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14.Oscillations

Physics Smart Booklet

Theory + NCERT MCQs + Topic Wise Practice
MCQs + NEET PYQs

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Oscillations
A repeated motion at equal interval of time is called “periodic motion”. If a particle moves back and forth
(or to and fro) over the same path, then its motion is said to be oscillatory or vibratory.
When oscillatory motion of a particle can be expressed in terms of sine and cosine functions, it is said to be a harmonic
motion. All harmonic motions are oscillatory.
Simple harmonic motion is a special case of harmonic motion in which restoring force acting on the particle is
directly proportional to its displacement from the equilibrium position, i.e., F  x
or F = −kx, where k is called the force constant. Negative sign shows that the force is always directed towards the
equilibrium position.
d2 x  k 
 F = ma = −kx or +   x=0 ...(1)
dt 2 m
This equation is a second order linear homogeneous differential equation. It is called the differential equation of
simple harmonic motion. The figure shows the representation of a SHM.
displacement

position A

−A

Conditions of simple harmonic motion

1. There must be a stable equilibrium position.
2. The motion must be symmetrical about the equilibrium position.
3. The restoring force must be proportional to the displacement and should be opposite to direction of displacement.
4. Velocity should be a continuous function of time.
Basic terms related to SHM
1. Amplitude (A)
It is the maximum displacement of an oscillating body from its equilibrium position or mean position.
2. Period (T)
It is the time taken by an oscillating particle to complete one oscillation.
3. Frequency (f)
It is the number of oscillations made in one second by the oscillating particle.
4. Angular frequency ()
It is the rate of change of angle.  = 2/T = 2f
Equation of SHM
Solutions of the differential equation (1) are the equations of SHM.
In general, the equation of motion may be represented by any of the following equations.
x = A sin(t + )
x = A cos(t + )
x = a sin t + b cos t

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We use x = A sin(t + ) as the general equation of motion, where  is called the phase constant and
(t + ), the phase of the particle.
Simple harmonic motion as a projection of circular motion
Consider a particle P moving on a circle of radius A with a constant angular speed . Let us take the centre of the
circle as the origin and two perpendicular diameters as the X and Y-axis at time t. Drop perpendicular PQ on X-axis
and PR on Y-axis. The x and y-coordinates of the particle at time t are
x = OQ = OP cos t
or, x = A cos t … (1)
and y = OR = OP sin t
or, y = A sin t … (2)
Equation (1) shows that the foot of perpendicular Q executes a simple harmonic motion on Y
P
the X-axis. The amplitude is A and the angular frequency is . Similarly, equation (2) shows R
t
that the foot of perpendicular R executes a simple harmonic motion on the Y-axis. The X
O Q
amplitude is A and the angular frequency is . The phases of the two simple harmonic
  
motions differ by [remember cos t = sin  t + 
2  2
Thus, the projection of uniform circular motion on a diameter of the circle is a simple harmonic motion.
Displacement, velocity and acceleration as a function of time
v
Suppose we start measuring time when the body is in equilibrium.
Then  = 0. The instantaneous velocity of the body, A
dx
v is given by v = = A cos t.
dt
O t
The figure shows the plot of velocity as a function of time.
dx   −A a
v= = A cos t or v = A sin  t + 
dt  2 2A
The acceleration of the body is given by,
dv
a= = −2A sin t O t
dt
a = 2A sin (t + ) −2 A
The figure shows a plot of acceleration v/s time.

1. The maximum displacement is A, the maximum velocity is A and the maximum acceleration is 2A.

 2. The velocity is

2
ahead of the displacement and the acceleration is

2
ahead of the velocity or  ahead of the

displacement.
Velocity and acceleration as a function of displacement
a = −2x
dv dv dx dv
Since a = = =v , we can write, v
dt dx dt dx
dv
v = − 2x  vdv = − 2xdx A
dx
v2 x2
On integrating, we get, = −2 + C (where C is a constant)
2 2
x
2 A 2 −A O A
At x = 0, v = vmax = A  C =
2
v= A2 − x2
Illustrations

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1. A particle executes simple harmonic motion with a period of 3s and amplitude 5 cm. At = 0, it is at position x = 2.5
cm going along the positive x-axis. The equation for the displacement of the particle at any instant t is given by
   2 
(A) x = 5sin  t +  (B) x = 5sin  t + 
3 3 3 3 
 2   2 
(C) x = 5sin  t +  (D) x = 5sin  t + 
 3 3  3 6
Ans (D)
In general, the equation for displacement as function of time is given by x = A sin (t + )
2 2
Here A = 5 cm and  = =
t 3
x 25 
at t = 0; x = A sin    = sin −1 = sin −1 =
A 5 6
 2 
 The equation is x = 5sin  t + 
 3 6
 
2. The simple harmonic motion of a particle is represented by an equation x = 10sin 10t +  and the motion starts at
 3
t = 0. When does the particle first come to rest?
   
(A) s (B) s (C) s (D) s
30 60 40 80
Ans (B)
  dx  
x = 10 sin 10t +  ; v = = 100 cos 10t + 
 3  dt  3
     
v = 0 when 10t + =  10t = − =  t = s
3 2 2 3 6 60
3. A particle executes simple harmonic motion with an amplitude 1.414 m and a time period of 24 s. The time taken by
the particle to cover half the amplitude is
(A) 4 s (B) 3 s (C) 2 s (D) 1 s
Ans (C)
A  2 
x = Asin t; = A sin t   t  =  t = 3
2 6 T 6
T
t  = = 2s
12
4. A particle is performing SHM in a straight line. If the velocity of the particle are v1 and v2 at displacements x1 and x2
respectively, the angular frequency of the particle is
v12 + v 22 v12 − v 22 v12 − v 22 v12 − v 22
(A) (B) (C) (D)
x 22 + x12 x 22 + x12 x12 − x 22 x 22 − x12
Ans (D)
v =  A 2 − x 2  v1 =  A 2 − x12 and v 2 =  A 2 − x 2
or v12 = 2 (A 2 − x12 ) and v 22 = 2 (A 2 − x 22 )
v12 − v 22 v12 − v 22
 v1 − v 2 = (x 2 − x1 )   =   =
2 2 2 2 2 2

x 22 − x12 x 22 − x12

5. Which of the following relationships between acceleration a and displacement x of a particle represent the simple
harmonic motion?
(A) a = 2x (B) a = − 25 x (C) a = 4x2 (D) a = − 100 x2

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Ans (B)
In SHM, acceleration and displacement are related as a = −2 x 
Hence (B) is the correct choice.

