Download Chapter 9:Simple Harmonic Motion

PHYSICS
CHAPTER 9
CHAPTER 9:
Simple Harmonic Motion
(5 Hours)
1
PHYSICS
CHAPTER 9
SIMPLE HARMONIC MOTION




9.1 Simple Harmonic Motion (SHM)
9.2 Kinematics of Simple Harmonic Motion
9.3 Graphs of Simple Harmonic Motion
9.4 Period of Simple Harmonic Motion
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PHYSICS
CHAPTER 9
Learning Outcome:
9.1 Simple harmonic motion (1 hour)
At the end of this chapter, students should be able to:

Explain simple harmonic motion (SHM) as periodic
motion without loss of energy.

Examples of linear SHM system are simple
pendulum, frictionless horizontal and vertical spring
oscillations

Introduce and use SHM according to formulae:
2
d x
2
a  2   x
dt
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CHAPTER 9
9.1 Simple harmonic motion
9.1.1 Simple harmonic motion (SHM)

is defined as a periodic motion without loss of energy in
which the acceleration of a body is directly proportional to
its displacement from the equilibrium position (fixed point)
and is directed towards the equilibrium position but in
opposite direction of the displacement.
OR mathematically,
2
d x
a   x  2
dt
2
where
(9.1)
a : accelerati on of the body
ω : angular ve locity(ang ular frequency)
x : displaceme nt from the equilibriu m position, O
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 always constant thus

The angular frequency,

The negative sign in the equation 9.1 indicates that the
direction of the acceleration, a is always opposite to the
direction of the displacement, x.
The equilibrium position is a position at which the body would
come to rest if it were to lose all of its energy.
Equation 9.1 is the hallmark of the linear SHM.
Examples of linear SHM system are simple pendulum,
horizontal and vertical spring oscillations as shown in Figures
9.1a, 9.1b and 9.1c.




a  x

a
Fs
m
x
O
Figure 9.1a
x
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CHAPTER 9
x

Fs
m
O
x

a
m
 
Fs a
Figure 9.1b
x
O
x
Figure 9.1c
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9.1.2 Terminology in SHM
Amplitude (A)
is defined as the maximum magnitude of the displacement
from the equilibrium position.
 Its unit is metre (m).
Period (T)
 is defined as the time taken for one cycle.
 Its unit is second (s).
 Equation :
1

T
f
Frequency (f)
 is defined as the number of cycles in one second.
 Its unit is hertz (Hz) :
1 Hz = 1 cycle s1 = 1 s1
 Equation :

  2f OR f 
2
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CHAPTER 9
Equilibrium Position
-- a point where the acceleration of the body undergoing oscillation is
zero.
-- At this point, the force exerted on the body is also zero.
Restoring Force
-- the force which causes simple harmonic motion to occcur. This force is
proportional to the displacement from equilibrium & always directed
towards equilibrium.
Fs   k x
Equation for SHM
-- Consider a system that consists of a block of mass, m attached to the end
of a spring with the block free to move on a horizontal, frictionless
surface.
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-- when the block is displaced to one side of its equilibrium position &
released, it moves back & forth repeatedly about a maximum values of
displacement x.
-- Maximum value of x is called amplitude, A
-- It can be negative (–) or positive (+).
-- the spring exerts a force that tends to restore the spring to its equilibrium
position.
-- Given by Hooke’s law :
Fs   k x
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CHAPTER 9
Fs   k x
-- Fs is known as restoring force.
-- Applying Newton’s 2nd Law to the motion of the block :
Fnet  ma
k xma
k
a  
x
m
(
k
: constant v alue )
m
-- denote ratio k/m with symbol ω2 :
a   2 x
[ equation for SHM ]
-- any system that satify this equation is said to exhibit Simple Harmonic
Motion ( SHM )
-- from above equation, we find that:
a  x
-- the acceleration, a is proportional to the displacement of the block & its
direction is opposite the direction of the displacement.
PHYSICS
CHAPTER 9
Learning Outcome:
9.2 Kinematics of SHM (2 hours)
At the end of this chapter, students should be able to:

Write and use equation of displacement for SHM,
x  A sin t

Derive and apply equations for :

velocity,
dx
2
v
dt
  A  x 2

acceleration,

kinetic energy,

potential energy,
dv d 2 x
a
 2   2 x
dt dt
K

1
m 2 A2  x 2
2
1
U  m 2 x 2
2

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9.2 Kinematics of SHM
9.2.1 Displacement, x

Uniform circular motion can be translated into linear SHM and
obtained a sinusoidal curve for displacement, x against angular
displacement, graph as shown in Figure 9.6.
S
x

A
x1
N
A
O
M

 1
P 0  1 
2
Figure 9.6
T

3
2
2
 (rad )
A
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
CHAPTER 9
At time, t = 0 the object is at point M (Figure 9.6) and after time t
it moves to point N , therefore the expression for displacement,
x1 is given by
x1  A sin 1 where 1     and   t
x1  A sin t   
In general the equation of displacement as a function of time
phase
in SHM is given by
displacement from
(9.4)
equilibrium position x  A sin t 



Initial phase angle
(phase constant)
amplitude
angular time
frequency
 The S.I. unit of displacement is metre (m).
Phase
 It is the time-varying quantity t   .
 Its unit is radian.


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Initial phase angle (phase constant),

It is indicate the starting point in SHM where the time, t = 0 s.

If  =0 , the equation (9.4) can be written as

where the starting point of SHM is at the equilibrium position,
O.
For examples:
x  A sin t 
a. At t = 0 s, x = +A
x  A sin t   
A  A sin 0   

  rad
A O A
2


Equation : x  A sin  t   OR x  A cost 
2

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b. At t = 0 s, x = A
x  A sin t   
 A  A sin 0   

3

rad OR  rad
2
A O A
2
3  x  A sin  t   

Equation :


x  A sin  t   OR
2

2 

OR
x   A cost 
c. At t = 0 s, x = 0, but v = vmax

 vmax
A
Equation :
O A
x  A sin t    OR x   A sin t 
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9.2.2 Velocity, v

From the definition of instantaneous velocity,
dx
and x  A sin t   
v
dt
d
v   A sin( t   ) 
dt
d
v  A sin( t   ) 
dt
v  A cos(t   )
(9.5)

Eq. (9.5) is an equation of velocity as a function of time in SHM.

The maximum velocity, vmax occurs when cos(t+)=1 hence
vmax  A
(9.6)
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CHAPTER 9

The S.I. unit of velocity in SHM is m s1.

If  = 0 , equation (9.5) becomes
v  A cos t
Relationship between velocity, v and displacement, x
 From the eq. (9.5) :
(1)
v  A cos(t   )


From the eq. (9.4) :
x  A sin t   
x
sin t    
A
From the trigonometry identical,
sin 2   cos 2   1and 
cost    
 t   
1  sin t   
2
(2)
(3)
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
CHAPTER 9
By substituting equations (3) and (2) into equation (1), thus
x
v  A 1   
 A
2
2


x
2
2
v   A  A  2 
A 
v  A  x
2
2
(9.7)
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9.2.3 Acceleration, a

From the definition of instantaneous acceleration,
dv
and v  A cost   
a
dt
d
a   A cos(t   ) 
dt
d
a  A cos(t   ) 
dt
a   A sin( t   )
2


(9.8)
Eq. (9.8) is an equation of acceleration as a function of time in
SHM.
The maximum acceleration, amax occurs when sin(t+)=1
hence
(9.9)
a A 2
max

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CHAPTER 9

The S.I. unit of acceleration in SHM is m s2.

If  = 0 , equation (9.8) becomes
a   A sin t
2
Relationship between acceleration, a and displacement, x
 From the eq. (9.8) :
(1)
a   2 A sin( t   )


From the eq. (9.4) :
x  A sin t   
(2)
By substituting eq. (2) into eq. (1), therefore
a   x
2
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
CHAPTER 9
Caution :
 Some of the reference books use other general equation for
displacement in SHM such as
x  A cost   

The equation of velocity in term of time, t becomes
dx
v
  A sin( t   )
dt

(9.10)
(9.11)
And the equation of acceleration in term of time, t becomes
dv
a
  A 2 cos(t   )
dt
(9.12)
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9.2.4 Energy in SHM
Potential energy, U

