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9. SIMPLE HARMONIC MOTION AND OSCILLATIONS
(5 hours)
9.1 Simple Harmonic Motion
What is oscillation or periodic motion?
The back-and-forth motion of an object about a fixed point is called oscillation
For examples: the motion of the piston in a car engine, the swing bob of a
pendulum, the motion of a mass suspended from a spring.
F
Fig 9.1(a)
Equilibrium line
Fig 9.1(b)
F
Fig 9.1(c)
x
Horizontal spring oscillation
In Fig 9.1(a), when the mass is pulled, the spring is extended. As the mass is
displaced to the right, the restoring force of the spring acting on the mass is
directed to the left. Hence when it is released, the mass will move back to the left.
In Fig 9.1(b), when the mass is at the point x=0, the restoring force of the spring
=0, since F=kx. This point is called equilibrium point.
However, the mass continues its motion to the left after passing through the
equilibrium point because of its inertia.
In Fig 9.1(c), the spring is compressed as the mass continues its motion to the
left. The restoring force of the spring is now acting to the right.
From Fig 9.1(a) and (b), it shows that the directions of the restoring force, F is
always directed toward the equilibrium point.
The back-and-forth motion of the mass attached to the spring is called simple
harmonic motion.
1
Equations for SHM
By Hooke’s Law, the restoring force, F on the spring varies with the displacement
x as follow
F = - kx
where k = spring constant ......... (9.1)
The restoring force F is proportional directly to the displacement x. The negative sign
indicates that the force, F and the displacement, x always have opposite signs.
The mass, which is performing simple harmonic motion has an acceleration, a
given by
d 2x
a= 2
dt
Using Newton’s Law of motion, F=ma
From equation (9.1) F=ma = - kx where m = mass attached on a spring.
Therefore
kx
kx
d 2x
a= or
=...................................(9.2)
2
m
m
dt
k
Since
is a constant, the acceleration, a is proportional to the
m
displacement, x.
If the mass is oscillating with an angular velocity or angular frequency,, the
acceleration of the object is given by
a = - 2 x. ................................
(9.3)
The maximum acceleration, amax= 2A.
Definition of SHM
If the force, F acting on an object is proportional to the displacement, x of the
object from its equilibrium point, the object is performing simple harmonic
motion. This force must always act in a direction towards the equilibrium point
of the object.
F= - kx
Or
An object is in simple harmonic motion when its acceleration, a is directly
proportional to its displacement, x from its equilibrium point or a fixed point and
its direction of motion must always towards this point.
a= -kx .
2
Harmonic Vibrators with Small Oscillation
The objects which experienced the oscillating simple harmonic motions are also
called as Harmonic Vibrators With Small Oscillation. The examples are:
