speed of sound in air is 330 m s–1, then the fundamental resonant frequency is A 1 375 Hz C 4 125 Hz B 2 750 Hz D 5 500 Hz 3. Standing waves are set up in four tubes. The fi gures below represent the displacement of the air molecules along the axis of the tube from their respective mean positions. The length of each tube is L and the end correction is c. The speed of sound in air is v. Which one of the following fi gures represents the standing wave of frequency v (L + 2c) ? A B C D 4. A pipe P closed at one end and another pipe Q opened at both ends have the same fundamental frequency. What is the ratio of the length of pipe P to pipe Q? A 1 : 1 C 2 : 1 B 1 : 2 D 2 : 3 5. A stationary sound wave is set up in a pipe opened at both ends. The sound emitted is the fundamental frequency. P, Q and R are particles along the axis of the pipe. P• Q• R• Which statement about the vibrations of P, Q and R is not correct? A Q is stationary. B P and R are vibrating in phase. C Vibrations of P and R along the axis of the pipe. D P and R have the same amplitude of vibration. 6. An opened pipe has a fundamental frequency of 300 Hz. The fi rst overtone of a closed pipe has the same frequency as the fi rst overtone of the opened pipe. What is the length of each pipe? (Speed of sound in air = 330 m s–1)
94 21 Physics Term 3 STPM Chapter 21 Sound Waves 7. (a) What is the frequency of the fundamental note emitted by a pipe opens at both ends of length 1.6 m? (Speed of sound in air = 320 m s–1) (b) What is the effect, if any, on the frequency of the note emitted when (i) the atmospheric pressure increases? (ii) the temperature of the air increases? 8. The fi gure shows a small loudspeaker S which is connected to an audio frequency oscillator at the end P of a pipe opened at both ends. The frequency of the oscillator is varied from 0 to 1 500 Hz. If the speed of sound in air is 330 m s–1, at what frequencies will resonance occur? P S 21.3 Intensity Level of Sound Students should be able to: • defi ne and calculate the intensity level of sound Learning Outcome 1. The intensity of sound at a point from the source is defi ned as the energy per second received by a unit area placed perpendicularly to the direction of propagation of sound. The unit of sound intensity is W m–2. 2. It is observed that normal human ear is able to detect sound of a minimum intensity 10–12 W m–2 for a reference frequency of 1 000 Hz. 3. Sound of intensity greater than 1 W m–2 can cause damage to the ear. 4. The intensity level of sound = log10 I I0 I = intensity of sound I0 = lowest sound intensity detectable which is 10–12 W m–2 5. The unit for intensity level of sound is bel (B), but the decibel (dB) is more commonly used. 1 dB = 10–1 B 6. Using the equation, intensity level of sound = log10 I I0 when I = I0 = 10–12 W m–2, the intensity level = 0 7. If the intensity level is 60 dB or 6 B, the sound intensity I is given by 6 B = log10 I I0 I I0 = 106 I = 106 × (10–12) = 1.0 × 10–6 W m–2 2007/P1/Q15, 2010/P1/Q13, 2013/P3/Q5, 2013/P3/Q5, 2016/P3/Q6