Consider the oscillation of a spring as a SHM hence the
potential energy for the spring is given by
1 2
U  kx and k  m 2
2
1
2 2
(9.13)
U  m x
2

The potential energy in term of time, t is given by
1
U  m 2 x 2 and x  A sin t   
2
1
2 2
2
U  m A sin t   
2
(9.14)
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Kinetic energy, K
 The kinetic energy of the object in SHM is given by
1 2
K  mv and v   A2  x 2
2
1
2
2
2
(9.15)
K  m A  x 
2

The kinetic energy in term of time, t is given by
1 2
K  mv and v  A cost   
2
1
2 2
2
K  m A cos t   
2
(9.16)
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Total energy, E
 The total energy of a body in SHM is the sum of its kinetic
energy, K and its potential energy, U .
E  K U

From the principle of conservation of energy, this total energy is
always constant in a closed system hence
E  K  U  constant

The equation of total energy in SHM is given by


1
1
2
2
2
E  m A  x  m 2 x 2
2
2
1
2 2
(9.17)
E  m A
2
OR
1 2
E  kA
2
(9.18)
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CHAPTER 9
Example 9.1 :
An object executes SHM whose displacement x varies with time t
according to the relation


x  5.00 sin  2t  
2

where x is in centimetres and t is in seconds.
Determine
a. the amplitude, frequency, period and phase constant of the
motion,
b. the velocity and acceleration of the object at any time, t ,
c. the displacement, velocity and acceleration of the object at
t = 2.00 s,
d. the maximum speed and maximum acceleration of the object.
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Solution :
a. By comparing


x  5.00 sin  2t   with x  A sin t   
2

thus
i.
ii.
iii. The period of the motion is
iv. The phase constant is
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Solution :
b. i. Differentiating x respect to time, thus
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Solution :
b. ii. Differentiating v respect to time, thus
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Solution :
c. For t = 2.00 s
i. The displacement of the object is
ii. The velocity of the object is
OR
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Solution :
c. For t = 2.00 s
iii. The acceleration of the object is
OR
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Solution :
d. i. The maximum speed of the object is given by
ii. The maximum acceleration of the object is
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Example 9.2 :
The length of a simple pendulum is 75.0 cm and it is released at an angle
8 to the vertical. Frequency of the oscillation is 0.576 Hz. Calculate the
pendulum’s bob speed when it passes through the lowest point of the
swing.
(Given g = 9.81 m s2)
Solution :
8
L
A
m
A
O
A
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Solution : L  0.75 m;   8
At the lowest point, the velocity of the pendulum’s bob is
maximum hence
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Example 9.3 :
A body hanging from one end of a vertical spring performs vertical
SHM. The distance between two points, at which the speed of the
body is zero is 7.5 cm. If the time taken for the body to move
between the two points is 0.17 s, Determine
a. the amplitude of the motion,
b. the frequency of the motion,
c. the maximum acceleration of body in the motion.
Solution :
a. The amplitude is
7.5 102
A
 3.75 102 m
2
A
7.5 cm O
A
b. The period of the motion is
m
t  0.17 s
T  2t  20.17
T  0.34 s
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Solution :
b. Therefore the frequency of the motion is
c. From the equation of the maximum acceleration in SHM, hence
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CHAPTER 9
Example 9.4 :
An object of mass 450 g oscillates from a vertically hanging light
spring once every 0.55 s. The oscillation of the mass-spring is
started by being compressed 10 cm from the equilibrium position
and released.
a. Write down the equation giving the object’s displacement as a
function of time.
b. How long will the object take to get to the equilibrium position
for the first time?
c. Calculate
i. the maximum speed of the object,
ii. the maximum acceleration of the object.
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Solution : m  0.450 kg; T  0.55 s
a. The amplitude of the motion is A  10 cm
The angular frequency of the oscillation is
10 cm
m
t 0
and the initial phase angle is given by
0
10 cm
x  A sin t   
Therefore the equation of the displacement as a function of time is
OR
37
PHYSICS
Solution : m  0.450 kg; T
CHAPTER 9
 0.55 s
b. At the equilibrium position, x = 0
OR
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CHAPTER 9
Solution : m  0.450 kg; T  0.55 s
c. i. The maximum speed of the object is
ii. The maximum acceleration of the object is
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Example 9.5 :
An object of mass 50.0 g is connected to a spring with a force
constant of 35.0 N m–1 oscillates on a horizontal frictionless surface with
an amplitude of 4.00 cm and ω is 26.46 rads-1 . Determine
a. the total energy of the system,
b. the speed of the object when the position is 1.00 cm,
c. the kinetic and potential energy when the position is 3.00 cm.
Solution :
m  50.0 10 3 kg; k  35.0 N m 1 ; A  4.00 10 2 m
a. By applying the equation of the total energy in SHM, thus
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CHAPTER 9
b) The speed of the object when x = 1.00 ×10–2 m
c) The kinetic energy of the object
and the potential energy of the object
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CHAPTER 9
Example 9.6 :
An object of mass 3.0 kg executes linear SHM on a smooth horizontal
surface at frequency 10 Hz & with amplitude 5.0 cm. Neglect all resistance
forces. Determine :
(a) total energy of the system
(b) The potential & kinetic energy when the displacement of the object is 3.0 cm.
Solution:
Given : m = 3.0 kg
A = 5 cm = 0.05 m
f = 10 Hz → knowing : ω = 2πf
(b )
PHYSICS
b)
To calculate Kinetic energy :
CHAPTER 9
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CHAPTER 9
Exercise 9.1 :
1. A mass which hangs from the end of a vertical helical spring is
in SHM of amplitude 2.0 cm. If three complete oscillations take
4.0 s, determine the acceleration of the mass
a. at the equilibrium position,
b. when the displacement is maximum.
ANS. : U think ; 44.4 cm s2
2. A body of mass 2.0 kg moves in simple harmonic motion. The
displacement x from the equilibrium position at time t is given by