i)
the crystal quartz in the watch,
ii)
the current in an oscillating circuit,
iii)
the vibration of the atoms in the molecules or solid and
iv)
the simple pendulum.
9.2 Kinematics of Simple Harmonic Motion
Simple Harmonic motion and Circular motion
In figure 9.2, when the particle, P moves from the point R along the circle with
radius A in the anti-clockwise direction, its shadow, Q is moving
simultaneously towards the equilibrium point O along the x-axis.
x
Figure 9.2
If OQ=x, then the equation for
the displacement of Q can be written as
x = A cos θ
or
x = A cos  t
since cos θ =x/A
and the angular velocity,  = θ /t
At the time t=0 .Q is located at the maximum displacement,
where x=A( amplitude)
3
(9.4)
The displacement, x verses time t graph can be shown as in figure 9.3
x=Acost
Figure 9.3
To show that the acceleration of SHM, a= -2 x
By differentiating equation 9.4, the equation for the velocity,
dx
v=
= -  A sint
(9.5)
dt
Differentiating 9.5 again, the equation for the acceleration, a, is given by
d 2 x dv
=
= -2A cos t
dt
dt 2
Substitute x=A cos t
a=
Then
a=-2 x
(9.6)
Graph of acceleration versus displacement
a
Figure 9.4
Figure 9.4 shows the graph of acceleration versus displacement, x for the
equation 9.6.
The same graph can be obtained if the equation 9.1, is plotted with restoring
force F versus displacement, x.
F = - kx= - 2x a
The gradient of the graph= 2. Hence the angular velocity can be determined.
The x- intercept = xo= its amplitude of oscillation.
4
General solutions for a=-2 x
For the differential equation,
d 2x
dt 2
= -2 x, it can be shown mathematically
that the solutions are
or
x= A sin( t +ф )
x=A cos( t +ф )
where ф is the angular phase.
From equation 9.7.a, at t=0 and x=0, then
0 = A sin ( 0 + ф )
 x = A sin  t
and
ф =0
But for equation 9.7.b at t=0, and x=A
A = A cos ( 0 + ф )
 x = A cos  t
and
ф =0
Hence if ф=0, the displacement equations are
x = A sin  t
(9.8.a)
or
x = A cos  t
(9.8.b)
Graphs of (a) displacement, (b) velocity, and (c) acceleration
Figure 9.5 (a) displacement, (b) velocity, and (c) acceleration
5
(9.7.a)
(9.7.b)
Figure 9.5 (a), (b), (c) demonstrate how displacement, velocity and the acceleration
change with time. From the graphs it is observed that
i)
When t=0, displacement, x=0, the maximum velocity vmax = 2 A and
acceleration, a=0.
ii)
When t=T/4, displacement, x=A( maximum), the velocity v =0, and
acceleration, a= - 2 A, ( its negative maximum value).
iii)
When t=T/2, displacement, x=0 , the velocity v =-  A( its negative
maximum value) and acceleration, a =0.
The periodic time, T of SHM
,
The time for a body to move in a complete oscillation in a SHM is called
Period, T.
The number of oscillations in one second is called frequency, f .
n 1
f =  , where n=number of oscillations. For one oscillation, n=1
t T
and the time taken=T, the period
2
But =
, where  is the angular velocity.
T
1 

