5.

Oscillations
a definite interval of time is called periodic
Can you recall? motion. A body performing periodic motion
goes on repeating the same set of movements.
1. What do you mean by linear motion
and angular motion? The time taken for one such set of movements
2. Can you give some practical examples is called its period or periodic time. At the end
of oscillations in our daily life? of each set of movements, the state of the body
3. What do you know about restoring is the same as that at the beginning. Some
force? examples of periodic motion are the motion
4. All musical instruments make use of of the moon around the earth and the motion
oscillations, can you identify, where? of other planets around the sun, the motion of
5. Why does a ball floating on water
electrons around the nucleus, etc. As seen in
bobs up and down, if pushed down and
released? Chapter 1, the uniform circular motion of any
object is thus a periodic motion.
5.1 Introduction: Another type of periodic motion in
Oscillation is a very common and which a particle repeatedly moves to and
interesting phenomenon in the world of Physics. fro along the same path is the oscillatory or
In our daily life we come across various vibratory motion. Every oscillatory motion is
examples of oscillatory motion, like rocking periodic but every periodic motion need not be
of a cradle, swinging of a swing, motion of the oscillatory. Circular motion is periodic but it is
pendulum of a clock, the vibrations of a guitar not oscillatory.
or violin string, up and down motion of the The simplest form of oscillatory periodic
needle of a sewing machine, the motion of the motion is the simple harmonic motion in which
prongs of a vibrating tuning fork, oscillations every particle of the oscillating body moves
of a spring, etc. In these cases, the motion to and fro, about its mean position, along a
repeated after a certain interval of time is a certain fixed path. If the path is a straight line,
periodic motion. Here the motion of an object the motion is called linear simple harmonic
is mostly to and fro or up and down. motion and if the path is an arc of a circle,
Oscillatory motion is a periodic motion. In it is called angular simple harmonic motion.
this chapter, we shall see that the displacement, The smallest interval of time after which the to
velocity and acceleration for this motion can be and fro motion is repeated is called its period
represented by sine and cosine functions. These (T) and the number of oscillations completed
functions are known as harmonic functions. per unit time is called the frequency (n) of the
Therefore, an oscillatory motion obeying such periodic motion.
functions is called harmonic motion. After
studying this chapter, you will be able to Can you tell?
understand the use of appropriate terminology
to describe oscillations, simple harmonic Is the motion of a leaf of a tree blowing in
motion (S.H.M.), graphical representations the wind periodic?
of S.H.M., energy changes during S.H.M., 5.3 Linear Simple Harmonic Motion
damping of oscillations, resonance, etc. (S.H.M.):
5.2 Explanation of Periodic Motion: Place a rectangular block on a smooth
Any motion which repeats itself after frictionless horizontal surface. Attach one end
109
of a spring to a rigid wall and the other end to f   kx --- (5.1)
the block as shown in Fig. 5.1. Pull the block where, k is a constant that depends upon the
of mass m towards the right and release it. The elastic properties of the spring. It is called the
block will begin its to and fro motion on either force constant. The negative sign indicates
side of its equilibrium position. This motion is that the force and displacement are oppositely
linear simple harmonic motion. directed.
If the block is displaced towards left from
its equilibrium position, the force exerted by
the spring on the block is directed towards the
right and its magnitude is proportional to the
displacement from the mean position. (Fig.
5.1(c))
Thus, f = - kx can be used as the equation
of motion of the block.
Now if the block is released from the
rightmost position, the restoring force exerted
by the spring accelerates it towards its
equilibrium position. The acceleration (a) of
the block is given by,
f k
a     x --- (5.2)
m m
Fig. 5.1 (a), (b) and (c): Spring mass oscillator. where, m is mass of the block. This shows
that the acceleration is also proportional to
Remember this
the displacement and its direction is opposite
For such a motion, as a convention, we shall to that of the displacement, i.e., the force and
always measure the displacement from the acceleration are both directed towards the
mean position. Also, as the entire motion mean or equilibrium position.
is along a single straight line, we need not As the block moves towards the mean
use vector notation (only ± signs will be position, its speed starts increasing due to
enough). its acceleration, but its displacement from
Fig. 5.1(b) shows the equilibrium position the mean position goes on decreasing. When
in which the spring exerts no force on the the block returns to its mean position, the
block. If the block is displaced towards the displacement and hence force and acceleration
right from its equilibrium position, the force are zero. The speed of the block at the mean
exerted by the spring on the block is directed position becomes maximum and hence its
towards the left [Fig. 5.1(a)]. On account of its kinetic energy attains its maximum value.
elastic properties, the spring tends to regain its Thus, the block does not stop at the mean
original shape and size and therefore it exerts a position, but continues to move beyond the
restoring force on the block. This is responsible mean position towards the left. During this
to bring it back to the original position. This process, the spring is compressed and it exerts
force is proportional to the displacement but its a restoring force on the block towards right.
direction is opposite to that of the displacement. Once again, the force and displacement are
If x is the displacement, the restoring force f is oppositely directed. This opposing force
given by, retards the motion of the block, so that the
110
speed goes on reducing and finally it becomes
zero. This position is shown in Fig. 5.1(c). In Activity
this position the displacement from the mean
position and restoring force are maximum. Some experiments described below can be
This force now accelerates the block towards performed in the classroom to demonstrate
the right, towards the equilibrium position. The S.H.M. Try to write their equations.
process goes on repeating that causes the block
to oscillate on either side of its equilibrium (a) A hydrometer is
(mean) position. Such oscillatory motion along immersed in a glass jar
a straight path is called linear simple harmonic filled with water. In the
motion (S.H.M.). Linear S.H.M. is defined as equilibrium position
the linear periodic motion of a body, in which it floats vertically in
force (or acceleration) is always directed water. If it is slightly
towards the mean position and its magnitude depressed and released, it bobs up and down
is proportional to the displacement from the performing linear S.H.M.
mean position. (b) A U-tube is filled with a sufficiently long
column of mercury. Initially when both the
Use your brain power
If there is friction between a block and
the resting surface, how will it govern the
motion of the block?
arms of U tube are exposed
Remember this to atmosphere, the level of
A complete oscillation is when the object mercury in both the arms
goes from one extreme to other and back to is the same. Now, if the level of mercury
the initial position. in one of the arms is depressed slightly
The conditions required for simple harmonic and released, the level of mercury in each
motion are: arm starts moving up and down about the
1. Oscillation of the particle is about a equilibrium position, performing linear
fixed point. S.H.M.
2. The net force or acceleration is always
directed towards the fixed point. 5.4 Differential Equation of S.H.M. :
3. The particle comes back to the fixed In a linear S.H.M., the force is directed
point due to restoring force. towards the mean position and its magnitude
Harmonic oscillation is that oscillation is directly proportional to the displacement
which can be expressed in terms of a single of the body from mean position. As seen in
harmonic function, such as x  a sin t or Eq. (5.1),
x  a cos t f = - kx
Non-harmonic oscillation is that oscillation where k is force constant and x is displacement
from the mean position.
which cannot be expressed in terms of single
According to Newton’s second law of motion,
harmonic function. It may be a combination
f = ma ∴ ma = - kx --- (5.3)
of two or more harmonic oscillations such
dx
as x = a sin ω t + b sin2 ω t , etc. The velocity of the particle is, v=
dt
111
 dv d2x d2x
and its acceleration, a = = 2 ∴ 2   x
2
--- (5.6)
dt dt dt
Substituting it in Eq. (5.3), we get d2x
d2x But a = 2 is the acceleration of the particle
m 2   kx dt
dt performing S.H.M.
d2x k ∴ a   2 x --- (5.7)
 2  x  0 --- (5.4)
dt m This is the expression for acceleration in terms
k of displacement x.
Substituting   2 , where ω is the d2x
angular frequency, m From Eq. (5.6), we have 2   2 x
dt
d2x d  dx 
∴   2
x
dt 2
2x  0 --- (5.5) dt  dt 
Eq. (5.5) is the differential equation of linear dv
   2 x
S.H.M. dt
dv dx
   2 x
Can you tell? dx dt
dv 
Why is the symbol ω and also the term v   2 x
dx
angular frequency used for a linear motion?  v dv   2 x dx
Example 5.1 A body of mass 0.2 kg Integrating both the sides, we get
v dv   x dx
2
performs linear S.H.M. It experiences
a restoring force of 0.2 N when its v2 2x2
  C , --- (5.8)
displacement from the mean position is 4 2 2
cm. Determine (i) force constant (ii) period
of S.H.M. and (iii) acceleration of the where C is the constant of integration.
body when its displacement from the mean Let A be the maximum displacement
position is 1 cm. (amplitude) of the particle in S.H.M.
Solution: (i) Force constant, When the particle is at the extreme
k = f / x position, velocity (v) is zero.
= (0.2)/ 0.04 = 5 N/m Thus, at x   A,v  0
(ii) Period T  2 /  Substituting in Eq. (5.8), we get
m 0.2  2 A2
0 C
= 2 2 = 0.4π s 2
k 5
(iii) Acceleration  2 A2
C   --- (5.9)
k 5 2
a   2 x   x    0.04  1 m s 2
m 0 .2 Using C in Eq. (5.8), we get
5.5  Acceleration (a), Velocity (v) and v2  2 x 2  2 A2
 
