Tutorial Solution (DP014_Chapter 1)

Tutorial Solution

CHAPTER 1
SIMPLE HARMONIC MOTION

DP024 2020/2021

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QUESTION 1: Solution 1:
Which of the following statements is true
about the magnitude of the acceleration for SHM is defined as a periodic motion without loss
an object undergoing simple harmonic of energy in which the acceleration of a body is
motion? directly proportional to its displacement from the
equilibrium position (fixed point) and is directed
A. Directly proportional to displacement towards the equilibrium position but in opposite
B. Inversely proportional to displacement direction of the displacement.
C. Directly proportional to velocity
D. Inversely proportional to velocity Answer :

→A

Directly proportional to displacement.

Hint:
- Check definition of SHM

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QUESTION 2: Solution 2:

For a particle undergoing simple harmonic
motion, which of the following graphs
represents the variation of acceleration, a with
its displacement, x?

graph v-x Graph for a-
t or v-t or x-t
Hint: Answer :
- Refer graph a-x graph for a-t
→A or v-t or x-t

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QUESTION 3: Solution 3:
Which of the following is FALSE
about the period of a pendulum in a Formula of period of SHM:
simple harmonic motion?
A. Directly proportional to square
= 2
root of length of string
B. Inversely proportional to square A. ∝

root of gravitational acceleration. 1
C. Inversely proportional to B. ∝

frequency. C. 2 = 2 2 1 1
D. Directly proportional to mass = = 2 = ∝

Hint: D. ∝ ∝
- Check formula of period for SHM (include Correct answer :
msry20202021
for pendulum and spring system. From spring system : = 2

Answer :

→D

QUESTION 4: Solution 4:

A body oscillates with simple harmonic From the question:
= 2 sin( 4 )
motion along the x-axis. Its displacement then compare with: = sin( )

varies with time, t according to the following

equation
= 2 sin( 4 )

where x is in cm and t in second. so,

The amplitude of the motion is A – amplitude =
2
A. 0.97 cm

B. 1.11 cm Use calculator : = 1.57 cm
2
C. 1.57 cm

D. 1.42 cm

Hint:
- Refer SHM equation.

Answer :

→C

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QUESTION 5: Solution 5:
= −1
What is the SI unit for frequency From the question: = 2
A. s 2
B. rad s-1
C. s-1 = and = 2 = 2
D. m
So, 2 −1
Hint: = 2 =
- Check frequency formula.
1 = −1
=
msry20202021
and T = second (s)

Unit for frequency is second-1 or s-1.

Answer :

→C

QUESTION 6: Solution 6:

a) Define simple harmonic motion and give a) Define simple harmonic motion
one example.
SHM is defined as a periodic motion without loss of energy in
b) Define which the acceleration of a body is directly proportional to its
i. amplitude, A displacement from the equilibrium position (fixed point) and is
ii. period, T directed towards the equilibrium position but in opposite
iii. frequency, f direction of the displacement.

and give one example = pendulum, grandfather clock, swing.

Hint: b)
- Check definition. i) Amplitude = maximum magnitude of the

Answer : displacement from the equilibrium position, O.
ii) Period = the time taken to complete one
→
cycle/rotation/revolution
iii) Frequency = the number of cycles completed in one

second

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QUESTION 7: Solution 7:
Equation of SHM: = sin( )
The displacement of a particle undergoing a
simple harmonic motion is given as x = 8 sin Compare with = 8 sin(4 )
4πt where x is measured in meter and t in
second. Determine where x is measured in meter and t in second

i. The amplitude i) amplitude, A = 8 m (unit compulsory !!)
ii. The angular velocity
iii. The period of oscillation ii) Angular velocity = angular frequency =
iv. The displacement when t = 0.2 s
= 4 rad s-1
Hint:
- CLO3. iii) 2 ( iv) The displacement when t = 0.2 s
- Use equation and formula Use : = 8 sin(4 )
= sub. t = 0.2 s → = 8 sin(4 (0.2))
Answer : 2 **Calculator in mode [RAD]
→ 8 m, rad s-1 , 0.5 s, 4.70 m
= 4 = 4.70
2

= 4
1

= 2 = 0.5

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QUESTION 8: Solution 8 (a):