x  6.0 sin 2 t  
6

where x is in metres and t is in seconds. Determine
a. the amplitude, period and phase angle of the SHM.
b. the maximum acceleration of the motion.
c. the kinetic energy of the body at time t = 5 s.

ANS. : 6.0 m, 1.0 s,
rad ; 24.02 m s2; 355 J
3
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3. A horizontal plate is vibrating vertically with SHM at a frequency
of 20 Hz. What is the amplitude of vibration so that the fine sand
on the plate always remain in contact with it?
ANS. : 6.21104 m
4. A simple harmonic oscillator has a total energy of E.
a. Determine the kinetic energy and potential energy when the
displacement is one half the amplitude.
b. For what value of the displacement does the kinetic energy
equal to the potential energy?
ANS. :
3 1
2
E , E;
A
4 4
2
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CHAPTER 9
Learning Outcome:
9.3 Graphs of SHM (1 hour)
At the end of this chapter, students should be able to:

Sketch, interpret and distinguish the following graphs:

displacement - time

velocity - time

acceleration - time

energy - displacement
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9.3 Graphs of SHM
9.3.1 Graph of displacement-time (x-t)

From the general equation of displacement as a function of time
in SHM,
x  A sin t  




If  = 0 , thus x  A sin t 
The displacement-time graph is shown in Figure 9.7.
x
Period
A
Amplitude
0
T
4
T
2
3T
4
T
t
A
Figure 9.7
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PHYSICS

CHAPTER 9
For examples:
a. At t = 0 s, x = +A

Equation: x  A sin 
 t 
Graph of x against t:

 OR x  A cost 
2

x
A
0
T
4
T
2
3T
4
T
t
A
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CHAPTER 9
b. At t = 0 s, x = A

Equation: x  A sin  t

OR


3 
  OR x  A sin  t  
2
2 

x   A cost 
Graph of x against t:
x
A
0
T
4
T
2
3T
4
T
t
A
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c. At t = 0 s, x = 0, but v = vmax
Equation:
x  A sin t    OR x   A sin t 
Graph of x against t:
x
A
0
T
4
T
2
3T
4
T
t
A
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CHAPTER 9
How to sketch the x against t graph when   0
Sketch the x against t graph for the following expression:

π

x  2 cm sin  2t  
2

From the expression,

2

T  1s
rad s 
T
Sketch the x against t graph for equation x  2 sin 2t 

the amplitude, A  2 cm
the angular frequency,   2
1
x (cm)
2 T
4
t (s )
0
0 .5
1
2
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PHYSICS
CHAPTER 9


Because of

Sketch the new graph.
T
   rad  t  hence shift the y-axis to the
2
4 right by T
4
x (cm)
2
0
RULES
0 .5
1
t (s )
2
If  = negative value
shift the y-axis to the left
If  = positive value
shift the y-axis to the right
52
PHYSICS
CHAPTER 9
9.3.2 Graph of velocity-time (v-t)

From the general equation of velocity as a function of time in
SHM,
v  A cos t  



If  = 0 , thus v  A cost 

The velocity-time graph is shown in Figure 9.8.
v
A
0
T
4
T
2
3T
4
T
t
 A
Figure 9.8
53
PHYSICS

CHAPTER 9
From the relationship between velocity and displacement,
v   A2  x 2
thus the graph of velocity against displacement (v-x) is
shown in Figure 9.9.
v
A
A
0
A
x
 A
Figure 9.9
54
PHYSICS
CHAPTER 9
9.3.3 Graph of acceleration-time (a-t)

From the general equation of acceleration as a function of time
2
in SHM,

a   A sin t   
2
If  = 0 , thus a   A sin t 

The acceleration-time graph is shown in Figure 9.10.
a
A 2
0
T
4
T
2
3T
4
T
t
 A 2
Figure 9.10
55
PHYSICS

CHAPTER 9
From the relationship between acceleration and displacement,
a   2 x
thus the graph of acceleration against displacement (a-x) is
shown in Figure 9.11.
a
A 2
A
0
A
x
 A 2
Figure 9.11

The gradient of the a-x graph represents
gradient , m   2
56
PHYSICS
CHAPTER 9
9.3.4 Graph of energy-displacement (E-x)

From the equations of kinetic, potential and total energies as a
term of displacement

1
K  m 2 A2  x 2
2

1
1
2
2
; U  m x and E  m 2 A2
2
2
thus the graph of energy against displacement (a-x) is shown
in Figure 9.12.
E
Figure 9.12
1
E  m 2 A2  constant
2
1
U  m 2 x 2
2
1
K  m 2 A2  x 2 
2
x
57
PHYSICS

CHAPTER 9
The graph of Energy against time (E-t) is shown in Figure
9.13.
Energy
E
1
m 2 A2
2
1
U  m 2 A2 sin 2 t 
2
1
K  m 2 A2 cos 2 t 
2
t
Figure 9.13
58
PHYSICS
CHAPTER 9
Example 9.7 :
The displacement of an oscillating object as a function of time is
shown in Figure 9.14.
x (cm )
15.0
0
0.8
t (s)
 15.0
Figure 9.14
From the graph above, determine for these oscillations
a. the amplitude, the period and the frequency,
b. the angular frequency,
c. the equation of displacement as a function of time,
d. the equation of velocity and acceleration as a function of time.
59
PHYSICS
CHAPTER 9
Solution :
a. From the graph,
Amplitude,
Period,
Frequency,
b. The angular frequency of the oscillation is given by
c. From the graph, when t = 0, x = 0 thus   0
By applying the general equation of displacement in SHM
x  A sin t   
60
PHYSICS
CHAPTER 9
Solution :
d. i. The equation of velocity as a function of time is
ii. and the equation of acceleration as a function of time is
61
PHYSICS
CHAPTER 9
Example 9.8 :
 4.00
a (m s 2 )
0.80
0
4.00 x (cm )
 0.80
Figure 9.15
Figure 9.15 shows the relationship between the acceleration a and
its displacement x from a fixed point for a body of mass 2.50 kg at
which executes SHM. Determine
a. the amplitude,
b. the period,
c. the maximum speed of the body,
d. the total energy of the body.
62
PHYSICS
CHAPTER 9
Solution : m  2.50 kg
2
a. The amplitude of the motion is A  4.00 10 m
2
b. From the graph, the maximum acceleration is amax  0.80 m s
By using the equation of maximum acceleration, thus
OR The gradient of the a-x graph is
63
PHYSICS
CHAPTER 9
Solution : m  2.50 kg
c. By applying the equation of the maximum speed, thus
d. The total energy of the body is given by
64
PHYSICS
CHAPTER 9
Example 9.9 :
x ( m)
0.2
0
1
2
3 4 5
t (s )
 0.2
Figure 9.16
Figure 9.16 shows the displacement of an oscillating object of
mass 1.30 kg varying with time. The energy of the oscillating object
consists the kinetic and potential energies. Calculate
a. the angular frequency of the oscillation,
b. the sum of this two energy.
65
PHYSICS
CHAPTER 9
Solution : m  1.30 kg
From the graph,
Amplitude,
Period,
a. The angular frequency is given by
b. The sum of the kinetic and potential energies is
66
PHYSICS
CHAPTER 9
Exercise 9.2 :
a(ms -2 )
2
0
0.2
0.4 0.6 0.8 1.0
t (s )
2
1. The graph shows the SHM acceleration-time graph of a 0.5 kg
mass attached to a spring on a smooth horizontal surface. By
using the graph determine
(a) the spring constant
(b) the amplitude of oscillation
(c) the equation of displacement x varies with time, t.
x  0.032 cos 2.5t
ANS: 30.8 Nm-1, 0.032 m,
67
Summary :
PHYSICS
amax
 max
vmax
amax
 max
v   A2  x 2
0
A
0
 A 2
T
4
0
 A
0
a   2 x T
A 0
1 2
K  mv 2
2
vmax
1 2 3T
0 A
U  kx
4
2
amax
 max
A
t
x
v
a
CHAPTER
9
O
A
T
A
0
A
2
0
 A 2
K
U
0
1 2
kA
2
1
mA2 2
2
0
0
1 2
kA
2
1
mA2 2
2
0
0
1 2
kA
2
68
PHYSICS
CHAPTER 9
Learning Outcome:
9.4 Period of simple harmonic motion (1 hour)
At the end of this chapter, students should be able to:

Derive and use expression for period of oscillation, T for
simple pendulum and single spring.
(i) simple pendulum: T  2
(ii) single spring:
T  2
l
g
m
k
69
PHYSICS
CHAPTER 9
A. Simple pendulum oscillation
 Figure 9.2 shows the oscillation of the simple pendulum of
length, L.

L

T
x m P
O
mg sin  mg cos 

Figure 9.2

mg
70
PHYSICS




CHAPTER 9
A pendulum bob is pulled slightly to point P.
The string makes an angle,  to the vertical and the arc length,
x as shown in Figure 9.2.
The forces act on the bob are mg, weight and T, the tension in
the string.
Resolve the weight into


: mg cos 
: mg sin
the tangential component
the radial component
The resultant force in the radial direction provides the
centripetal force which enables the bob to move along the arc
and is given by
mv2


T  mg cos  

r
The restoring force, Fs contributed by the tangential
component of the weight pulls the bob back to equilibrium
position. Thus F  mg sin 
s
71
PHYSICS
CHAPTER 9


The negative sign shows that the restoring force, Fs is
always against the direction of increasing x.
For small angle, ;


sin    in radian
arc length, x of the bob becomes straight line (shown in
Figure 9.3) then
 L
x
sin    
L
thus
x
Fs  mg 
L
x
Figure 9.3
72
PHYSICS

CHAPTER 9
By applying Newton’s second law of motion,
 F  ma  F
s
Thus

a  x
By comparing
Thus
Simple pendulum executes
linear SHM
g
a    x
L
g
 
L
2
mgx
ma  
L
g
a    x
L
and
with
a   2 x
2

T
73
PHYSICS
CHAPTER 9
Therefore
L
T  2
g
where
(9.2)
T : period of the simple pendulum
L : length of the string
g : gravitatio nal accelerati on

The conditions for the simple pendulum executes SHM are



the angle,  has to be small (less than 10).
the string has to be inelastic and light.
only the gravitational force and tension in the string acting
on the simple pendulum.
Simulation 9.1
74
PHYSICS
CHAPTER 9
B. Spring-mass oscillation
Vertical spring oscillation

F
x1
O
m

mg
Figure 9.4a
Figure 9.4b

F1
x

a
O
m

mg
Figure 9.4c
75
PHYSICS
CHAPTER 9

Figure 9.4a shows a free light spring with spring constant, k
hung vertically.