Displacement (x) of S.H.M. : 2 2 2
We can obtain expressions for the 
 v 2   2 A2  x 2 
acceleration, velocity and displacement of
a particle performing S.H.M. by solving the  v   A2  x 2 --- (5.10)
differential equation of S.H.M. in terms of This is the expression for the velocity of a
displacement x and time t. 2 particle performing linear S.H.M. in terms of
d x displacement x .
From Eq. (5.5), we have 2x  0 dx
dt 2
Substituting v = in Eq. (5.10), we get
dt
112
 dx
  A2  x 2 In the cases (i) and (ii) above, we have used
dt
dx the phrase, “if the particle starts S.H.M…...”

   dt More specifically, it is not the particle that
A2  x 2
starts its S.H.M., but we (the observer)
Integrating both the sides, we get
dx start counting the time t from that instant.
 A2  x 2   dt The particle is already performing its
motion. We start recording the time as per
x our convenience. In other words, t = 0 (or
sin 1     t   --- (5.11)
 A initial condition) is always subjective to the
Here f is the constant of integration. To observer.
know f , we need to know the value of x at
any instance of time t, most convenient being Expressions of displacement (x), velocity (v)
t = 0. and acceleration (a) at time t:
 x  Asin  t    --- (5.12) From Eq. (5.12), x  Asin  t   
dx
This is the general expression for the v   A cos  t   
dt
displacement (x) of a particle performing
dv
linear S.H.M. at time t. Let us find expressions a    A 2 sin  t   
dt
for displacement for two particular cases.
Case (i) If the particle starts S.H.M. from the Example 5.2: A particle performs linear
mean position, x = 0 at t = 0 S.H.M. of period 4 seconds and amplitude
x
Using Eq. (5.11), we get   sin 1    0 or 4 cm. Find the time taken by it to travel a
 A
Substituting in Eq. (5.12), we get distance of 1 cm from the positive extreme
x   A sin  t  --- (5.13) position.
This is the expression for displacement at any Solution: x  Asin  t   
instant if the particle starts S.H.M. from the Since particle performs S.H.M. from
π
mean position. Positive sign to be chosen if it positive extreme position, f = and
2
starts towards positive and negative sign for from data
starting towards negative. x  A  1  3cm
 2 
Case (ii) If the particle starts S.H.M. from the  3 4 sin  t   
extreme position, x   Aat t  0  T 2
3 2 π
x 
  sin 1    or
3 ∴  cos t = cos  t
 A 2 2 4 4 2
c
Substituting in Eq. (5.12), we get    
∴ t  41.4   41.4  
 t = 0.46s
0

   3  2  180 
x  A sin   t  or x  A sin   t  
 2  2   180 
∴ x   A cos t  --- (5.14)  Or ,  2 t  41.4
 t  0.46 s 
 
This is the expression for displacement at any
Example 5.3: A particle performing
instant, if the particle starts S.H.M. from the
linear S.H.M. with period 6 second is
extreme position. Positive sign for starting
at the positive extreme position at t = 0.
from positive extreme position and negative
The particle is found to be at a distance
sign for starting from the negative extreme
of 3 cm from this position at time t = 7s,
position.
before reaching the mean position. Find the

113
 amplitude of S.H.M. 
 x   Asin  ∴ xmax   A
Solution: x  Asin( t   ) 2
Thus, at the extreme position the displacement
Since particle starts (t = 0) from positive
of the particle performing S.H.M. is maximum.
extreme position, f = π/2 and x  A  3
  2) Velocity: According to Eq. (5.10) the
 x  A sin   t   magnitude of velocity of the particle performing
 2
 2  S.H.M. is v   A2  x 2
 A  3  Asin t    At the mean position, x  0  v max   A .
 T 2
A3  2  Thus, the velocity of the particle in S.H.M.
  sin   7   is maximum at the mean position.
A  6 2
A3  7   At the extreme position, x    A  v min  0.
  sin    Thus, the velocity of the particle in S.H.M.
A  3 2
A3     1 is minimum at the extreme positions.
  sin     cos  3) Acceleration: The magnitude of the
A 3 2 3 2
acceleration of the particle in S.H.M is ω 2 x

 2 A  6  A
At the mean position x= 0 , so that the
 A  6cm
acceleration is minimum. ∴ a min 0 .
Example 5.4: The speeds of a particle At the extreme positions x   A , so that the
performing linear S.H.M. are 8 cm/s and acceleration is maximum amax = ω A
2