 
So f =
T 2
2
Hence the function (x = A sin  t) is repeated at each periodic time T.
The maximum velocity of a SHM
Differentiating the displacement equation, x = A cos  t, we obtain the
velocity, v = - A sin t
But sin2 t + cos2  t = 1
x
1
 sin t=  1  cos2 t  1  ( )2 
A2  x 2
A
A
Substitute sin t into the velocity equation
v=±  A2  x 2
(9.9)
when x=0, the maximum velocity, vmak = ± A
(9.10)
For SHM, the object has its maximum velocity when it is at the equilibrium
point. Fig 9.6 shows the variation of the velocity with the displacement
v
+A
x
- A
Fig. 9.6
6
Example 9.1
The equation of motion for a particle oscillating in simple harmonic motion is
given as : x= 5 sin 3t., where x is the displacement in cm.
a)
what is its amplitude,
b)
Find its period of oscillation,
c)
what is its displacement at time t=0.2s
d)
Find its maximum velocity,
e)
Sketch the graph of displacement against time.
Solution:
a) The amplitude, A= 5 cm.
b) From the equation given, the angular velocity,
2
=3=
T
Hence the period, T=(2)/3=2.09s
180o
c) x= 5 sin 3(0.2)= 5 sin 0.6x

= 5sin 34.38o
= 2.82 cm.
d) The maximum velocity , vmax=A
= 3x(5x10-2)
= 0.15ms-1.
e)
Displacement, x
t
Example 9.2
The graph below shows the forces acting on a particle of mass, 2kg.
F/N
20N
0.1m
-0.1m
0
a) What type of motion is the particle
following? Give reason for your answer.
b) What is its amplitude?
c) Find i) its angular velocity,
ii) its period,
iii) its maximum velocity.
-20N
7
Solution:
a) The particle is following a simple harmonic motion. It is because the force is
proportion to its displacement.
b) The amplitude is 0.1m
c) From the equation F= -m2x
i)
The slope of the graph= m2
2x 2 = 20/0.1
  = 10 rads-1.
The angular velocity is 10 rads-1
2
ii) Using =
= 10
T
The period, T=0.63s
iii) The maximum velocity,
vmax=A=10 x 0.1=1.0 ms-1.
9.3 Systems of simple Harmonic Motion
The energy of a SHM
Total mechanical energy:
The total mechanical energy of the system of SHM consists of kinetic
energy, K and the potential energy, U. The energy is conservative if there is
no external force acted on it.
Total mechanical energy, E= K + U
(9.11)
Kinetic energy:
For a simple pendulum which oscillates as a SHM, the kinetic energy, K for the
bob of mass m and its velocity v, is given by
K= ½ mv2
Its velocity, v= A2  x 2
Hence
K = ½ m[2(A2 – x2 )]
(9.12)
The value of K is maximum when x=0, i.e. when the bob of the pendulum is at
the equilibrium point of the SHM.
Hence the kinetic energy at the equilibrium point is
K = ½ m2A2
(9.13)
The variation of K with x is shown in figure 9.7. P and Q are the intersection
points for the graphs K and U. At these points K = U.
8
Potential energy:
The potential energy of a mass suspended from a spring is equal to the amount
of potential energy stored by the spring which is extended by x.
U= ½ kx2 where k is the spring constant.
From the equation of motion, F=ma= -kx and for SHM, a= - 2x
k
2=
 k=m2
m
Then, U=½ m2 x2.
When the mass suspended is at its equilibrium point, x=0, no energy is stored
in the spring, the total energy of the system, E is given by its kinetic energy, i.e.
E = ½ m2A2
(9.14)
The total energy E at any displacement = K +U .
Therefore from E = K + U, substitute E and K , we have
½ m2A2 =½ m[2(A2 – x2 )] + U
Then
U =½ m2x2
(9..15)
From Fig.9.7, the potential energy, U of the SHM is maximum, when x= ±
A(Amplitude). U is minimum i.e. zero when the mass is situated at its
equilibrium point.
Energy
Total Energy
E = K + U=½ m2A2
U =½ m2x2
K=½ m[2(A2 – x2 )]
X
Figure 9.7
9
K and U as a function of time:
From equation 9.12, if x= A sin  t, the kinetic energy,
K=½ m[2(A2 – A2sin 2t )]
=½ m2A2(1 – sin 2t )
=½ m2A2(1 – cos 2t )
(9.16)
The potential energy,
U =½ m2x2
=½ m2 sin 2t
(9.17)
The variations of K and U with time, is shown in the figure 9.8 where T is
the period time for one oscillation.
The total energy, E is constant and it does not change with time, t.
E=½ m2A2
U=½ m2 sin 2 t
K =½ m2cos 2 t
t
Figure9.8
Example 9.3
A steel strip, clamped at one end, vibrates with a frequency of 20 Hz. On the free
end, a small object of mass 2g is attached. If the amplitude of the oscillation is
5mm,
find ( a ) the velocity of the object when it passes through the zero position,
( b ) the acceleration of the object at its maximum displacement,
( c ) the maximum kinetic and potential energy of the object.
Solution
The oscillation of the steel strip is a SHM, with y as the displacement.
( a ) The velocity , v,=  A 2  y 2 .
When the end of the strip passes through the zero position
y=0; and the maximum speed vm is given by
vm= A,
But = 2  f = 2  x20 , and A=0.005m,