6. A piston executes simple harmonic motion in a vertical direction with a period 1 s. The maximum value of amplitude
for which a block placed on the piston does not just leave the piston is
(A) 0.25 cm (B) 0.25 m (C) 2.5 cm (D) 2.5 m
Ans (B)
For the block not to leave the piston, the maximum acceleration should be just equal to the acceleration due to gravity
g g gT 2
i.e., a max = g   A = g  A = 2 = = = 0  25m
2

  2 
2
4 2
 
 T 

7. A particle undergoes a SHM. The graph of velocity v/s displacement is

(A) a straight line (B) a circle (C) an ellipse (D) a hyperbola
Ans (C)
v2 = 2 (A2 – y2) is an equation of ellipse.
8. The time taken by a particle executing SHM of period T to move from the mean position to half the maximum
displacement is
T T T T
(A) (B) (C) (D)
2 4 8 12
Ans (D)
A 2 T
y = A sin t i.e., = A sin t  t=
2 T 12

9. If a watch with wound spring is taken to the moon, it

(A) runs faster (B) runs slower (C) does not work (D) shows no change
Ans (D)
Time period of stretched spring is independent of g. So there will be no change in period.

10. A block is resting on a piston which is moving vertically with SHM of period 1 s. At what amplitude of motion will
the block and piston separate?
(A) 0.2 m (B) 0.25 m (C) 0.3 m (D) 0.35 m
Ans (B)
Weight kept on the system will separate from the piston when the maximum force just exceeds the weight of the
body. Hence, m2y = mg
g 9 8
or y= 2 = = 0.25 m
 (2) 2

11. A particle undergoing SHM has velocities of 10 cm s−1 and 8 cm s−1 at the displacements 4 cm and 5 cm from the
mean position. Its period is
 3
(A) (B) (C)  (D) 2
2 2
Ans (A)
2 (A 2 − x 2 )
102 = 2 (A 2 − 42 )
82 = 2 (A 2 − 52 )
2
102 − 82 = 2 (25 − 16)   = 2 rad s −1  T = =


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12. A particle is performing SHM given by the equation x = A sin t. After crossing the mean position, it takes a time t1
to cover the first half of the amplitude and t2 to cover the next half. Then the ratio t1 : t2 is
(A) 1 : 1 (2) 1 : 3 (C) 1 : 2 (D) 2 : 1
Ans (C)
x = A sin t, when t = t1
A A  T
x=  = Asin t1  t1 =  t1 =
2 2 6 12
T
T T T t 6 1
t2 = − =  1 = 12 = =
4 12 6 t2 t 12 2
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Energy conservation
The kinetic and potential energies of an oscillating body vary with time as well as with position. But, the total
mechanical energy of the body always remains conserved at all positions and at all instants during the oscillation.
(i) Energy variation with respect to time
The instantaneous kinetic energy is,
1 1
K = mv2 = m[A cos t]2
2 2
1
K = m2A2 cos2 t
2
2 x 2
The instantaneous potential energy is, yU = −  Fdx =  m2 xdx = m +C
2
x2
At x = 0, U = 0  C = 0 Thus, U = −m2
2
1
Since, x = A sin t, U = m2 A2 sin2 t
2
 The total mechanical energy of the body is
1 1
E = K + U = m 2A2[cos2 t + sin2 t] = m2A2
2 2
Thus the total energy is a constant.

(ii) Energy variation with respect to displacement

Kinetic energy at a position x is given by,
1 1 1
K = mv2 = m2(A2 − x2) and the potential energy is given by, U = m 2x2
2 2 2
The total mechanical energy is given by
1 1 1
E = K + U = m2(A2 − x2) + m2x2 = m2A2 = constant.
2 2 2

Illustrations

13. The motion of a particle varies with time according to the relation, y = a (sin t + cos t). Then,
(A) the motion is not SHM
(B) the motion is SHM with an amplitude a
(C) the motion is SHM with an amplitude 2a (D) the motion is SHM with an amplitude a 2

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Ans (D)
y = a (sin t + cos t)
 1 1    
= a 2  sin t + cos t  = a 2  sin t cos 4 + cos t sin 4 
 2 2  
 
= a 2 sin  t + 
 4
 The motion is SHM with amplitude a 2 .

14. A body of mass 5 g is executing SHM about a fixed point, with amplitude of 10 cm and maximum velocity 100 cm
s–1. Its velocity will be 50 cm s–1 at a distance of
(A) 5 cm (B) 5 2 cm (C) 5 3 cm (D) 10 2 cm
Ans (C)
vmax = 100 cm s−1, i.e., A = 100
100 100
= = = 10 rad s–1
A 10
Now, v2 = 2(A2 – y2)
502 = 102 (102 – y2)
y = 75 = 5 3 cm.

15. The maximum acceleration of a body moving in SHM is a0 and maximum velocity is v0. The amplitude is given by
v02 a2 1
(A) (B) a0v0 (C) 0 (D)
a0 v0 a 0 v0
Ans (A)
Maximum acceleration, a0 = 2A
and maximum velocity, v0 = A  v02 = A22
a 1 v2
On dividing, 02 = or A = 0
v0 A a0

16. A particle undergoes SHM of period T. The time taken to complete 3/8th oscillation starting from the mean position
is
3 5 5 7
(A) T (B) T (C) T (D) T
8 8 12 12
Ans (C)
The time to complete 3/8th oscillation = time to complete 1/4th oscillation + time to complete 1/8th oscillation from
T
the extreme. The time to complete 1/4th oscillation =
4
th
The time taken to complete 1/8 oscillation from extreme position is obtained from the equation,
a
y = = a cos t
2
1 2 1
 cos t =  cos t=
2 T 2
T T T 5T
t= s  The total time = + =
6 4 6 12
17. Which of the following graphs does not represent a S.H.M?