6 cm/s at respective displacements of 6 cm

and 8 cm. Find its period and amplitude. Can you tell?
Solution:

v   A2  x 2  1. State at which point during an oscillation
the oscillator has zero velocity but
8 
∴ 
A 2
 62  4  A 2
 36  positive acceleration?
or 
6  A 2
 82  3  A 2
 64  2. During which part of the simple
harmonic motion velocity is positive
∴ A = 10cm but the displacement is negative, and

v1    A2  x12  vice versa?
2 2 3. During which part of the oscillation the
 8 
T

 10 2  6 2  8 
T
8
two are along the same direction?
∴ T = 6.284 s
Example 5.5: The maximum velocity of a
Extreme values of displacement (x), velocity particle performing S.H.M. is 6.28 cm/s. If
(v) and acceleration (a): the length of its path is 8 cm, calculate its
1) Displacement: The general expression for period.
displacement x in S.H.M. is x  Asin  t    Solution:
At the mean position,  t    = 0 or π v max  6.28 
cm
 2 
cm
and A= 4 cm
∴ xmin = 0. s s
2
Thus, at the mean position, the v max  A  A
T
displacement of the particle performing 2
S.H.M. is minimum (i.e. zero). ∴ 2  4
 3π T
At the extreme position,  t     or ∴ T = 4s
2 2
 x  Asin  t   
114
 Example 5.6: The maximum speed of a This result shows that the particle is at the
2
particle performing linear S.H.M is 0.08 same position after a time . That means,

m/s. If its maximum acceleration is 0.32 m/ the particle completes one oscillation in time
s2, calculate its (i) period and (ii) amplitude. 2 . It can be shown that t  T  2 is the
Solution:  
a max A 2 2
 0.32  2
minimum time after which it repeats.
  
(i) v max A T 0.08 T 2
Hence its period T is given by T 
T  1.57 s 
From Eq.(5.4) and Eq.(5.5)
2  A  2cm k force per unit displacement
(ii) v max  A  A 2  
T m mass
= acceleration per unit displacement
5.6: Amplitude(A), Period(T) and Frequency 2
(n) of S.H.M. : T 
acceleration per unit displacement 
5.6.1 Amplitude of S.H.M.:
m
Also, T = 2π --- (5.15)
k
5.6.3 Frequency of S.H.M.:
Fig. 5.2 S.H.M. of a particle. The number of oscillations performed by
Consider a particle P performing S.H.M. a particle performing S.H.M. per unit time is
along the straight line MN (Fig. 5.2). The called the frequency of S.H.M.
centre O of MN is the mean position of the In time T , the particle performs one
particle. 1
oscillation. Hence in unit time it performs
The displacement of the particle as given oscillations. T
by Eq. (5.12) is x  A sin  t    Hence, frequency n of S.H.M. is given by
The particle will have its maximum 1  1 k
displacement when sin  t     1, i.e., n    --- (5.16)
T 2 2 m
when x   A . This distance A is called the
Combination of springs: A number of
amplitude of S.H.M.
springs of different spring constants can be
The maximum displacement of a particle
combined in series (Figure A) or in parallel
performing S.H.M. from its mean position is
(Figure B) or both.
called the amplitude of S.H.M.
Series combination (Figure A): In this case,
5.6.2 Period of S.H.M.: all the springs are connected one after the
The time taken by the particle performing other forming a single chain. Consider
S.H.M. to complete one oscillation is called the an arrangement of two such springs of
period of S.H.M. spring constants k1 and k2. If the springs are
Displacement of the particle at time t is massless, each will have the same stretching
given by x  A sin  t    force as f. For vertical arrangement, it
2 will be the weight mg. If e1 and e2 are the
After a time t   t   the displacement will
   respective extensions, we can write,
be
f f
  2   f  k1e1  k 2 e2 
 e1  and e2 
x '  Asin   t   k1 k2
     The total extension is
 x '  Asin  t  2    1 1
e  e1  e2  f   
 x '  Asin  t     k1 k 2  .

115
If ks is the effective spring constant (as if Let f 1  k1e,f 2  k 2 e, be the individual
there is a single spring that gives the same restoring forces.
total extension for the same force), we can If kp is the effective spring constant, a
write, single spring of this spring constant will be
f 1 1 1 1 1
e   f       stretched by the same extension e, by the
ks  k1 k 2  k s k1 k 2
same stretching force f.
For a number of such (massless) springs, in  f  k p e  f 1  f 2   k1e  k 2 e 
series, 1  1  1    1   k p  k1  k 2   ki
k s k1 k 2
i  k 
 i
For only two massless springs of For m such identical massless springs of
spring constant k each, in parallel, k p = mk
spring constant k each, in series,
kk Product 5.7 Reference Circle Method:
ks  1 2  Figure 5.3 shows a rod rotating along a
k1  k 2 Sum
For n such identical massless springs, in vertical circle in the x-y plane. If the rod is
k illuminated parallel to x-axis from either side
series, k s = by a linear source parallel to the rod, as shown
n
in the Fig. 5.3, the shadow (projection) of the
rod will be produced on the y-axis. The tip of
this shadow can be seen to be oscillating about
the origin, along the y-axis.
Fig. A

Fig 5.3: Projection of a rotating rod.

We shall now prove that motion of
Fig. B the tip of the projection is an S.H.M. if the
Parallel combination (Figure B): In such a corresponding motion of the tip of the rod
combination, all the springs are connected is a U.C.M. For this, we should take the
between same two points, one of them is the projections of displacement, velocity, etc. on
support and at the other end, the stretching any reference diameter and confirm that we
force f is applied at a suitable point. get the corresponding quantities for a linear
Irrespective of their spring constants, each S.H.M.
spring will now have the same extension e. Figure 5.4 shows the anticlockwise
The springs now share the force such that in uniform circular motion of a particle P, with
the equilibrium position, the total restoring centre at the origin O. Its angular positions are
force is equal and opposite to the stretching decided with the reference OX. It means, if the
force f. particle is at E, the angular position is zero, at
116
F it is 90° = π2 , at G it is 180° = πc, and so on. If
c
Projection of velocity: Instantaneous velocity
it comes to E again, it will be 360° = 2πc (and of the particle P in the circular motion is the
 tangential velocity of magnitude rω as shown
not zero). Let r = OP be the position vector of
this particle. in the Fig. 5.5.
Its projection on the reference diameter
will be v y  r cos   r cos  t    . This
is the expression for the velocity of a particle
performing a linear S.H.M.
Projection of acceleration: Instantaneous
acceleration of the particle P in circular
motion is the radial or centripetal acceleration
of magnitude rω 2 , directed towards O. Its
projection on the reference diameter will be
a y  r 2 sin   r 2 sin  t  
    2 y .
Fig 5.4: S.H.M. as projection of a U.C.M.
Again, this is the corresponding
At t = 0, let the particle be at P0 with
acceleration for the linear S.H.M.
reference angle φ . During time t, it has
From this analogy it is clear that projection
angular displacement ω t . Thus, the reference
of any quantity for a uniform circular motion
angle at time t is    t    . Let us choose
gives us the corresponding quantity of linear
the diameter FH along y-axis as the reference
S.H.M. This analogy can be verified for any
diameter and label OM as the projection of
 diameter as the reference diameter. Thus, the
r = OP on this.
projection of a U.C.M. on any diameter is an
Projection of displacement: At time t, we
S.H.M.
get the projection or the position vector
5.8 Phase in S.H.M.:
OM = OP sin   y  r sin  t    . This is the
Phase in S.H.M. (or for any motion) is
equation of linear S.H.M. of amplitude r. The
basically the state of oscillation. In order to
term ω can thus be understood as the angular
know the state of oscillation in S.H.M., we
velocity of the reference circular motion. For
need to know the displacement (position), the
linear S.H.M. we may call it the angular
direction of velocity and the oscillation number
frequency as it decides the periodicity of the
(during which oscillation) at that instant of
S.H.M. In the next section, you will come to
time. Knowing only the displacement is not
know that the phase angle    t    of the
enough, because at a given position there are
circular motion can be used to be the phase of
two possible directions of velocity (except
the corresponding S.H.M.
the extreme positions), and it repeats for
successive oscillations. Knowing only velocity
is not enough because there are two different
positions for the same velocity (except the
mean position). Even after this, both these
repeat for the successive oscillations.
Hence, to know the phase, we need a
quantity that is continuously changing with
time. It is clear that all the quantities of linear
Fig 5.5: Projection
of velocity.
S.H.M. (x, v, a etc) are the projections taken