vm= 2  x20x0.005= 0.628 ms-1
( b ) The acceleration = -2y = -2 A at the maximum displacement,
 acceleration, a
= (2  x20)2 x 0.005
10
= 78.96 ms-2.
( c ) m=2g=2x10-3kg, vm= 0.628 ms-1.
 maximum K = ½ mvm2 = ½ x(2x10-3 ) x 0.6282 = 3.94 x10-4 J
Maximum U = maximum K = 3.94 x10-4 J
Example 9.4
A helical spring with a particle of mass 200g suspended from its free end, is
extended by 2.00 cm. If the mass is oscillating in vertical plane,
a) Prove that the mass is in simple harmonic motion, and
b) Find i) the period of the motion.
ii) the frequency of its oscillation.
iii) the maximum velocity of the particle
iv) the maximum kinetic energy of the particle.
Solution:
a)
The downward force on the spring= the weight of the mass=mg
The restoring force of the spring=kx ,
where k is a constant of the spring in Nm-1.
Thus, mg =kx = k x (2.00x10-2)…………......……..(i)
If the mass is x m below its original position at some instant and is moving
downwards, then the extension=( x + 2.00x10-2 )m,
the net downward force =mg – k( x + 0.02)
= mg – kx0.02 – kx = - kx
Using F=ma = - kx
The acceleration, a = -
k
x
m
Compare with the equation of SHM, a= -2x , then the mass suspended from
the helical spring is oscillating with SHM.
b)(i)
k
g
=
m 0.02
From equation (i),
g
g
x = - 2x, where 2=
.
0.02
0.02
 Acceleration, a =
Period T =
2

(ii) Frequency, f=
=
0.02 / g = 4.52x10-2 s
1
1

= 22 Hz.
T 4.52 x10 2
11
(iii) The maximum velocity,
vmax=A=2fA=2x22 x 0.02=2.76 ms-1
( iv ) The maximum kinetic energy,
K= ½ mv2 = ½ (200x10-3)x(2.76)2
= 0.76J
To determine the period for system of the SHM system
(a) Motion of a mass attached to a horizontal spring.
F
Equilibrium Line
F
F
x
Figure 9.9 Motion of mass attached to the spring on smooth surface.
The figure 9.9 shows the motion of a mass attached to the spring on a smooth,
frictionless surface.
When the spring is stretched by amount x, from its equilibrium point, the
restoring force of the spring, F will act in the opposite direction .The same thing
happens when the spring is compressed.
According to Hooke’s Law
F =- kx ( k = spring constant )
(9.18)
The force, F which acted on the mass is proportional directly to the displacement,
x and always acting towards its equilibrium point.
The negative sign indicates that F is always in the opposite direction to the
displacement.
From Newton’s second Law of motion, equation 9.18 can be written as
12
F = - kx = ma
Hence the acceleration, a.= -
k
x
m
(9.19)
Compare with the equation of SHM: a= -2 x
k
2=
m
The angular velocity,  =
The period,
T= 2π
k
m
(9.20)
m
k
(9.21)
Example 9.5
A mass of 200g is attached to a horizontal spring on a frictionless surface. The
mass is pulled 2.00cm to the right from its equilibrium point and then releases. It
oscillates in SHM. If the spring constant is 100Nm-1, find the frequency of the
mass.
Solution:
Using Hooke’s Law: F= - kx and
the equation for SHM: F= -m2 x
m2 =k
k
=
m
The frequency is given by , = 2πf

1 k
1
100
f=


2 2 m 2 200 x103
= 3.56 Hz.
13
(b) Simple Pendulum
Figure 9.10 Simple Pendulum
In figure 9.10, the weight of the bob, mg can be resolved into two
perpendicular components.
The component tangential to the circular path =mg sin θ is the restoring force,
F, which acted on the bob to bring it back to its equilibrium position.
F = mg sin θ
In NOP
(9.22)
sin θ =
x
l
From Newton’s second Law
x
l
[The negative sign indicates F and x act in opposite directions]
x
Hence, the acceleration, a = - g
(9.23)
l
F = ma = - mg sin θ= - mg
Compare with the equation for SHM: a = - 2 x
g
2=
l
g
=
(9.24)
l
and
the period , T =2π
l
g
(9.25)
14
Example 9.6
A student uses a simple pendulum of length 80.0cm to determine the
gravitational acceleration, g. If there are 20 oscillations in 35.9s, find:
i) The value of g, the acceleration due to gravity,
ii) The period of oscillation if the experiment is done in the moon,
where its gravitational field strength is only 1/6 of that of the earth.
Solution:
i) The period, T=35.9/20=1.795s
l
Using T =2π
g
l
T2 = 4 π2
g
2
l
2 80 x10
)