An object of mass, m is tied to the lower end of the spring as
shown in Figure 9.4b. When the object achieves an equilibrium
condition, the spring is stretched by an amount x1 . Thus
F 0

F W  0
 kx1  W  0
W  kx1
The object is then pulled downwards to a distance, x and
released as shown in Figure 9.4c. Hence
 F  ma
F1  W  ma and F1  k x1  x 
 k x1  x   kx1   ma
k
a    x
m
then
a  x
Vertical spring oscillation executes
linear SHM
76

a
PHYSICS
Horizontal spring oscillation
 Figure 9.5 shows a spring is
initially stretched with a
displacement, x = A and then
released.
 According to Hooke’s law,

CHAPTER 9 Fs
m
Fs  kx

The mass accelerates toward
equilibrium position, x = 0 by
the restoring force, Fs hence
Then

Fs
m
a  x
T
t
4

a
T
t
2

Fs  0
m
Fs  ma
ma  kx
k
a    x
m
executes
linear SHM
Figure 9.5
t 0

Fs  0
m
3T
t 4
Fs

a
m
x  A
x 0
t T
x  A77
PHYSICS

CHAPTER 9
k
2
By comparing a    x with a   x
m
2
k
2
and  


Thus
T
m
where
Therefore
(9.3)
m T : period of the spring
oscillatio n
T  2
k m : mass of the object
k : spring constant (force constant)

The conditions for the spring-mass system executes SHM are
 The elastic limit of the spring is not exceeded when the
spring is being pulled.
 The spring is light and obeys Hooke’s law.
 No air resistance and surface friction.
78
PHYSICS
CHAPTER 9
Example 9.10 :
A certain simple pendulum has a period on the Earth surface’s of
1.60 s. Determine the period of the simple pendulum on the
surface of Mars where its gravitational acceleration is 3.71 m s2.
(Given the gravitational acceleration on the Earth’s surface is
g = 9.81 m s2)
2
2
Solution : TE  1.60 s; g E  9.81 m s ; g M  3.71 m s
The period of simple pendulum on the Earth’s surface is
But its period on the surface of Mars is given by
79
PHYSICS
CHAPTER 9
Solution : TE  1.60 s; g E  9.81 m s 2 ; g M
By dividing eqs. (1) and (2), thus
 3.71 m s 2
80
PHYSICS
CHAPTER 9
Example 9.11 :
A mass m at the end of a spring vibrates with a frequency of
0.88 Hz. When an additional mass of 1.25 kg is added to the mass
m, the frequency is 0.48 Hz. Calculate the value of m.
Solution : f1  0.88 Hz; f 2  0.48 Hz; Δm  1.25 kg
The frequency of the spring is given by
After the additional mass is added to the m, the frequency of the
spring becomes
81
PHYSICS
CHAPTER 9
Solution : f1  0.88 Hz; f 2  0.48 Hz;
By dividing eqs. (1) and (2), thus
Δm  1.25 kg
82
PHYSICS
CHAPTER 9
Exercise 9.3 :
1. An object of mass 2.1 kg is executing simple harmonic motion,
attached to a spring with spring constant k = 280 N m 1. When
the object is 0.020 m from its equilibrium position, it is moving
with a speed of 0.55 m s 1. Calculate
a. the amplitude of the motion.
b. the maximum velocity attained by the object.
ANS. : 5.17x10 2 m; 0.597 m s 1
2. The length of a simple pendulum is 75.0 cm and it is released
at an angle 8° to the vertical. Calculate
a) the period of the oscillation,
b) the pendulum’s bob speed and acceleration when it passes
through the lowest point of the swing.
(Given g = 9.81 m s 2)
ANS.: 1.74s; 0.378ms-1
83
PHYSICS
3.
CHAPTER 9
The acceleration of free fall on the Moon is 1/6 the acceleration
of free fall on the earth. If the period of a simple pendulum on
the earth is 1.0 second, what would its period be on the Moon.
ANS: 2.45 s
84
PHYSICS
CHAPTER 9
THE END…
Next Chapter…
CHAPTER 10 :
Mechanical waves
85