117
on a diameter, of the respective quantities for
the reference circular motion. The angular − 3
is at A , heading to the mean position.
displacement    t    can thus be used as 2
the phase of S.H.M. as it varies continuously Determine the phase angle.
with time. In this case, it will be called as the Solution:
c c
 3    
phase angle. A sin 1  A 1      or 2  
2  3  3
Special cases:
(i) Phase θ = 0 indicates that the particle From negative side, the particle is heading to
is at the mean position, moving to the the mean position. Thus, the phase angle is
positive, during the beginning of the first in the fourth quadrant for that oscillation.
oscillation. Phase angle   360 0 or 2 c  
c

1   2  
is the beginning of the second oscillation,  3
and so on for the successive oscillations. As it is the third oscillation, phase
(ii) Phase   180 0 or  c indicates that during  
  2  2  1    4   2  
its first oscillation, the particle is at the  3
mean position and moving to the negative.   17 
c

 6   
Similar state in the second oscillation will 3  3 
have phase    360  180  or 2    ,
0 c

and so on for the successive oscillations. 5.9. Graphical Representation of S.H.M.:

c
  (a) Particle executing S.H.M., starting from
(iii) Phase     90 or  indicates
0
that
2 mean position, towards positive:
the particle is at the positive extreme As the particle starts from the mean position
position during first oscillation. Fig (5.6), towards positive, f = 0
For the second oscillationc it will be  displacementx  A sin  t
  Velocity v = Aω cosω t
   360  90  or 2   , and so on
0

 2 Acceleration a   A 2 sin  t
for the successive oscillations.
 3 
c
(t) 0 T/4 T/2 3T/4 T 5T /4
(iv) Phase   270 or0
 indicates that the
 2  π 3π 5π
particle is at the negative extreme position during (q )
* 0 2 π 2 2π 2
the first oscillation. For the second oscillation c
 3  (x) 0 A 0 -A 0 A
it will be    360  270  or 2 
0
, and
 2  (v) Aω 0 - Aω 0 Aω 0
so on for the successive oscillations.
(a) 0 -Aω2 0 Aω2 0 -Aω2
Example 5.7: Describe the state of * phase q = wt + f
oscillation if the phase angle is 11100. Conclusions from the graphs:
Solution: 1110 0  3  360 0  30 0 • Displacement, velocity and acceleration of
3 × 360 0 plus something indicates 4th S.H.M. are periodic functions of time.
A
oscillation. Now, A sin 30 =
0
• Displacement time curve and acceleration
2
Thus, phase angle 11100 indicates that time curves are sine curves and velocity
during its 4th oscillation, the particle is at time curve is a cosine curve.
+A/2 and moving to the positive extreme. • There is phase difference of π/2 radian
between displacement and velocity.
Example 5.8: While completing its third
• There is phase difference of π/2 radian
oscillation during linear S.H.M., a particle
between velocity and acceleration.
118
• There is phase difference of π radian
between displacement and acceleration.
• Shapes of all the curves get repeated after
2π radian or after a time T.
(a)

(a)

(b)

(b)

(c)

Fig. 5.7: (a) Variation of displacement with

time, (b) Variation of velocity with time,
(c) (c) Variation of acceleration with time.
5.10 Composition of two S.H.M.s having
Fig. 5.6: (a) Variation of displacement with same period and along the same path:
time, (b) Variation of velocity with time, Consider a particle subjected
(c) Variation of acceleration with time. simultaneously to two S.H.M.s having the
same period and along same path (let it be
((b) Particle performing S.H.M., starting
along the x-axis), but of different amplitudes
from the positive extreme position.
and initial phases. The resultant displacement
As the particle starts from the positive extreme at any instant is equal to the vector sum of its