4

T2
1.7952
= 9.80 ms-2.
g= 4 π2 (
ii) From

l
g
2
T g = constant
(T2 g) moon = (T2 g) earth
T2 = 4 π2
T2 (1/6)g = (1.795)2 g
T= 4.40s
(c) Oscillation of liquid in a U tube.
Left
Right
water
Liquid
Figure 9.11
By blowing air gently down the right arm, the liquid there will be lower then
the one on the left with a displacement, x. However, the liquid will
immediately return to the right arm.
The levels of the liquid in the tube will oscillate for a short time before finally
coming to rest. The motion is one example of SHM.
15
If the equilibrium level for the liquid in both arms of the tube is S. The heights
of the liquid in both arms are h.
When the liquid is displaced in the left arm of the tube, the excess pressure in
the left arm is given
P = 2x ρ g
where ρ is the density of the liquid and g is the acceleration
due to gravity
The force F which causes the liquid to oscillate is
F= 2x ρ g A
where A= cross section area of the tube.
From Newton Second law, F =ma,
Hence
ma = 2x ρ g A
(9.26)
Where m= mass of the liquid in the tube i.e.
m = 2h ρ A
By replacing m into equation 9.26, it becomes
(2h ρ A) a = -2x ρ g A.
[The minus sign is used because the force and the displacement are always in
opposite direction.]
Hence
The acceleration, a = -
g
x
h
Compare with the equation of a SHM, a = -2 x,
2 x=

g
x
h
(9.27)
 = √ ( g/h)
The period, T =2π √ ( h/g)
(9.28)
16
(d) Torsion Pendulum
Figure .9.12
In the figure 9.12, a rod is suspended at the center of gravity by a wire NO. If
the rod is rotated by a small angle and then released, it will oscillate back and
forth about the equilibrium axis OP. This oscillating motion is an example of
SHM.
When the rod is rotated by a small angle θ, a restoring torque, Г will cause the
rod to return to its equilibrium position OP
The restoring torque is Г =k θ,
where k is torque per unit radian
(9.29)
Torque is given also as
=
Г =I,
d 2
,
dt 2
where I and  are the moment of inertia and the angular acceleration
respectively.
Hence
Consequently,
Г= I
I
d 2
dt 2
d 2
=kθ
dt 2
17
k
d 2
=- θ
2
I
dt
and
(9.30)
Compare with the equation of a SHM for angular displacement,
d 2
=
= -2 θ
2
dt
The angular velocity
 = √ (k/I)
and
the Period, T =2π
since T =
(9.31)
I
k
(9.32)
2