The displacement of a particle oscillating in From equation : x = 4 sin 10πt
simple harmonic motion is given as:
a) vmax →
x = 4 sin 10πt
(4 10 ) Hint:
where x is measured in cm and t in second. = = Notes velocity of SHM
Find
Hint:
a) The maximum velocity of the particle = Calculator in DEG mode

b) The velocity of the particle when x = 3 cm = 4(10 ) 10

c) Sketch a graph of velocity, x, against time, = 40 10
t for this particle
vmax when cos10 = 1
so, the equation →

Hint: = 40
- CLO3. so answer v = 40 cm s-1 or 125.66 cm s-1
- Use equation and formula
or 1.26 m s-1

Answer : or just use :
= =4(10π) = 40π cm s-1
→

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Solution 8 (b): Solution 8(c): QUESTION 8:

(b) Use formula: (c) Sketch a graph of velocity, x, against time, t for this
particle
v =  A2 − x2
Form equation : x = 4 sin 10πt Hint : Compare
Form equation : x = 4 sin 10πt equation and
where x is measured in m and t in second study SHM
= 10 rad s-1 graph for x-t
A = 4 cm → 0.04 m
x = 3 cm → 0.03 m use

subs. Into : = ± 2 − 2 2
=

Hint: = (10 ) (0.04)2−(0.03)2
= (10 )(0.0265)
Calculator in = 0.83 −1
DEG mode

T(period) Hint:
Don’t forget to label axis with
unit, amplitude and time.

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QUESTION 9: Solution 9 (a):

One end of a tuning fork oscillates in simple From question :
harmonic motion of amplitude 0.50 mm and SHM, Amplitude,A = 0.50 mm, Period,T = 0.001 s
period of oscillation of 0.001 s. Determine
(a)
a) The maximum acceleration
From equation : = − 2
b) The acceleration when the displacement is
0.10 mm will be maximum when = .

c) Deduce an expression for the where A is amplitude.
displacement, x, in terms of the time, t.
Assume that when t = 0 s, the end is at the so, we get : = − 2 Hint:
equilibrium position. First, find ꞷ : -ve sign can be ignored.

Hint: 2 then use:
- CLO3. = = 2
- Use equation and formula
2 substitue:
= 0.001

= 2000 π rad s-1 = (2000 )2(0.50 × 10−3)

Answer : = 19739.20 −2
= 1.974 × 104 −2
→
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Solution 9 (b): Solution 9 (c): QUESTION 9:

(b) The acceleration when the displacement is (c) Deduce an expression for the displacement, x, in terms of
0.10 mm? the time, t. Assume that when t = 0 s, the end is at the
: equilibrium position.
From equation : = − 2
Form equation : x = Asinꞷt
= 2
substitue: From question SHM: Hint:
amplitude, A = 0.50 mm and = 2000 π rad s-1
= (2000 )2(0.10 × 10−3) No need to
= 19739.20 −2 Subtitue into equation: calculate
= 1.974 × 104 −2
= 0.50 × 10−3 2000
Hint:
Calculator in where x is measured in m and t in second
DEG mode

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QUESTION 10: Solution 10 (a):
The displacement of a particle in simple The displacement :
harmonic motion is described by the
equation: x = 5.0 cos (2πt+ π/3) Form equation : x = 5.0 cos (2πt+ π/3)
where x in meter, t in seconds and
(2πt + π/3) in radian. where x in meter, t in seconds
Determine at t = 4.0 s
when x at t = 4.0 s 25
a) The displacement 3
x = 5.0 cos (2πt+ π/3)
b) The velocity x = 5.0 cos (2π(4.0)+ π/3)

c) The acceleration x = 5.0 cos (26.1799)

x = (5.0)(0.50)

x = 2.50 m Hint:
Calculator in RAD mode

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 10:

Solution 10 (b) : Method 1 Method 2
The velocity 10 (b) Use formula:
Form equation : x = 5.0 cos (2πt+ π/3)

where x in meter, t in seconds

Derived equation x at t = 4.0 s Form equation : x = 4 sin 10πt
x = 5.0 cos (2πt+ π/3)
= 2 rad s-1
= = (5.0 cos (2πt+ π/3) A = 5.0 m
= cosθ x = 2.5 m

subs. Into : = ± 2 − 2
= - 5.0(2π) sin (2πt+ π/3) = − θ
= (2 ) (5.0)2−(2.5)2
then, = - (10π) sin (2π(4.0)+ π/3) = (2 )(4.3301)
= ± 27.21 −1
= - (10π)(0.8660)