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(A) (B) (C) (D)

Ans (A)
Only in graph (A), when x is +ve F is also +ve and when x is −ve F is also −ve. Hence F does not represent a restoring
force.
18. A 2.0 kg mass is oscillating on a spring with spring constant 100 N m−1. The velocity of the mass m is 3 ms −1 when
the spring is stretched by 5 cm. The amplitude of oscillation is
(A) 10 6 cm (B) 0.25 cm (C) 25 cm (D) 5 cm
Ans (C)
From the law of conservation of energy
1 2 1 2 1 1 
 mv + kx  =  mv2 + kx 2 
 2 2 extreme position  2 2 at any position
1 1 1
0 + (100)A 2 = (2) ( 3) 2 + (100)(0.05) 2 , on simplifying we get A = 25 cm.
2 2 2
19. Which of the following equations represent a S.H.M.?
(A) y = 5sin 2t + 3cos t (B) y = 4sin 4t + 3cos 4t
(C) y = 6sin t + 6cos 2t (D) y = 2sin2 t
Ans (B)
Check which of the given choices, after twice differentiation gives the same function with a negative
constant. Choice (B) satisfies this criteria.
y = 4sin 4t + 3cos 4t
dy
= 4(4)cos 4t + 3(−4)sin 4t
dt
d2 y
= 4(4)(−4)sin 4t + 3(−4)(4) cos 4t
dt 2
= −162 y

20. A simple pendulum of length l has a time period T for small oscillation. A fixed obstacle is
placed directly below the point of suspension, so that only the lower quarter of the string
continues oscillations. The pendulum is released from rest at a certain point O. The period
of oscillation assuming small angles will be
T
(A) T (B)
4
3T T
(C) (D) (1 + 3)
4 2
Ans (C)
1 l l/ 4  1 T  3T
T =  2 + 2  =  T +  =
2 g g  2 2 4

A B
21. A particle of mass m is placed in a potential field U(x) =
− , where A and B are positive constants. The particle
x2 x
performs SHM about the mean position x0 which is equal to
2A A A B
(A) (B) (C) (D)
B 2B B A
Ans (A)
A B dU 2A B
U(x) = 2 − ; F = − = −
x x dx x 3 x 2

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2A
F = 0 at x 0 = is the mean position.
B
 
22. A point moves along x-axis according to the equation x = A sin2  t −  . Find the amplitude, period and velocity
 4
as a function of x.
Solution
 
x = A sin2  t − 
 4
  1
= A (sin t cos – cos t sin )2 = A (sin t – cos t)2
4 4 2
A
= (1 – sin 2t) … (1)
2
A
Maximum displacement occurs when sin 2t = 0  Amplitude =
2
 
2 
Period is given by T =  
 
dx A
= (2 cos 2t) = A cos 2t = A 1 − sin 2 2t
dt 2
2
 2x   2x 
= A 1 − 1 −  (because from (1), 1 −  = sin 2t)
 A  A
 2x  2x   x  x 
= A  2 −   = 2A 1 −   = 2 (A − x)x
 A A
   A  A 

23. A particle executes a SHM of time period T. Find the time taken by the particle to go directly from its mean position
to half the amplitude.
Solution
Let the equation of motion be x = A sin t
At t = 0, x = 0  Particle is at its mean position.
A
The particle will be at x = at a time t,
2
A 1 
where = A sin t or sin t = or t =
2 2 6
A
(Minimum value of t is taken because we want the time to go directly from x = 0 to x = )
2
  T
Thus, t = = =
6 
  12
2
6 
 T 

24. A particle is undergoing SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance
equal to half its amplitude from the equilibrium position is
(A) 3s (B) 2 s (C) 1 s (D) 0.5 s
Ans (C)
y = A sin t
A 2
= A sin t
2 12
1  t 
= sin t  = t=1s
2 6 6 6
Determination of period and frequency of SHM
The frequency or period of a body executing SHM can be determined in two ways.

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(i) Force (or torque) method

It is based on Newton’s second law of motion.
(ii) Energy method
It is based on the principle of conservation of energy.
Force method
Identify the types of oscillatory motion.

Translational SHM Rotational SHM

Step I The motion of the body may be translational The motion of the body may be rotational about an
about an equilibrium position and it equilibrium position, and it experiences a net
experiences a net restoring force. restoring torque.
Step II Identify the equilibrium position, i.e., where Identify the equilibrium position i.e., where  = 0.
F = 0.
Step III Displace the body through a displacement x Displace the body through an angular displacement 
from the equilibrium position and from the equilibrium position. Determine the net
determine the net restoring force acting restoring torque acting on it.  = −k, where k is
on it. F = −kx, where k is a constant. a constant.
Step IV Apply Newtons’ second law to form the Apply Newton's second law to form the differential
differential equation of SHM. equation of SHM.
d2 x d2 x d2 d2
F = ma = m 2 . Thus m 2 = −kx =I=I . Thus I = −k
dt dt dt 2 dt 2
2
d x k d2  k 
or +  x =0 or 2 +    = 0
dt 2 m dt I
Step V The coefficient of x is 2, so The coefficient of  is 2, so
m 1 k I 1 k
T = 2 and f = T = 2 and f =
k 2 m k 2 I
Energy method
Step I Assume an instantaneous state of the system which should not be the equilibrium or extreme position.
Step II Determine the instantaneous kinetic and potential energies of the system. Write down an expression for the
total mechanical energy of the system. E = K + U.
Step III Convert all the assumed variables in terms of the variable associated with the oscillating mass.
Step IV Differentiate the total mechanical energy of the system with respect to time and equate it to zero, (because
the total mechanical energy of an oscillatory body is constant). Make the appropriate approximation if required.
Step V Obtain the time period from the differential equation of SHM.
The spring mass system
(i) Series combination
1 1 1
The equivalent spring constant k, of two springs connected in series is = +
k k1 k2
where k1 and k2 are the individual spring constants.
(ii) Parallel combination
The equivalent spring constant, k of two springs connected in parallel is k = k1 + k2
1
(iii) The spring constant, k  , where l is the length of the spring. If a spring of length l and spring constant k is cut into
l
two pieces of equal lengths, then, each will have a spring constant of 2 k.