position Fig. (5.7),   displacements due to both the S.H.M.s at that
2
instant.
 displacement, x  Asin  t   / 2   Acos t
Equations of displacement of the two S.H.M.s
dx d  Acos t  along same straight line (x-axis) are
Velocity, v =    A sin  t 
dt dt x1 = A1 sin (ωt + φ1 ) and x2 = A2 sin (ωt + φ2 )
Acceleration,
The resultant displacement (x) at any instant
dv d   A sin  t   (t) is given by x = x1 + x2
a     A 2 cos  t 
dt dt x = A1 sin (ωt + φ1 ) + A2 sin (ωt + φ2 )
(t) 0 T/4 T/2 3T/4 T 5T /4 ∴ x = A1 sin ωt cos φ1 + A1cos ωt sin φ1
+ A2 sin ωt cos φ2 + A2 cos ωt sin φ2
π 3π 5π
( θ )* π 2π 3π A1, A2, φ1 and φ2 are constants and ωt is variable.
2 2 2
Thus, collecting the constants together,
(x) A 0 -A 0 A 0 x = (A1 cos φ1 + A2 cos φ2 ) sin ωt +
(v) 0 -Aω 0 Aω 0 -Aω (A1 sin φ1 + A2 sin φ2 ) cos ωt
As A1, A2, φ1 and φ2 are constants, we can
(a) -Aω2 0 Aω2 0 -Aω2 0
combine them in terms of another convenient
*(Phase θ = ωt + φ) constants R and δ as
119
 R cos δ = A1 cos f1 + A2 cos f2 --- (5.17)
Activity
and R sin δ = A1 sin f1 + A2 sin f2 --- (5.18)
∴ x = R (sin ωt cos δ + cos ωt sin δ) Tie a string horizontally tight between
∴ x = R sin (ωt + δ) two vertical supports. To this string, tie
This is the equation of an S.H.M. of the three pendula, two of them (A and B) of
same angular frequency (hence, the same equal lengths. Third one (C) need not have
period) but of amplitude R and initial phase δ. the same length, but not very different.
It shows that the combination (superposition) Oscillate the pendula A and B in a plane
of two linear S.H.M.s of the same period and perpendicular to the horizontal string. It
occurring a long the same path is also an will be observed that pendulum C also
S.H.M. starts oscillating in the same plane, with the
same period as those of A and B.
Resultant amplitude,
With this system and procedure, we are
R   R sin     R cos  
2 2
imposing two S.H.M.s of the same period.
Substituting from Eq. (5.17) and Eq. (5.18), we The resultant energy transfers through
get the strings into the third pendulum C and
R2 = A12 + A22 + 2A1A2cos( f1 - f2 ) it starts oscillating. Special cases (i), (ii)
and (iii) above can be verified by making
 R  A12  A22  2 A1 A2 cos 1  
 2  - (5.19)
suitable changes.
Initial phase (δ) of the resultant motion:
Dividing Eq. (5.18) by Eq. (5.17), we get 5.11: Energy of a Particle Performing
S.H.M.:
R sin A sin1  A2 sin2 
 1 While performing an S.H.M., the particle
R cos A1cos1  A2 cos2 possesses speed (hence kinetic energy) at all
A1sin1  A2 sin2  the positions except at the extreme positions.
∴ tan δ = In spite of the presence of a restoring force
A1cos1  A2 cos2
(except at the mean position), the particle
occupies various positions. This is an
1  A1sin1  A2 sin2  
∴   tan  A cos  A cos  --- (5.20) indication that work is done and the system
 1 1 2 2 
has potential energy (elastic - in the case of
Special cases: (i) If the two S.H.M.s are in a spring, gravitational - for a pendulum,
phase, ( f1 - f2 ) = 0°, ∴ cos ( f1 - f2 ) = 1. magnetic - for a magnet, etc.). Total energy of
 R  A12  A22  2 A1A2    A1  A2  . Further, the particle performing an S.H.M. is thus the
sum of its kinetic and potential energies.
if A1 = A2 = A, we get R = 2A Consider a particle of mass m, performing
(ii) If the two S.H.M.s are 90° out of phase, a linear S.H.M. along the path MN about the
( f1 - f2 ) = 90° ∴cos ( f1 - f2 ) = 0. mean position O. At a given instant, let the
particle be at P, at a distance x from O.
∴ R  A12  A22 Further, if A1= A2 = A, we
get, R = 2 A
(iii) If the two S.H.M.s are 180° out of phase,
( f1 - f2 ) = 180° ∴cos ( f1 - f2 ) = -1 Fig. 5.8: Energy in an S.H.M.
∴ R  A12  A22  2 A1 A2 ∴ R  A1  A2 Velocity of the particle in S.H.M. is given
as v   A2  x 2  A cos  t    ,
Further, if A1 = A2 = A, we get R = 0
120
 where x is the displacement of the particle constant, the total energy of the particle at any
performing S.H.M. and A is the amplitude of point P is constant (independent of x and t). In
S.H.M. other words, the energy is conserved in S.H.M.
Thus, the kinetic energy, If n is the frequency of S.H.M.,   2 n .
1 1 Using this in Eq. (5.24), we get
2
 2
 
Ek  m 2 A2  x 2   k A2  x 2 --- (5.21) 1
E  m  2 n  A2  2 2 n 2 A2 m
2

This is the kinetic energy at displacement x. 2

At time t, it is A2 --- (5.25)
 2 2 m 2
1 1 T
Ek  mv 2  mA2 2 cos 2  t    Thus, the total energy in S.H.M. is directly
2 2
proportional to (a) the mass of the particle
1
 kA2 cos 2  t    --- (5.22) (b) the square of the amplitude (c) the square
2
of the frequency (d) the force constant, and
Thus, with time, it varies as cos 2 θ .
inversely proportional to square of the period.
The restoring force acting on the particle
at point P is given by f = - kx where k is the force Can you tell?
constant. Suppose that the particle is displaced
further by an infinitesimal displacement dx To start a pendulum swinging, usually you
against the restoring force f. The external work pull it slightly to one side and release.
done (dW) during this displacement is • What kind of energy is transferred to the
dW  f  dx    kx  dx   kxdx mass in doing this?
• Describe the energy changes that occur
The total work done on the particle to
when the mass is released.
displace it from O to P is given by
x x
1 • Is/are there any other way/ways to start
W  dW  kx dx  kx 2 the oscillations of a pendulum? Which
0 0
2
energy is supplied in this case/cases?
This should be the potential energy (P.E.)
Special cases: (i) At the mean position, x = 0
Ep of the particle at displacement x.
and velocity is maximum.
1 1 1
 EP  kx 2   m 2 x 2 --- (5.23) Hence E   Ek max  m 2 A2 and potential
2 2 2
At time t, it is
1 1
 
energy Ep
min
0
EP  kx 2  kA2 sin 2  t    (ii) At the extreme positions, the velocity of the
2 2
particle is zero and x   A
1
 mA2 2 sin 2  t    --- (5.23a) 1
Hence E   Ep   m 2 A2 and kinetic
2 max 2
Thus, with time, it varies as sin 2 θ . energy  Ek min  0
The total energy of the particle is the sum As the particle oscillates, the energy
of its kinetic energy and potential energy. changes between kinetic and potential. At the
 E  Ek   Ep mean position, the energy is entirely kinetic;
Using Eq. (5.21) and Eq. (5.23), we get while at the extreme positions, it is entirely
1 1
 
E   m 2 A2  x 2  m 2 x 2
2 2
potential. At other positions the energy is
partly kinetic and partly potential. However,
1 1 1
E  m 2 A2   kA 2  m  v max  ---(5.24) the total energy is always conserved.
2

2 2 2 (iii) If K . E .= P. E .,

This expression gives the total energy
1 A
of the particle at point P. As m, ω and A are
2
  1
m 2 A2  x 2   mω 2 x 2 x 
2 2
121
 A E The distance between the point of
Thus at x  , the K.E. = P.E. = for a suspension and centre of gravity of the bob
2 2
particle performing linear S.H.M. (point of oscillation) is called the length of the
A 1 11  E pendulum. Let m be the mass of the bob and
(iv) At x  , P.E.  kx 2   kA 2  
2 2 42  4 T' be the tension in the string. The pendulum
 K.E.  3  P.E  remains in equilibrium in the position OA,
A with the centre of gravity of the bob, vertically
Thus, at x  , the energy is 25% potential below the point of suspension O. If now the
2
and 75% kinetic. pendulum is displaced through a small angle
The variation of K.E. and P.E. with θ (called angular amplitude) and released, it
displacement in S.H.M. is shown in Fig. (5.9) begins to oscillate on either side of the mean
(equilibrium) position in a single vertical
plane. We shall now show that the bob
performs S.H.M. about the mean position for
small angular amplitude θ .
Rigid support

Fig. 5.9: Energy in S.H.M.