= - 27.21 m s-1

Hint:
Calculator in RAD mode

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QUESTION 10:

Solution 10 (c) : Method 1 Method 2 : The acceleration
10 (c)
a) The acceleration Form equation : = − 2
Form equation : = - (10π) sin (2πt+ π/3)
= 2 rad s-1
where v in m s-1, t in seconds x = 2.5 m

Derived equation v at t = 4.0 s subs. Into :
= - (10π) sin (2πt+ π/3) = − 2
= −(2 )2(2.5)
= = (− (10π) sin (2πt+ π/3)) = −98.696
= ± 98.70 −2


then, = - 10.0π (2π) cos (2π(4.0)+ π/3)
= - (20.0π2)(0.5) = cosθ


= − θ

= - 98.696 m s-2

Hint:
Calculator in RAD mode

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QUESTION 11: Solution 11 (a):
From the graph:
(a)

A = 2 m , vmax = 6 m s-1 , f = ?

Use formula =

= Remember:
Formula not given in
PSPM

= 6 = 3 −1 msry20202021
2
The graphs in FIGURE 1.1 show the
variation of displacement, x with time, t for Then use, = 2
an object in simple harmonic motion.
Determine the frequency of the object if the
maximum velocity is 6 ms-1 = 2

= 3 = 0.477 −1 @ Hz

2

Hint: Answer :
- CLO3.
- Use equation and formula →

QUESTION 11: Solution 11 (b):
From the graph:
(b)
A = 4 m , vmax = 10 m s-1 , f = ?
Determine the frequency of the motion
from graph in FIGURE 1.2 above if the Use formula =
maximum velocity is 10 ms-1
=



= 10 = 2.5 −1

4

Then use, = 2


= 2

= 2.5 = 0.398 −1 @ Hz

2

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 12: Solution 12 :

From the graph:

F = - 4 N , A = 0.1 m , m = 2 kg

(i) Use formula = = − 2

= − 2 and x = A

= − 2

Find −4 = −(2) 2(0.1)

The graph in FIGURE 1.3 shows the = 20
relationship between the force, F and its = 4.47 −1
displacement, x for an object of mass 2 kg
which is in simple harmonic motion. Then use, = 2 = 2 = 1.406 s
Determine 4.47

i. The period of oscillation (ii) =
ii. The maximum speed of the object. = (0.1)(4.47)
= 0.447 −1
Hint: Answer :
- CLO3. msry20202021
- Use equation and formula →

QUESTION 13: Solution 13:

A block of mass 100 g is suspended From the graph:
from a vertical helix spring of spring m = 100 g , k = 20 N m-1 , A = 2 cm
constant 20 N/m. The block is pulled to
a distance of x = 2 cm from its (a) Use formula ,
equilibrium position of x = 0 cm.
Determine = , = − = − 2

a) The period of oscillation − = − 2 and x = A
= 2
b) The maximum velocity of the block

= = 20 = 14.142 rad s-1

0.1

Then, use = 2 = 2 = 0.444 s
14.142

+x

(b) = O
= (0.02)(14.142)
Hint: Answer : = 0.283 −1 −x m 
- CLO3. Fs a
- Use equation and formula →
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QUESTION 14: Solution 14:
From the question:
An object of mass 200 g is suspended from a m = 200 g, x = 2.0 cm = A ,
vertical helix spring. The spring is then
extended by 2.0 cm. If the object is (a) = = … =
oscillating in a vertical plane, determine = , =

a) The angular frequency of the oscillation. From = − 2 , then = = 9.81 = 22.15 rad s-1
0.02
b) The period of oscillation.
(b) Use = 2 = 2 = 0.284 s
c) The maximum velocity of the oscillating 22.15
object

d) The maximum acceleration of the object

(c) =
= 0.02 22.15 = 0.443 −1

(d) = − 2A
= 22.15 2 0.02 = 9.81 −2
Answer :
Hint: msry20202021
- CLO3. →
- Use equation and formula

QUESTION 15: Solution 15:

A body oscillates with Simple Harmonic (a) = 3 4 ℎ =
Motion described by the following expression
So, = 4 −1
= 3 4
Use , 2 2 = 1.571
Where x and t is displacement in meter and = = 4
time in second, respectively.
(a) Calculate the oscillating period. (b) (i) = = (3 4 )
(b) At t = 2 s,
Hint:
(i) calculate the speed of the body. = 3(4) 4 Calculator in RAD mode
(ii) state direction of the body’s motion
(iii) calculate the acceleration of the body. = 12 4

then, substitute t = 2 s into = 12 4

= 12 4(2)
= −1.746 −1

(ii) direction : to the left (negative sign)

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 15 : Solution 15 (continue):

A body oscillates with Simple Harmonic (b)
Motion described by the following expression (iii) = = (12 4 )

= 3 4

Where x and t is displacement in meter and = −12 4 4
time in second, respectively. = −48 4
(a) Calculate the oscillating period. then, substitute t = 2 s into = −48 4
(b) At t = 2 s,
= −48 4(2)
(i) calculate the speed of the body. = −47.49 −2
(ii) state direction of the body’s motion
(iii) calculate the acceleration of the Hint:
body. Calculator in RAD mode

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 16: Solution 16:

(a) Define period and frequency of a simple (a)
harmonic motion (SHM).
Period - the time taken to complete one cycle / rotation /
(b) Write an equation that relates the period and revolution
frequency of a SHM.
Frequency - the number of cycles completed in one second
(c) The acceleration, a of SHM in terms of
displacement, x is given by ∝ − . What (b) = 1 where T : period
is the meaning of the negative sign? f : frequency

(c) The negative sign in the equation indicates that the
direction of the acceleration, a is always opposite to the
direction of the displacement, x.

Hint: Answer :
- CLO1.
→
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QUESTION 17: Solution 17 :

A 80 g mass performs SHM according to the (i) = 25 10 ℎ =
equation
A = 25 mm
= 25 10
Where x in mm and t in s. 2 2 = 0.2
= = 10
(i) What is the amplitude and period of the
motion? (ii) = = (25 10 )

(ii) Write the SHM equation for a velocity and
acceleration.
= 25(10 ) 10
(iii)Sketch and label the graph of x-t, a-x
(iv)Calculate the velocity of the mass when its = 250 10

displacement is 15 mm. where v in mm s-1 and t in s

= = (250 10 )



= − 250 10 10
= −2500 2 10

where v in mm s-2 and t in s

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 17: Solution 17 :
(iii) Graph of x-t
A 80 g mass performs SHM according to the
equation

= 25 10

Where x in mm and t in s.

(i) What is the amplitude and period of the Graph of a-x
motion?
( m s−2)
(ii) Write the SHM equation for a velocity and 2500 2
acceleration.

(iii)Sketch and label the graph of x-t, a-x

(iv)Calculate the velocity of the mass when its
displacement is 15 mm.

−25 0 25 ( m)

Hint: Answer : −2500 2 msry20202021
- CLO3.
- Use equation and formula →

QUESTION 17: Solution 17 :

A 80 g mass performs SHM according to the (iv) = ± 2 − 2
equation
x = 15 mm = 15 × 10−3 m
= 25 10 = 10 −1
Where x in mm and t in s. A = 25 mm = 25 × 10−3 m

(i) What is the amplitude and period of the = (10 ) (25 × 10−3)2−(15 × 10−3)2
motion?
= (10 ) 4 × 10−4
(ii) Write the SHM equation for a velocity and
acceleration. = 0.628 −1

(iii)Sketch and label the graph of x-t, a-x
(iv)Calculate the velocity of the mass when its

displacement is 15 mm.

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 18: Solution 18 :
l = 150 cm = 1.5 m
A simple pendulum has a length of 150 cm and
performs simple harmonic motion. (a) = 2

(a) Calculate the period for the oscillation.
= 2 1.5
(b) What happen to the period of the oscillation
if the mass of pendulum is double? 9.81

= 2.46

(b) If the mass of pendulum is double?

From the formula,

= 2 , T is directly proportional to where g is


not constant value. So, T is not depend (independent) with mass. T

is remain constant.