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(iv) A spring of mass m0 and spring constant k is equivalent to a massless spring of spring constant k attached to a block
m m0
of mass 0  T = 2
3 3k
Pendulums
(i) Simple pendulum
l l
For small oscillations, T = 2
g
(ii) Simple pendulum of length comparable to the radius of earth Simple pendulum
1
T = 2
1 1 
g + 
l R  O
d
(iii) Physical pendulum  CM
I
T = 2
mgd
I = moment of inertia of the physical pendulum about O Physical pendulum
m = mass of the physical pendulum.
(iv) Torsional pendulum
I
T = 2 , where I = moment of inertia of the body about the vertical axis,
k
k = torsional constant of the wire.

Torsional pendulum
Superposition of SHMs
Consider two SHMs' performed by a particle, due to two forces, acting along the same direction. The SHM’s are
represented by x1 = A1 sin t and x2 = A2 sin(t + ).
The resultant position of the particle is given by x = x1 + x2
= A1 sin t + A2 sin(t + )
This can be reduced to the form
x = A sin(t + ) … (1)
where A = A + A + 2A1A 2 cos 
2
1
2
2 … (2)
A 2 sin 
and tan  = … (3)
A1 + A 2 cos 
The equation (1) shows that the resultant of two SHMs along the same direction is itself a simple harmonic motion.
The equation (2) gives the resultant amplitude and the equation (3) gives the phase difference between the resultant
and the first SHM.

Illustrations

25. Two pendulums start swinging simultaneously. By the time first pendulum makes 20 oscillations,
2nd pendulum makes 15 oscillations. The ratio of their lengths is
9 16 4 3
(A) (B) (C) (D)
16 9 3 4
Ans (A)
f1 l 20 l 225 9
= 2 =  1 = =
f2 l1 15 l2 400 16

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26. Length of simple pendulum is l which consists of a metallic bob of mass m carrying charge q. Pendulum is placed in
an electric field of strength E directed vertically downwards. Period of oscillation of the pendulum is
l l l l
(A) 2 (B) 2 (C) 2 (D) 2
g qE qE 2g
g− g+
m m
Ans (C)
mg + qE qE
g eff = =g+
m m
l l
T = 2 = 2
g eff qE
g+
m

27. A simple pendulum is of length l has a maximum angular displacement . If m is mass of the bob, its maximum
kinetic energy is
1 mg 1 l 1
(A) (B) m (C) mg l sin  (D) mgl(1 − cos )
2 l 2 g 2
Ans (D)
Change in PE as it moves from equilibrium position to maximum angular displacement
= mgl − mgl cos  = maximum kinetic energy
28. A simple pendulum of length l has been set up inside a railway wagon sliding down a frictionless inclined plane
having an angle of inclination  = 30 with the horizontal. The period of oscillation as recorded by an observer inside
the wagon is
2l 2l l 3l
(A) 2 (B) 2 (C) 2 (D) 2
3g g g 2g
Ans (A)
On the inclined plane, the effective acceleration due to gravity,
3
g/ = g cos 30 = g 
2
l 2l
 T = 2 / = 2
g 3g

29. Two particles are executing SHM of same amplitude and frequency along the same straight line path. They pass each
other when going in opposite directions, each time their displacement is half of their amplitude. The phase difference
between them is
5 2  
(A) (B) (C) (D)
6 3 3 6
Ans (B)
y = A sin (t + )
A A
When y = , = A sin(t + )
2 2
1  5
or (t + ) = sin–1   = or
2 6 6
 5
 The phases of the two particles are and radian.
6 6
5  2
 Phase difference = – =
6 6 3

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30. A body is attached to one end of an inextensible string and the other end is connected to the prong of a tuning fork
which vibrates in a vertical straight line with a frequency n. Show that the string will not remain tight during the
g
motion unless, n2 < where A is the amplitude of the motion.
42 A
Solution
2
The maximum acceleration of the upper end executing SHM is 2a and its period is T =

1 4  2
Also T =  2 = 2 = 42n2  Maximum acceleration = 42n2 A
n T
The maximum acceleration of the particle is g which is possible when the string is not tight. Hence the string will not
remain tight if the acceleration of the upper end is greater than g.
42n2A > g
g
i.e., the string will not remain tight unless n2 <
42 A
31. (a) One end of an ideal spring is fixed to a wall at O and other end of spring is connected to a
block of mass m = 1 kg. The axis of the spring is parallel to x-axis. The mass m is executing SHM.
Equation of position of block is x = (10 + 3sin 10t) where t is in seconds, x is in cm. Calculate the
force constant of the spring.
(b) Now another identical block, moving towards origin with velocity 0.6
m s–1 collides elastically with the block performing SHM at t = 0. m = 1 kg m
k1
Calculate, the O
(i) new amplitude of oscillations.
(ii) equation of position of the block executing SHM
Solution
x = 10 + 3sin 10t
At t = 0, x = x0 = 10
 Displacement of the block from the equilibrium position, s = (x – x0)
= 3sin 10t cm = 0.03sin 10t m
Standard equation is S = A sin t.
 A = 0.03 m,  = 10 s–1  Force constant, k = m2 = 1  102 = 100 N/m.
dx d
Velocity of the block = v = = (0.03 sin 10t) = 0.3cos 10t.
dt dt
Velocity at t = 0 is v0 = 0.3 m s–1.
Velocity of the other block just before collision = 0.6 m s–1.
As the masses of colliding particles are equal and the collision is elastic, the velocities of the particles get
interchanged. Thus after the collision, the velocity of the block executing SHM becomes,
v = 0.6 m s–1 to the left. This is the velocity at the mean position.
v/ 0.6
 New amplitude, A/ = 0 = = 0.06 m = 6 cm.
 10
Equation of position of the block will be,