Example 5.9: The total energy of a particle
of mass 200 g, performing S.H.M. is 10-2 J.
Find its maximum velocity and period if the
amplitude is 7 cm.
Solution:
1 1
E  m 2 A2 E  m  vmax  
2
Fig.5.10: Simple pendulum.
2 2 In the displaced position (extreme
2E position), two forces are acting on the bob.
∴ v max =
m (i) Force T' due to tension in the string, directed
2  10 2 along the string, towards the support and
 v max   0.3162 m / s (ii) Weight mg, in the vertically downward
0.2
2 2 A direction.
v max  A  A  T    1.391s At the extreme positions, there should not be
T v max
any net force along the string. The component
5.12 Simple Pendulum: of mg can only balance the force due to
An ideal simple pendulum is a heavy tension. Thus, weight mg is resolved into two
particle suspended by a massless, inextensible, components;
flexible string from a rigid support. (i) The component mg cos θ along the string,
A practical simple pendulum is a small which is balanced by the tension T ' and
heavy (dense) sphere (called bob) suspended (ii) The component mg sin θ perpendicular
by a light and inextensible string from a rigid to the string is the restoring force acting on
support. mass m tending to return it to the equilibrium
position.
122
 ∴Restoring force, F = - mg sin θ --- (5.26) (b) The period of a simple pendulum is
As θ is very small ( θ <10°), we can write inversely proportional to the square root
sin θ ≅ θ c  F  –mg of acceleration due to gravity.
x
From the Fig. 5.10, the small angle  ,   (c) The period of a simple pendulum does not
L depend on its mass.
x
∴ F   mg --- (5.27) (d) The period of a simple pendulum does
L
As m, g and L are constant, F ∝- x not depend on its amplitude (for small
Thus, for small displacement, the amplitude).
restoring force is directly proportional to the These conclusions are also called the 'laws of
displacement and is oppositely directed. simple pendulum'.
Hence the bob of a simple pendulum 5.12.1 Second’s Pendulum:
performs linear S.H.M. for small amplitudes. A simple pendulum whose period is two
From Eq. (5.15), the period T of oscillation of seconds is called second’s pendulum.
a pendulum from can be given as, L
Period T  2
2 g

 2π L
=  Forasecond ' s pendulum , 2  2 s
acceleration per unit displacement  g
x where Ls is the length of second’s pendulum,
Using Eq. (5.27), F   mg
L having period T = 2s.
x g
∴ ma   mg Ls  2  --- (5.30)
L 
x Using this relation, we can find the length
∴ a   g a   g  g  inmagnitude 
L x L L of a second’s pendulum at a place, if we know
Substituting in the expression for T, we get, the acceleration due to gravity at that place.
L Experimentally, if Ls is known, it can be used
T  2 --- (5.28)
g to determine acceleration due to gravity g at
The Eq. (5.28) gives the expression for the that place.
time period of a simple pendulum. However,
while deriving the expression the following Example 5.10: The period of oscillations of
assumptions are made. a simple pendulum increases by 10%, when
(i) The amplitude of oscillations is very its length is increased by 21 cm. Find its
small (at least 20 times smaller than the initial length and initial period.
length). l
(ii) The length of the string is large and Solution: T  2
g
(iii) During the oscillations, the bob moves 100 l1
along a single vertical plane. ∴ 110 = l
2
Frequency of oscillation n of the simple
pendulum is 10 l1
∴ 
1 1 g --- (5.29) 11 l1  0.21
n 
T 2 L
From the Eq. (5.28), we can conclude the ∴ 1.21l1  l1  0.21
 l1  1m
following for a simple pendulum. l 1
∴ Period T  2  2
(a) The period of a simple pendulum is g 9.8
directly proportional to the square root of
= 2.007 s (π = 3.142)
its length.
123
 Activity Example 5.11: In summer season, a
pendulum clock is regulated as a second’s
When you perform the experiment to pendulum and it keeps correct time. During
determine the period of simple pendulum, it winter, the length of the pendulum decreases
is recommended to keep the amplitude very by 1%. How much will the clock gain or
small. But how small should it be? And why?
To find this it would be better to lose in one day. (g = 9.8 m/s2)
measure the time period for different angular Solution: In summer, with period Ts = 2 s,
amplitudes. the clock keeps correct time. Thus, in a day
L of 86400 seconds, the clock’s pendulum
Let T0  2 be the period for (ideally)
g should perform 86400 = 43200 oscillations,
very small angular amplitude and Tθ be 2
the period at higher angular amplitude to keep correct time.
Lw 1% lessthansummer  0.99 Ls
θ . Experimentally determined values of the
L
T T  2
ratio θ are as shown in the table below. g
T0
Tw Lw T
θ 20° 45° 50° 70° 90° T  L 
   w  0.99
Ts Ls 2
Tθ 
 Tw  1.99s

T0 1.02 1.04 1.05 1.10 1.18
With this period, the pendulum will now
It shows that the error in the time 86400
perform = 43417 oscillations per
period is about 2% at amplitude of 20°, 5% 1.99
at amplitude of 50°, 10% at amplitude of day. Thus, it will gain 43417 - 43200 = 217
70° and 18% at amplitude of 90°. Thus, the oscillations, per day.
recommended maximum angular amplitude Per oscillations the clock refers to 2 second.
is less than 20°. It also helps us in restricting Thus, the time gained, per day = 217 × 2
the oscillations in a single vertical plane. = 434 second = 7 minutes, 14 second.
Conical pendulum Simple pendulum
1 Trajectory and the plane of the motion of Trajectory and the plane of motion of the
the bob is a horizontal circle bob is part of a vertical circle.
2 K.E. and gravitational P.E. are constant. K.E. and gravitational P.E. are interconverted
and their sum is conserved.
3 Horizontal component of the force due to Tangential component of the weight is the
tension is the necessary centripetal force governing force for the energy conversions
(governing force). during the motion.
4 Period, Period,
L cos  L
T  2 T  2
g g
5 String always makes a fixed angle with the With large amplitude, the string can be
horizontal and can never be horizontal. horizontal at some instances.
6 During the discussion for both, we have ignored the stretching of the string and the energy
spent for it. However, the string is always stretched otherwise it will never have tension
(except at the extreme positions of the simple pendulum). Also, non-conservative forces
like air resistance are neglected.
124
5.13: Angular S.H.M. and its Differential Angular S.H.M. is defined as the oscillatory
Equation: motion of a body in which the torque for
Figure 5.11 shows a metallic disc attached angular acceleration is directly proportional
centrally to a thin wire (preferably nylon or to the angular displacement and its direction is
metallic wire) hanging from a rigid support. If
opposite to that of angular displacement.
the disc is slightly twisted about the axis along
The time period T of angular S.H.M. is given
the wire, and released, it performs rotational
by, 2
motion partly in clockwise and anticlockwise T 
(or opposite) sense. Such oscillations are called 
2
angular oscillations or torsional oscillations. 
This motion is governed by the restoring angularaccelerationperunit
torque in the wire, which is always opposite angulardisplacement
to the angular displacement. If its magnitude 5.13.1 Magnet Vibrating in Uniform
happens to be proportional to the corresponding Magnetic Field:
angular displacement, we can call the motion If a bar magnet is freely suspended in the
to be angular S.H.M. plane of a uniform magnetic field, it remains
in equilibrium with its axis parallel to the
direction of the field. If it is given a small
angular displacement θ (about an axis passing
through its centre, perpendicular to itself and
to the field) and released, it performs angular
oscillations Fig. (5.12).

Fig. 5.11: Torsional (angular) oscillations.