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 19: Solution 19(a):

(a) First, find ꞷ : = − 2
2 = 1.571 2(0.14)
= ±0.346 −2
=
2 (m s−2)

= 4 0.346
= 1.571 rad s-1

a) FIGURE 1.4 shows a displacement versus −0.14 0 0.14 (m)
time graph of a body that oscillates in simple
harmonic motion. Sketch and label the graph −0.346
of acceleration versus displacement.
(b)
b) Find the maximum velocity of a body
=
= 0.14 1.571 = 0.22 −1

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 20: Solution 20(a):

(a) What is meant by simple harmonic (a) SHM is defined as a periodic motion without loss of
motion (SHM)? energy in which the acceleration of a body is directly
proportional to its displacement from the equilibrium
(b) A body connected to a light position (fixed point) and is directed towards the
vertical spring performs simple harmonic equilibrium position but in opposite direction of the
motion with an amplitude of 2.0 cm and a displacement.
period of 0.25 s. Calculate the acceleration
of the body when it is at 0.5 cm below the
equilibrium position.

(b) From the question:
A = 2.0 cm = 0.02 m , T = 0.25 s , x = 0.5 cm = 0.005 m

2 = − 2
= = 25.133 2(0.005)
= ±3.16 −2
2 = 3.16 −2
= 0.25

= 25.133 rad s-1

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 21: Solution 21:
From the question:
A = 4 cm , m = 50 g, k = 0.2 N m-1

(a) (i) A = 4 cm or 0.04 m

(ii) Use = 2 and =


FIGURE 1.5 shows a displacement x versus time = 2
t graph of a body of mass 50 g attached to a spring
and moving in simple harmonic motion. The
spring has a spring constant of 0.2 N m-1. =2 0.05 = 3.142
0.2
(a) Determine the
(i) amplitude of oscillation. (b) =
(ii) period of the oscillation.
= 0.04 0.2 = 0.08 −1
(b) Calculate the maximum velocity of the body. 0.05

(c) Sketch a graph of acceleration a against (c) = − 2 (m s−2)
displacement x for the motion. = − 2 2(0.04) 0.16
= 0.16 −2
Hint: Answer : 0 0.04 (m)
- CLO3. −0.04
- Use equation and formula → msry20202021
−0.16

QUESTION 22: Solution 22:
(a) State two examples of simple harmonic
motion. (a) - frictionless horizontal spring

(b) An object undergoes simple harmonic - frictionless vertical spring
motion with amplitude 0.2 m. Calculate the - Simple pendulum
total distance the particle has covered at the
end of 1.5 oscillations (b)

Hint: Answer : A = 0.2 m
- CLO3. For 1.5 oscillations, the object move forth and back in 6
- Use equation and formula → times. So,

Total distance = 6 x 0.2 m = 1.2 m

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QUESTION 23: Solution 23:

From the graph:
When x = -0.1 m , F = 4 N , m – 1.20 kg

(a) amplitude, A = 1.0 m

(b) Use formula = − to find k,

= −

The figure 1.6 shows the relationship between the 4 = − (−1.0)
force, F and its displacement, x from a fixed point = 4 = 4 −1
for a body of mass 1.20 kg which is in simple
harmonic motion. Find 1.0

a) the amplitude of the body. Then, use = 2 and =

b) the period of the body.
= 2 =2 1.20 = 3.44
c) the maximum speed of the body. 4


(c) =

= (1.0)( 4)

1.20

Hint: Answer : = 1.83 −1
- CLO3.
- Use equation and formula → msry20202021

QUESTION 24: Solution 24:

From the graph:
x = 5 m , a = -8 m s-2

FIGURE 1.7 shows a graph of acceleration, a (a) = − 2
against displacement, x of an oscillating object.
where ∶ ∶
(a) Write a mathematical expression showing
the relationship between a and x. ∶
∶
(b) Determine the frequency of the motion.
(b)

Use formula = − 2
−8 = − 2(5)

= 8 = 1.26 −1
= 2 5

Then use,

= = 1.26 = 0.201 Hz

2 2

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →

QUESTION 25: Solution 25 (a):

A block of mass 200 g is suspended from From the graph:
a vertical helix spring of spring constant m = 200 g , k = 40 N m-1 , A = 4 cm
40 N m-1. The block is pulled a distance
x=4 cm from its equilibrium position at (a) Use formula ,
x=0. Determine
= , = − = − 2
a) the angular frequency of the oscillation
− = − 2 and x = A
b) the period of the oscillation = 2

= = 40 = 14.142 rad s-1

0.2

(b) Use

= 2 = 2 = 0.444 s
14.142

Hint: Answer : msry20202021
- CLO3.
- Use equation and formula →