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x = 10 + A sin (t + ) = 10 + 6 sin (10t + )

32. Find the time period of oscillation of the pulley in the system shown. The moment of
inertia of the pulley about its axis is I and the string does not slip over the pulley. The
string and the springs are light.
m
k

Solution
Let the spring be unstretched in the equilibrium position and the mass come down by a small distance x from the
mean position when in oscillation.
The total energy of the system,
1 1 1
E = mv2 + kx2 + I2
2 2 2
where v is the velocity of the block,  is the angular velocity of the pulley.
dE
Since the total energy is constant, =0
dt
1 dv 1 dx 1 d
0 = m2v + k.2x + I 2
2 dt 2 dt 2 dt
2 2
d x v d x I a
0 = mv 2 + kxv + I  = m 2 + kx +
dt r dt r r
2 2 2
d x I dx  I d x
= m 2 + kx + 2 2 =  m + 2  2 + kx
dt r dt  r  dt
d2 x kr 2 mr 2 + I
 + =0  T = 2
dt 2 mr 2 + I kr 2

33. An object of mass m is suspended from a vertical spring of spring constant k. Find the period of oscillations if the
mass is displaced slightly from its equilibrium position.
Solution
Case (i)
Let the spring be unstretched in the equilibrium position.
Net restoring force in the displaced position; F = –kx
From Newton’s second law,
d2 x
F = ma = m 2
dt
d2 x x
we get, m 2 = –kx
dt
2
d x
m 2 + kx = 0
dt
d2 x  k 
+   x=0
dt 2 m
Comparing with the standard equation,
d2 x k
2
+ 2x = 0, we get  =
dt m
2 m
T= = 2
 k
Aliter
We have k = m2

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k 2 k m
= or = or T = 2
m T m k
Case (ii)
Let the spring be stretched by x1 in the equilibrium position.
In equilibrium, mg = kx1
If the block is further displaced by x,
the resultant force is F = mg – k(x + x1)
d2 x x1
i.e., m 2 = mg – k(x + x1)
dt x
= mg – kx – kx1
= mg – kx – mg = – kx
2
d x k
 2 + x=0
dt m
k m
= T = 2
m k
Alternate method
We can use the energy conservation to solve this problem.
Let x be the instantaneous displacement of the block and let v be its instantaneous velocity.
The instantaneous total energy of the system is
1 1
E = mv2 + kx2
2 2
The displacement of the block and the extension of the springs are the same.
dE
Since the total energy remains constant, =0
dt
1 dv 1 dx
 0 = m2v + k.2x  mva + kxv = 0
2 dt 2 dt
d2 x k m
or m 2 + kx = 0   =  T = 2
dt m k

34. Determine the time period of SHM of a simple pendulum shown.

O
l

m
Solution
Force method
Step I: The system performs rotational SHM.
Step II: When the bob is below the point of suspension, pendulum is in its equilibrium position.
Step III: The net restoring torque about the point O when bob is displaced through an angle 
 = –mgl sin 
If the angle  is small, then sin    O O
l

  = –mgl  l T
d 2
 m
Step IV: Using Newton’s law  = I  = (ml2) 2 m
dt
mg
d  g
2
 2 +   = 0
dt l
l
Step V: The time period is T = 2
g

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Energy method
Datum
Step I: Let  be the instantaneous angular displacement of the bob and U=0 O
let  be the instantaneous angular velocity of the bob about O.  l  l cos 

m
Step II: Assuming the datum for PE at the point of suspension, then PE of
the system is
1 1
U = –mg l cos  and KE of the system is K = I2 = (ml2)2
2 2
1
The total mechanical energy of the system is E = ml2 2 – mgl cos .
2
Step III: Displacement of the string is equal to the angular displacement of the bob.
dE 1  d   d 
Step IV: = ml2  2  – mgl  − sin   = 0
dt 2  dt   dt 
dE d  d 
or =l + g sin  = 0   = 
dt dt  dt 
For small angle , sin   
d2  g  l
 2 +   = 0; Step V: T = 2
dt l g

 
35. A particle is subjected to two SHMs, y1 = A1 sin t and y2 = A2 sin  t +  . Find
 6
(i) the displacement at t = 0 (ii) the maximum speed of the particle
(iii) the maximum acceleration of particle.
Solution
We can use the principle of superposition to determine the resultant SHM and hence determine the required quantities.
(i) At t = 0
y1 = A1 sin t = 0
  A
y2 = A2 sin  t +  = 2
 6 2
A A
 Resultant displacement at t = 0 is y = y1 + y2 = 0 + 2 = 2
2 2
(ii) The resultant of two motions is a SHM of same angular frequency .
The amplitude of resultant motion is

A= A12 + A 22 + 2A1A 2 cos   = A12 + A 22 + 3A1A 2
6
The maximum speed is vmax = A =  A12 + A 22 + 3A1A 2
(iii) The maximum acceleration is
amax = A2 = 2 A12 + A 22 + 3A1A 2 .

Damped simple harmonic motion

We know that the oscillations of a simple pendulum or a loaded spiral spring slowly die out. This is because of the
resistance offered by air to the motion of the bob and mass. The air drag dissipates the energy of motion, and hence
the oscillations gradually die out. Such oscillations are called damped oscillations.
For free oscillations the restoring force is given by (F = −kx). For damped oscillations, the restoring force is given
by (F = −kx + Fd), where Fd is the damping force. The damping force depends on the nature of the medium in which
the system is oscillating. It is generally proportional to the velocity of the moving system and acts opposite to the
direction of velocity. Hence, it can be written as Fd = −b, where b is called damping coefficient. It depends on the

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characteristics of the medium and size and shape of the oscillating system. Hence, the restoring force for damped
oscillations can be written as [F = −kx − b].
d2 x
For free oscillations, we have the differential equation m 2 + kx = 0
dt
k
Solving this we get x(t) = A cos(t + ) where  = gives the angular frequency of oscillation.
m
d2 x dx
For damped oscillations, the differential equation gets modified as m 2 + b + kx = 0
dt dt
bt
− k b2
The solution of this is of the form x(t) = Ae 2m
cos (t + ) , where  = −
m 4m 2
2
The motion is oscillatory with period T =

bt
−
The amplitude of oscillations exponentially decrease because of the factor e 2m
. The mechanical energy of the
1 x
undamped oscillator is kA 2 . For the damped oscillator energy is given by,
2
bt
1 −
E = kA 2 e m .
2
This equation shows that the energy of the system decreases O t
exponentially with time. The effect of damping is determined by the value of the
b
dimensionless ratio . If this ratio is very much less than one, the effect of
km
damping can be ignored.