Thus, for the angular S.H.M. of a body,
the restoring torque acting upon it, for angular
displacement θ , is
   or   c --- (5.31)
The constant of proportionality c is the
restoring torque per unit angular displacement.
If I is the moment of inertia of the body, the
torque acting on the body is given by,   I 
Where α is the angular acceleration. Using Fig. 5.12: Magnet vibrating in a uniform
magnetic field.
this in Eq. (5.31) we get, I   c
Let µ be the magnetic dipole moment
d 2
 I 2  c  0  --- (5.32) and B the magnetic field. In the deflected
dt
This is the differential equation for position, a restoring torque acts on the magnet,
angular S.H.M. From this equation, the that tends to bring it back to its equilibrium
angular acceleration α can be written as, position. [Here we used the symbol µ for the
d 2 c magnetic dipole moment as the symbol m is
  2 
dt I used for mass].
Since c and I are constants, the angular The magnitude of this torque is   µ Bsin
acceleration α is directly proportional to If θ is small, sin    c   µB
θ and its direction is opposite to that of the For clockwise angular displacement θ ,
angular displacement. Hence, this oscillatory the restoring torque is in the anticlockwise
motion is called angular S.H.M. direction.
125
 ∴ τ  I    µB Example 5.13: Two magnets with the same
where I is the moment of inertia of the bar dimensions and mass, but of magnetic
magnet and α is its angular acceleration. moments µ1 = 100 A m2 and µ 2 = 50 A m2
 µB  are jointly suspended in the earth’s magnetic
      --- (5.33)
 I  field so as to perform angular oscillations
Since µ, B and I are constants, Eq. (5.33)
in a horizontal plane. When their like poles
shows that angular acceleration is directly
are joined together, the period of their
proportional to the angular displacement and
angular S.H.M. is 5 s. Find the period of
directed opposite to the angular displacement.
angular S.H.M. when their unlike poles are
Hence the magnet performs angular S.H.M.
joined together.
The period of vibrations of the magnet is given
Solution:
by I
2 T  2
T µB
angularaccelerationperunit
With like poles together, the effective
angulardisplacement
magnetic moment is  µ1  µ2 
2
 I
 ∴ T1  2
  µ1  µ2  BH
I With unlike poles together, the effective
 T  2
µB
--- (5.34) magnetic moment is  µ1  µ2 
I
 T2  2
Example 5.12: A bar magnet of mass 120 g,  µ1  µ2  BH
in the form of a rectangular parallelepiped,
∴
T1

 µ1  µ2 
has dimensions l = 40 mm, b = 10 mm and T2  µ1  µ2 
h = 80 mm. With the dimension h vertical,
the magnet performs angular oscillations 5 1
∴ =  T2  75  8.660 s
in the plane of a magnetic field with period T2 3
π s. If its magnetic moment is 3.4 A m2,
5.14 Damped Oscillations:
determine the influencing magnetic field.
I I
Solution: T  2   2
B B
4I
B 

For a bar magnet, moment of inertia
 l 2  b2 
I M 
 12 
 1600  100 
 I  0.12    10 6
 12 
5
 1.7  10 Am 2

4  1.7  10 5
B   2  10 5 Wbm 2 or T
3.4

Fig. 5.13: A damped oscillator.

126
 If the amplitude of oscillations of an The solution is found to be of the form
 bt
oscillator is reduced by the application of an x  Ae 2m
cos  t    --- (5.36)
external force, the oscillator and its motion
are said to be damped. Periodic oscillations  Ae is the amplitude of the damped
 bt
2m

of gradually decreasing amplitude are harmonic oscillations.

called damped harmonic oscillations and
the oscillator is called a damped harmonic
oscillator.
For example, the motion of a simple
pendulum, dies eventually as air exerts a
viscous force on the pendulum and there may
be some friction at the support.
Figure 5.13 shows a block of mass m that
can oscillate vertically on a spring. From the
block, a rod extends to vane that is submerged Fig. 5.14: Displacement against time graph.
on a liquid. As the vane moves up and down, As shown in the displacement against time
the liquid exerts drag force on it, and thus on the graph (Fig 5.14), the amplitude decreases with
complete oscillating system. The mechanical time exponentially. The term cos  t   
energy of the block-spring system decreases shows that the motion is still an S.H.M.
2
with time, as energy is transferred to thermal k  b 
The angular frequency,    
energy of the liquid and vane. m  2 m 
The damping force (Fd) depends on the 2 2
Period of oscillation, T =  
nature of the surrounding medium and is  k  b 
2


directly proportional to the speed v of the vane m  2 m 
and the block The damping increases the period (slows down
 Fd   bv  the motion) and decreases the amplitude.
Where b is the damping constant and negative 5.15 Free Oscillations, Forced Oscillations
sign indicates that Fd opposes the velocity. and Resonance:
For spring constant k, the force on the Free Oscillations: If an object is allowed
block from the spring is Fs   kx . to oscillate or vibrate on its own, it does so
Assuming that the gravitational force with its natural frequency (or with one of its
on the block is negligible compared to Fd and natural frequencies). For example, if the bob
Fs , the total force acting on the mass at any of a simple pendulum of length l is displaced
time t is and released, it will oscillate only with the
F  Fd  Fs 1 g
 ma  Fd  Fs frequency n  which is called its
2 l
 ma   bv  kx natural frequency and the oscillations are
 ma  bv  kx  0 free oscillations. However, by applying a
d2x dx periodic force, the same pendulum can be
 m 2  b  kx  0  --- (5.35)
dt dt made to oscillate with different frequency. The
The solution of Eq. (5.35) describes the oscillations then will be forced oscillations
motion of the block under the influence of a and the frequency is driver frequency or forced
damping force which is proportional to the frequency.
speed.
127
 Consider the arrangement shown in the same natural frequency as that of the source
Fig. 5.15. There are four pendula tied to a absorbs maximum energy from the source. In
string. Pendula A and C are of the same length, such case, it is said to be in resonance with
pendulum B is of shorter length and pendulum the source (pendulum A). For unequal natural
D is of longer length. Pendulum A has a solid frequencies on either side (higher or lower), the
rubber ball as its bob and acts as the driver energy absorbed (hence, the amplitude) is less.
pendulum or the source pendulum. Other three If the activity is repeated for a set of pendula
pendula have hollow rubber balls as their bobs of different lengths and squares of their
and act as the driven pendula. As the pendula amplitudes are plotted against their natural
A and C are of the same lengths, their natural frequencies, the plot will be similar to that
frequencies are the same. Pendulum B has shown in the Fig. 5.16. The peak occurs when
higher natural frequency as its length is shorter the forced frequency matches with the natural
than that of pendulum A. Natural frequency of frequency, i.e., at the resonant frequency.
pendulum D is less than that of pendula A and
C.

Fig 5.15: Forced oscillations. Fig 5.16: Resonant frequency.