Example 9.7
A circular disc is suspended with a steel wire with torque per unit radian, k=100
Nm rad-1. When it is displaced with a small angle, it oscillates 10 times in
22.5s, calculate the moment of inertia of the disc.
Solution:
Using: T =2π
T2 = 4 π2
I
k
T= 22.5/10= 2.25s
I
k
kT 2 100 x(2.25)2
= 12.8 kgm2
I 2 
4
4 2
9.4 Free oscillation
A free oscillation is an oscillation that is free of any external force while the
system is oscillating.
No energy is lost externally and so the total energy of the system is always
constant.
The total energy, E is given by
E=½ m2xo2
(1.33)
Displacement, x
Time, t
Figure 9.13
18
In figure 9.13, the graph displacement versus time shows that the amplitudes
are always constant because the total energy, E is a constant.
Every free oscillation system possesses its natural frequency, which is
determined by certain factors.
Free oscillation is one example of prefect SHM. Other examples are simple
pendulum, loaded spring, liquid in U tube and the torsion pendulum.
1.5 Damped Oscillation
A damped oscillation occurs when there are external forces disturbing the
oscillation.
Part of the energy of the system is lost since work has to be done to overcome
these force. Therefore the amplitudes of the oscillation would not remain
constant but become progressively smaller. Thus a damped oscillation is not a
prefect SHM .
exponent
load
Water
water
Figure 1.13
Figure 9.14
-x0
exponent
Figure 9.15
19
In Figure 9.14, it shows a mass, which suspended from a spring, is oscillating
in water. There is always resistance due to the water that opposes the motion
of the mass.
The energy of the system is lost gradually as work has to be done to overcome
this resistance. The amplitude of the oscillation will decrease gradually and
finally become zero. Hence the oscillation is a damped oscillation
Figure 1.13 shows a damped oscillation. The amplitude is progressively
reduced with time, t.
From the equation, E=½ m2xo2 , we can deduce that the energy of the system
is also progressively decreasing and becomes zero finally .
There are three types of damped oscillations
i. under damped
ii. critically damped
iii. over damped
The under damped oscillation
It is oscillation where the amplitudes are decreasing with time until it becomes
zero. The mass which oscillates in the water is an example of such oscillation.
The critically damped oscillation
For a critically damped oscillation, there is a resistance that prevents the
system to vibrate. The time for its motion is very short.
Example of such damped oscillation is the shock absorber system of a vehicle.
The passenger is not affected much when the vehicle is moving along a bumpy
road.
Over damped oscillation
If the damping is very strong, the system does not vibrate at all. It is called
over damped motion. Such system takes a long time to come back to its
equilibrium position.
If the mass suspended from the spring is submerged in a very viscous liquid,
such as lubrication oil, the mass will move upward very slowly and then stops.
20
All these motion are show in figure 9.16.
Critical damping
Over damping
Under damping
Figure 9.16
9.6 Forced Oscillation and Resonance.
A
B
bob
Z
Figure 9.17
In order to keep a system, which is damped, in continuous oscillatory motion
an external periodic forces must be used. Such an oscillation is called forced
oscillation.
The frequency of this force is called the forcing frequency. It is not the natural
frequency of the system.
Figure 9.17 shows three pendulums X, Y, Z and a drive pendulum are
suspended from a string AB. The lengths of the pendulum X, Y and Z are
different, but the length of the drive pendulum is same as that of Y.
The three pendulums, X, Y, and Z are initially stationary. The drive pendulum
is set in oscillation by a small displacement.
21
When the drive pendulum is oscillating, X, Y and Z will start to oscillate with
different amplitudes. However only pendulum Y is oscillating with the same
amplitude as the drive pendulum.
Pendulum Y has the greatest amplitude among the three pendulums
suspended. Pendulum Y is in a state of resonance. Its frequency is the same
as that of the drive pendulum.
When the forcing frequency is same as that of the natural frequency of the
system, then resonance will take place as show in Figure 9.18.
Amplitude
Under damping
Over damping
Driver frequency
Figure 9.18
Resonance frequency
When the frequency of the drive pendulum is increased, the amplitude of the
forced pendulum also increased.
The amplitude of the forced pendulum becomes maximum, if the frequency of
the drive pendulum is the same as the natural frequency of the forced
pendulum.
Examples of Resonance
i)
When a lorry passes by at a certain speed, the road seems to vibrate
because of resonance. In this case, the frequency of the lorry equals the
natural frequency of the road.
ii)
An old bus seems to vibrate at a certain speed when its natural frequency
equals that of the engine. However, if the frequency of the engine is
changed by moving faster or slowly, the vibration will stop. Another way
to stop the vibration is to change the natural frequency of the bus by
reducing the number of passengers.
iii)
A house built nearby the airport will vibrate when an airplane flying over
it. The frequency of the noise from the engine of the airplane equals the
natural frequency of the house.
iv)
The wind blowing over the sea has a certain frequency. If this frequency is
same as the natural frequency of the bridge, the bridge will vibrate because
of resonance.
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SUMMARY
 SHM
i) If the force, F acting on an object is proportional to the displacement, x of
the object from its equilibrium point, the object is performing simple
harmonic motion.
This force must always act in a direction towards the equilibrium point of the
object.
F= - kx
Or
ii) An object is in simple harmonic motion when its acceleration, a, is directly
proportional to its displacement, x from its equilibrium point or a fixed point
and its direction of motion must always towards this point.
a= -kx .
 Acceleration of SHM, a= -2 x