Illustrations
36. A spring-mass system is performing damped oscillations. The mass of the block is 400 g and the force constant of
the spring is 90 N m−1. If the damping constant is 60 g s−1, the period of oscillation is
(A) 0.1 s (B) 0.2 s (C) 0.4 s (D) 0.3 s
Ans (C)
b = 0.06 kg s −1 ; km = 90  0.4 = 6 kg s−1
Since b  km , the effect of damping can be ignored.
m 0.4
Hence, T = 2 = 2  3.14 ~ 0.4 s
k 90

37. In the above problem the time required for the amplitude to drop to half of its initial value is given by
(A) 2.8 s (B) 4 s (C) 28 s (D) 40 s
Ans (A)
bt
−
Amplitude of the damped oscillation is given by A d = A 0 e 2m
A
Let at t = t  ; A d = 0
2
bt 
A −
Then, 0 = A 0 e 2m
2
bt 
bt 
 e 2m = 2 ; = ln 2
2m
2m  0.693 log 2 2 (0.4) (0.693) (0.3010)
 t = = = 2.8 s
b 0.06
38. In the above problem the time required for the mechanical energy of the system to drop to half its initial value is

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(A) 14 s (B) 1.4 s (C) 2 s (D) 20 s
Ans (B)
bt
1 −
E= KA 2 where A = A 0 e 2m
2
bt bt
1 − −
E= KA 02 e m = E 0 e m
2
E
At t = t  let E =
2
bt 
E −
Then, = E0e 2
2
bt 
bt 
em = 2  = ln 2
m
m (0.4)
t = ln 2 = (0.693) (0.3010) = 1.4 s
b (0.06)

Forced oscillations and resonance

When a simple pendulum is disturbed from its equilibrium position, it oscillates with a frequency determined by its
length and the acceleration due to gravity at the place. These oscillations are called free oscillations and its frequency
is called natural frequency.
When a periodic force acts on an oscillating system, the system starts oscillating not with its natural frequency (0)
but with the frequency of the applied periodic force (d). These oscillations are called forced or driven oscillations.
If the periodic force acting on the system is represented by the equation F(t) = Fd cos d t, the net force on the system
is given by F(t) = (−kx − b + Fd cos dt).
d2 x dx
This leads to the differential equation m 2
+b + kx = F0 cos d t
dt dt
The solution of this equation is of the form x(t) = A cos (dt + )
F0
The amplitude of oscillation is given by A = 1
 m 2 ( 2 − 2 )2 + 2 b 2  2
 d d 

We notice that the amplitude depends on both natural frequency 0 and driving frequency d. Depending on the
relative magnitudes of these frequencies, two cases arise.
Case (1) : When d is much different from 0

In this case ( 2 − d2 ) is much greater than d2 b 2 . Hence, the second term can be ignored.
2

F0
Then the amplitude becomes A =
m ( 2 − d2 )

Case (2) : When d is close to 0

In this case ( 2 − d2 ) is much smaller than d2 b 2 . Hence, the first term can be ignored. Now the amplitude is given
2

F0
by A =
d b

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If b is very small, the amplitude will be very large. In this case the system oscillates with very large amplitude. Such
oscillations are called resonant oscillations.
Applications of resonance

1. In the sonometer experiment, when the frequency of the tuning fork is equal to the frequency of the vibrating string,
the string vibrates with large amplitude and consequently a paper rider placed on it flies off.
2. In the resonance column experiment, when the frequency of the tuning fork is equal to the frequency of the air
column, the air column resonates with very large amplitude and consequently loud sound is heard.
3. When an earth quake strikes a city some buildings are found to collapse while others not. This is because buildings
have a set of natural-frequencies instead of a single natural frequency. When the frequency of the seismic waves
match with one of the natural frequencies the building is destroyed, otherwise the building is not destroyed.

NCERT LINE BY LINE QUESTIONS

1. The equation of motion is represented by y = sin t + cos t . The time period of
periodic motion is [NCERT Pg. 339]
 2 2 4
(1) (2) (3) (4)
   
2. The equation of motion of particle executing SHM is given as y = sin t . The position of
2

equilibrium is [NCERT Pg. 341]

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1
(1) y = 0 (2) y=1 (3) y = (4) y = −1
2
3. A body execute SHM according to equation . At t= 3/2s. what is speed of the particle?
[NCERT Pg. 339]
(1) 10 ms–1
(2) 20 ms–1
(3) 22 ms–1
(4) 44 ms–1

4. Acceleration versus time graph for a particle executing SHM is shown in figure below.
Corresponding position time graph will be [NCERT Pg. 344)

5. Two springs with spring constants K and 2K are attached to a block of mass m and with
fixed supports as shown. When mass is displaced from equilibrium position on either
side, it executes SHM. The frequency of oscillation is (NCERT Pg. 345]

1 3m 1 m 1 3m 1 3K
1) 2) 3) 4)
2 K 2 2K 2 2K 2 m
6. A particle executes SHM. Its time period is T. The kinetic energy of the particle is also periodic
with time period of [NCERT Pg. 346]
T
(1) T (2) 2T (3) (4) Infinity
2

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7. A block whose mass is 500 g is fastened to a spring. The spring has spring constant of 100 N/m.
The block is pulled to a distance of x = 10 cm from its equilibrium position state of x=0
from rest at t = 0. What is kinetic energy of block at x = 5 cm? [NCERT Pg. 347]
(1) 0.375 J (2) 0.19 J (3) 0.56 J (4) 0.76 J
8. A block of mass 2 kg is attached to a spring of spring constant 200 Nm–1 oscillates without friction
over a smooth horizontal surface. The block is displaced by 10 cm from equilibrium position and
released. What is maximum acceleration of block? [NCERT Pg. 348]
(1) 1 ms –2
(2) 2 ms –2
(3) 0.5 ms –2
(4) 1.5 –2