Pendulum A is now set into oscillations In the next Chapter on superposition of
in a plane perpendicular to the plane of paper. waves, you will see that most of the traditional
In the course of time it will be observed that musical instruments use the principle of
the other three pendula also start oscillating in resonance. In the topic AC circuits, the
the same plane. This happens due to transfer resonance in the L.C. circuits is discussed.
of the vibrational energy through the string.
Oscillations of pedulum A are free oscillations Internet my friend
and those of pendula B, C and D are forced
oscillations of the same frequency as that of 1. https://hyperphysics.phy-astr.gsu.edu/
A. The natural frequency of pendulum C is the hbase/shm.html
same as that of A, as its length is the same as 2. https://hyperphysics.phy-astr.gsu.edu/
that of A. hbase/pend.html
It can also be seen that among the pendula 3. h t t p s : / / e n . w i k i p e d i a . o r g / w i k i /
B, C and D, the pendulum C oscillates with simpleharmonicmotion
maximum amplitude and the other two with 4. https://opentextbc.ca/physicstextbook
smaller amplitudes. As the energy depends 5. https://physics.info
upon the amplitude, it is clear that pendulum C
has absorbed maximum energy from the source
pendulum A, while the other two absorbed
less. It shows that the object C having the
128
 Exercises

1. Choose the correct option. (D) The kinetic energy is equal to total
i) A particle performs linear S.H.M. energy at time T/4.
starting from the mean position. Its
amplitude is A and time period is T. At
the instance when its speed is half the
maximum speed, its displacement x is
3 2
(A) A (B) A
3 2. Answer in brief.
2
1 i) Define linear simple harmonic motion.
(C) A 2 (D) A ii) Using differential equation of linear
2 S.H.M, obtain the expression for (a)
ii) A body of mass 1 kg is performing linear
velocity in S.H.M., (b) acceleration in
S.H.M. Its displacement x (cm) at t S.H.M.
(second) is given by iii) Obtain the expression for the period of a
x = 6 sin (100t + π/4). Maximum kinetic simple pendulum performing S.H.M.
energy of the body is iv) State the laws of simple pendulum.
(A) 36 J (B) 9 J v) Prove that under certain conditions a
(C) 27 J (D) 18 J magnet vibrating in uniform magnetic
field performs angular S.H.M.
iii) The length of second's pendulum on the
3. Obtain the expression for the period of a
surface of earth is nearly 1 m. Its length magnet vibrating in a uniform magnetic
on the surface of moon should be [Given: field and performing S.H.M.
acceleration due to gravity (g) on moon 4.  Show that a linear S.H.M. is the
is 1/6 th of that on the earth’s surface] projection of a U.C.M. along any of its
(A) 1/6 m (B) 6 m diameter.
1 5. Draw graphs of displacement, velocity
(C) 1/36 m (D) m
6 and acceleration against phase angle,
iv) Two identical springs of constant k are for a particle performing linear S.H.M.
from (a) the mean position (b) the
connected, first in series and then in
positive extreme position. Deduce your
parallel. A metal block of mass m is conclusions from the graph.
suspended from their combination. The 6. Deduce the expressions for the kinetic
ratio of their frequencies of vertical energy and potential energy of a particle
oscillations will be in a ratio executing S.H.M. Hence obtain the
(A) 1:4 (B) 1:2 (C) 2:1 (D) 4:1 expression for total energy of a particle
v) The graph shows variation of performing S.H.M and show that the
total energy is conserved. State the
displacement of a particle performing
factors on which total energy depends.
S.H.M. with time t. Which of the 7.  Deduce the expression for period of
following statements is correct from the simple pendulum. Hence state the factors
graph? on which its period depends.
(A) The acceleration is maximum at 8. At what distance from the mean position
time T. is the speed of a particle performing
(B) The force is maximum at time 3T/4. S.H.M. half its maximum speed. Given
(C) The velocity is zero at time T/2. path length of S.H.M. = 10 cm.
[Ans: 4.33 cm]
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9. In SI units, the differential equation 18. The period of oscillation of a body of
2
d x mass m1 suspended from a light spring
of an S.H.M. is 2
 36 x . Find its
dt is T. When a body of mass m2 is tied to
frequency and period. the first body and the system is made to
[Ans: 0.9548 Hz, 1.047 s] oscillate, the period is 2T. Compare the
10. A needle of a sewing machine moves masses m1 and m2 [Ans: 1/3]
along a path of amplitude 4 cm with
19. The displacement of an oscillating
frequency 5 Hz. Find its acceleration
particle is given by x  asin t  bcos t
 1 
 30  s after it has crossed the mean where a, b and ω are constants. Prove
  that the particle performs a linear S.H.M.
position. [Ans: 34.2 m/s2]
with amplitude A  a 2  b 2 
11. Potential energy of a particle performing
20. Two parallel S.H.M.s represented by
linear S.H.M is 0.1 π2 x2 joule. If mass of
x1 = 5sin (4π t + π/3) cm and x2 = 3sin
the particle is 20 g, find the frequency of
(4πt + π/4) cm are superposed on a
S.H.M. [Ans: 1.581 Hz]
particle. Determine the amplitude and
12. The total energy of a body of mass 2 kg
epoch of the resultant S.H.M.
performing S.H.M. is 40 J. Find its speed
[Ans: 7.936 cm, 54° 23']
while crossing the centre of the path.
21. A 20 cm wide thin circular disc of mass
[Ans: 6.324 cm/s]
200 g is suspended to a rigid support
13. A simple pendulum performs S.H.M of
from a thin metallic string. By holding
period 4 seconds. How much time after
the rim of the disc, the string is twisted
crossing the mean position, will the
through 60o and released. It now performs
displacement of the bob be one third of
its amplitude. [Ans: 0.2163 s] angular oscillations of period 1 second.
14. A simple pendulum of length 100 cm Calculate the maximum restoring torque
performs S.H.M. Find the restoring force generated in the string under undamped
acting on its bob of mass 50 g when the conditions. (π3 ≈ 31)
displacement from the mean position is [Ans: 0.04133 N m]
3 cm. [Ans: 1.48 × 10 N]
-2 22. Find the number of oscillations
15. Find the change in length of a second’s performed per minute by a magnet is
pendulum, if the acceleration due to vibrating in the plane of a uniform field
gravity at the place changes from 9.75 of 1.6 × 10-5 Wb/m2. The magnet has
m/s2 to 9.80 m/s2. moment of inertia 3 × 10-6 kgm2 and
[Ans: Decreases by 5.07 mm] magnetic moment 3 A m2.
16. At what distance from the mean position [Ans:38.19 osc/min.]
is the kinetic energy of a particle 23. A wooden block of mass m is kept
performing S.H.M. of amplitude 8 cm, on a piston that can perform vertical
three times its potential energy? vibrations of adjustable frequency and
[Ans: 4 cm] amplitude. During vibrations, we don’t
17. A particle performing linear S.H.M. want the block to leave the contact
of period 2π seconds about the mean
with the piston. How much maximum
position O is observed to have a speed
frequency is possible if the amplitude of
of b 3 m / s , when at a distance b
vibrations is restricted to 25 cm? In this
(metre) from O. If the particle is moving
case, how much is the energy per unit
away from O at that instant, find the
mass of the block? (g ≈ π2 ≈ 10 m s -2)
time required by the particle, to travel a
[Ans: nmax = 1/s, E/m = 1.25 J/kg]
further distance b. [Ans: π/3 s]

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