The periodic time, T of SHM
The time for a body to move in a complete oscillation in a SHM is called

Period, T. T=
, where  is the angular velocity.
2
The maximum velocity of a SHM
v=±  A2  x 2

Kinetic energy:
K = ½ m[2(A2 – x2 )]


Potential energy:
Total energy :
SHM Systems
U=½ m2 x2
E=½ m2A2
System
Mass attached to a horizontal spring
Simple pendulum
Liquid in a u tube
Torsion pendulum

Period,T
m
T= 2π
k
T =2π
l
g
T =2π
l
g
T =2π
I
k
Free oscillation
A free oscillation is an oscillation that is free of any external force while the
system is oscillating.
No energy is lost externally and so the total energy of the system is always
constant.
23

Damped Oscillation
A damped oscillation occurs when there are external forces disturbing the
oscillation.
Part of the energy of the system is lost to overcome these force. Therefore the
amplitudes of the oscillation become progressively smaller
 Types of damped oscillations
i.
under damped
ii.
critically damped
iii.
over damped
 Forced oscillation: there are external periodic forces to keep the system in
continuous oscillatory motion.
 Resonance takes place when the forcing frequency is same as that of the
natural frequency of the system. The amplitude of oscillation is maximum.
TUTORAIL 9
Objective Questions
1.
A particle is performing simple harmonic motion with its displacement, x. If x is
given as x=30 sin 20t where t is the time in second, what is the frequency of the
system.


A.
Hz
C.
Hz
10
30
10
30
B.
Hz
D.
Hz


2. A particle is performing simple harmonic motion about a point O with its
amplitude a and period T. The displacement of the particle at time T/8 after it
passes through O is
a
A.
C. 2 a
2
a
3a
B.
D .
2
2
3. A particle is performing simple harmonic motion with its amplitude 2.0x10-3m
and period 0.10s. The maximum velocity of the particle in ms-1 is
A. 0.08
B. 0.13
C. 0.35
D. 0.47
4. Two masses P and Q are suspended with two wires separately. The moments of
inertia of both masses are the same. The masses are set into oscillations with
the same angular amplitudes. If the frequency of Q is twice that of P, what is
the ratio of the maximum kinetic energy of P to that of Q?
A. 1/16
B. ¼
C ½.
D. 2
24
5. A particle is performing simple harmonic motion. Which of the following
graphs represents the relationship between the force, F and the displacement of
the particle?
A
B
C
D
F
0
F
x
0
F
x
F
0
x
0
x
6. The values of the acceleration, a and displacement x for a particle performing
simple harmonic motion are shown in the table below:
a(mms-2)
x(mm)
16
-2
8
-1
0
0
What is the period of the oscillation?