9. Length of a simple pendulum whose time period is 2 second on earth surface will be nearly
[NCERT Pg. 351]
(1) 0.5 m (2) 1m (3) 1.5 m (4) 2 m
10. A block of mass 500 g and attached to one end of a spring of spring constant K = 450 Nm–
1
. The friction is also present which dissipate energy and damping constant of system is 25 g/s.
What is time taken for its amplitude of oscillation to drop to half of its initial value.
[NCERT Pg. 352)
(1) 18.73 s (2) 27.72 s (3) 32.2 s (4) 6.52 s
11. Which of the following example does not represents SHM? [NCERT Pg. 358]
(1) Oscillations of a spring block system
(2) Motion of ball bearing inside smooth curved bowl, when released slightly away from
equilibrium
(3) Motion of oscillating mercury column in vertical U-tube
(4) Rotation of earth about its axis
12. A spring having spring constant of 800 Nm–1 is mounted on a horizontal table as shown.
A mass of 2 kg is attached to free end of the spring. The mass is pulled sideways to
distance of 2.5 cm and released. How much time the mass takes from one extreme to other
(NCERT Pg. 359]

(1) 0.157s (2) 0.2s (3) 0.314 s (4) 0.782 s

13. The acceleration due to gravity on the surface of moon is 1.7 ms–2. What will be period of
oscillation of a simple pendulum on the surface of moon if its time period on the surface of earth
is 2s ?
(1) 4.8s (2) 2.8 s (3) 1.8s (4) 3.5 s
14. A particle executes SHM has maximum speed of 20 cm s and maximum acceleration of 40 cm
–1

s–2. The period of oscillation is [NCERT Pg. 361]

 
(1) s (2) s (3) s (4) 2 s
2 3
15. A spring balance has a scale that reads from 0 to 100 kg. The length of scale is 25 cm. A block
suspended from this balance when displaced and released oscillates with time period of 0.2 s.
What is mass of block approximately? [NCERT Pg. 359]
(1) 2 kg (2) 4 kg (3) 5 kg (4) 6 kg

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16. The graph between length of pendulum and square of its time period is shown below. The best
graph is [NCERT Pg. 351]

17. A collar of mass 4 kg is attached to a spring of spring constant 500 Nnr1. If collar is displaced
from equilibrium position by a distance of 2 cm and released, what is frequency of oscillation?

[NCERT Pg. 348J

(1) 5.4 Hz (2) 1.78 Hz (3) 9.36 Hz (4) 3.26 Hz

18. Two identical springs of spring constant K each are attached to block of mass m and fixed
supports as shown in figure (a). The period of oscillation was observed to be 7. If one more
identical spring is attached as shown in figure (b) then new period will be
[NCERT Pg. 345)

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2 3 1
1) T 2) T 3) 2T 4) T
3 2 3
19. A particle executes SHM, has potential energy which changes with position. If potential energy
at equilibrium position is assumed to be Zero. then potential energy versus position graph is best
represented by [NCERT Pg. 347]

20. The graph of a particle executing SHM is shown for two particles A and B. The ratio of
maximum accelerations of A to B is [NCERT Pg. 341)

1) 1: 1 2) 1:2 3) 2:1 4) 1:4

NCERT BASED PRACTICE QUESTIONS
1. The type of motion of earth about its axis is?
(1) Simple harmonic motion
(2) Periodic but not SHM
(3) Oscillatory
(4) None
2. General vibration of a polyatomic molecule about its equilibrium position is :-
(1) Simple harmonic motion
(2) Neither periodic nor SHM
(3) Superposition of SHMs
(4) None
3. A ball performs uniform circular motion on a horizontal plane then shadow of ball on the vertical wall
performs.
(1) Projectile motion
(2) Uniform circular motion
(3) Simple harmonic motion
(4) Non uniform circular motion
4. Four pendulum A.B.C. and D are suspended from the same elastic support as shown in figure. A and C are
of same length while B is smaller than A and D is larger than A. If A is given a transverse displacement.

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(1) D will vibrate with maximum amplitude

(2) C will vibrate with maximum amplitude
(3) B will vibrate with maximum amplitude
(4) All the four will oscillate with equalamplitude
5. A particle oscillating under a force F = − K x − bv is (K and b are constants), then oscillator is :-
(1) Simple harmonic oscillator
(2) Linear oscillator
(3) Damped oscillator
(4) Forced oscillator
6. Time period of oscillation of vertical spring block system is 2 seconds, physical quantity which will be
having time period 1sec is :-
(1) Velocity (2) Potential Energy
(3) Acceleration (4) Momentum
7. The equation of motion of a particle is x = acos(t)2 the motion is :-
(1) Periodic but not oscillatory
(2) Periodic and oscillatory
(3) Oscillatory but not periodic
(4) Neither periodic nor oscillatory
8. A body is doing SHM, which is not true :-
(1) Average total energy per cycle is equal to its maximum kinetic energy
(2) Average kinetic energy per cycle in equal to half of its maximum kinetic energy
2
(3) Mean velocity over a complete cycle I equal to times of its max. velocity

1
(4) Root mean square velocity is times of its maximum velocity
2
9. The displacement of a particle varies with time according to the relation y = a sint + bcost
(1) The motion is oscillatory but not SHM
(2) The motion is SHM with amplitude a + b
(3) The motion is SHM with amplitude a2+b2
(4) The motion is SHM with amplitude a 2 + b 2
10. In SHM when the magnitude of displacement is greatest, then :-
(1) Magnitude of velocity is least
(2) Magnitude of velocity is greatest
(3) Magnitude of acceleration is least
(4) Magnitude of force is least
11. In SHM phase difference between displacement and velocity is :-
(1) Zero (2)  (3) /2 (4) None of these
12. In damped oscillation mechanical energy decreases with time :-
(1) Linearly (2) Exponentially
(3) Not decreases (4) None of these
13. In the ideal case of zero damping the ampliltude of SHM at resonance is :-
(1) Minimum (2) Zero (3) Infinite (4) Maximum but not infinite
14. A displacement equation of a particle executes S.H.M is x = sin2wt, then :-