A. . s
2
2
B.
s
2
-8
1
C.
D.

2
-16
2
s

s
4
7. The acceleration due to gravity on earth is six times that of the moon. If the
period of a simple pendulum on earth is 1s, what is the period for the same
pendulum on moon?
A.1/6s
C. 6s
B 1/ 6 s
D.
6s
8. A mass attached to a spring is performing simple harmonic motion with constant
amplitude 5.0 cm on a smooth horizontal table. The maximum kinetic energy of
the mass is 20J. If the potential energy is 10J, what is the magnitude of the
displacement from its equilibrium position?
A. 1.41cm
B. 2.24cm
C. 2.50cm
D. 3.54cm
25
Structured Questions:
.
1. A particle is performing a simple harmonic motion according to the equation
given: y= 10 sin t, where, , is the angular velocity of the particle in rad s-1
and t is the time in second. If the period of the oscillation is 30s, find
a. The amplitude,
b. The maximum velocity of the particle,
c. The maximum acceleration of the particle,
d. The displacement, velocity and acceleration of the particle when t=
15s.
2. An object moving with simple harmonic motion has an amplitude of 2 cm and
a frequency of 20Hz. Calculate
a. the period of oscillation,
b. the acceleration at the middle and end of an oscillation,
c. the velocities at the corresponding instants in (ii).
3. A body of mass 200g is performing simple harmonic motion with amplitude of
20mm. The maximum force which acts upon it is 0.064N. Calculate
a. its maximum velocity, and
b. period of oscillation.
4. If the displacement of a particle in simple harmonic motion is given by
y=Asint, show that the velocity of the particle is
dy
  A2  y 2
v=
dt
5. The graph for a particle in SHM is shown below:
y(cm)
4
0
-4
0.6
1.2
t(s)
a) Determine
i) the amplitude,
ii) the frequency,
iii) the period and
b) Write the displacement equation in the form of y=yo sin (t + )
6.
a) The time taken for 20 oscillations of a simple pendulum is 36.2s.What is
the length of the simple pendulum in a place where the acceleration
due to gravity, g=9.81ms-2?
b) What is the frequency of the simple pendulum from (a) if it is located in a
lift which is accelerating upward with 2.00ms-2?
26
c) What will be the frequency of the simple pendulum if it is falling freely?
7. A particle of mass 0.2kg is performing SHM with amplitude, 0.5m and period,
4s. Determine
a) velocity,
b) kinetic energy
c) potential energy
at a point + 0.2m from its equilibrium position.
8. A uniform wooden rod floats upright in water with a length of 30 cm
immersed. If the rod is depressed slightly and then released, prove that its
motion is simple harmonic and calculate the period.
9. A particle of mass 2kg is moving in simple harmonic motion. The changes of
potential energy, U with its displacement, x is shown in the graph below
U(joule)
1.0
-0.2
0
0.2
x(m)
Determine from the graph:
a) the angular velocity of the particle,
b) the period of oscillation.
Essay Questions:
1.
a) Explain what is meant by simple harmonic motion.
b) Show that the vertical oscillations of a mass suspended by a light
helical spring are simple harmonic and describe an experiment with the
spring to determine the acceleration due to gravity, g.
c) A small mass rests on a horizontal platform which vibrates vertically
in simple harmonic motion with a period of 0.50 second. Find the
maximum amplitude of the motion which will allow the mass to
remain in contact with the platform throughout the motion.
2.
a)Explain briefly what is the meaning of:
i) free oscillation, ii) force oscillation and iii) resonance
State the conditions for resonance to occur.
b) Using the same axes, sketch a set of graphs to show the variations of
the amplitudes of an oscillating particle with the changes in the forced
frequency :
i. under damped oscillation
ii. critically damped oscillation, and
iii. overdamped oscillation.
27
3.
a) Give two practical examples of oscillatory motion which are simple
harmonic motions.
b) State the conditions necessary for them to perform the simple harmonic
motion
c ) A point- mass is moving with simple harmonic motion. Draw sketch graphs
on the same axes to show the variation with position of
i)
the potential energy,
ii)
the kinetic energy ,and
iii)
the total energy of the mass.
28