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(a) Mean position x = 0
(b) Mean position x = 1/2

(c) Time period

2
(d) Time period =

(1) a, d (2) b,c (3) a,c (4) b,d
15. Displacement vs time curve for a particle executing S.H.M. is shown in fig given below.
Choose the correct statements :-

(1) Phase of the oscillation is same at t = 0s and t = 2s

(2) Phase of the oscillation is same at t = 2s and t = 6s
(3) Phase of the oscillation is same at t = 1s and t = 7s
(4) Phase of the oscillation is same at t = 1s and t = 3 s
16. Which one of the following is not a periodic motion?
(1) Rotation of the earth about its axis
(2) A freely suspended bar magnet displaced from its N-S direction and released
(3) Motion of hands of a clock
(4) An arrow released from a bow oscillations
17. Which of the following displacement time graphs represent damped harmonic oscillations?

(1) (2)

(3) (4)
18. In case of forced oscillations of a body :-
(1) Driving force is a constant
(2) Driving force is to be applied only momentarily
(3) Driving force has to be periodic and continuous
(4) Driving force is not required
19. Graph below shows variation of amplitude of forced oscillations with respect to frequency of
driving force. Choose the correct option.

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(1) Amplitude is maximum at d = 0

(2) Peak amplitude is maximum for curve (a) because for curve (a) damping is minimum
(3) In case of resonance there occur maximum transfer of energy from driver to driven
(4) All of above
20. The natural frequency of building depend on :-
(a) Its height and other size parameter
(b) The nature of building material
The correct statement :-
(1) Only (a) (2) Only (b) (3) Both (a) & (b) (4) Neither (a) & (b)
21. Choose incorrect option :-

(1) Amplitude of both are same

(2) Both curve have same frequency
(3) Initial phase of both curve are same
(4) Time period of a is twice of time period of b
22. A child climbs up a step waited for some time on step and comes down and repeats the
process. Choose x-t graph for this process.

(1) (2)

(3) (4)
23. In SHM :-
(1) PE is stored due to elasticity of system
(2) KE is stored due to inertia of system
(3) Total energy of system remains conserved

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(4) All of above

 
24. The displacement of a particle is represented by the equation y = 3cos  − 2t  . The motion of the
4 
particle is
2
(a) simple harmonic with period


(b) simple harmonic with period

(c) periodic but not simple harmonic
(d) non-periodic
25. The displacement of a particle is represented by the equation y = sin 3 t . The motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period 2p/w
(d) simple harmonic with period p/w
26. The relation between acceleration and displacement of four particles are given below. Which, one of the
particle is exempting simple harmonic motion?
(a) ax = +2x (b) ax = +2x2
2
(c) ax = –2x (d) ax = –2x
27. Motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid
28. A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos  t and
y = a sin  t. The trajectory of motion of the particle will be
(a) an ellipse (b) a parabola (c) a circle (d) a straight line

29. Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and
the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius
vector of the rotating particle P is

30. The equation of motion of a particle is x = a cos(  t)2. The motion is

(a) periodic but not oscillatory
(b) periodic and oscillatory
(c) oscillatory but not periodic
(d) neither periodic nor oscillatory
31. A particle executing SHM maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s2. The period
of oscillation is
 
(a)  sec (b) sec (c) 2  sec (d) sec
2 t

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32. Assertion : The force acting on a particle moving along x-axis is F = – k(x + v0t), where k is a
constant.
Reason : To an observer moving along x-axis with constant velocity v0, it represents
S.H.M.
(1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(2) Assertion is correct, reason is correct; reason is not a correct explanation for
assertion
(3) Assertion is correct, reason is incorrect
(4) Assertion is incorrect, reason is correct.
33. Assertion : A particle executing simple harmonic motion comes to rest at the extreme
positions .
Reason : The resultant force on the particle is zero at these positions.
(1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(2) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(3) Assertion is correct, reason is incorrect
(4) Assertion is incorrect, reason is correct.
34. Assertion : The graph of total energy of a particle in SHM w.r.t. position is a straight line with
zero slope.
Reason : Total energy of particle in SHM remains constant throughout its motion.
(1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(2) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(3) Assertion is correct, reason is incorrect
(4) Assertion is incorrect, reason is correct.
35. Assertion: If amplitude of simple pendulum increases then the motion of pendulum is
oscillatory but not simple harmonic.
Reason: For larger amplitude q is large and then sin q ¹ q, so the motion is no longer SHM.
(1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(2) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(3) Assertion is correct, reason is incorrect
(4) Assertion is incorrect, reason is correct.
36. Assertion: Amplitude of a forced vibration can never reach infinity.
Reason: The driving frequency cannot be increased beyond a certain limit.
(1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(2) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(3) Assertion is correct, reason is incorrect
(4) Assertion is incorrect, reason is correct.

TOPIC WISE PRACTICE QUESTIONS

TOPIC 1: Displacement, Phase, Velocity, Acceleration and Energy in S.H.M.
1. The displacement of a particle in simple harmonic motion in one time period is
(a) A (b) 2 A (c) 4 A (d) Zero
2. Which of the following quantities are always negative in a simple harmonic motion ?
(a) F.r (b) v.r (c) a.r (d) Both (a) & (c)
3. A particle moving along the X-axis, executes simple harmonic motion then the force acting on it is given
by (Where, A and k are positive constants).
(a) – A kx (b) A cos (kx) (c) A exp (– kx) (d) Akx

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4. A particle is subjected to two mutually perpendicular simple harmonic motions such that its x and y
 
coordinates are given by x = 2 sin t ; y 2sin  t +  . The path of the particle will be
 4
(a) a straight line (b) a circle (c) an ellipse (d) a parabola
5. A point mass oscillates along the x-axis according to the law x = x 0 cos ( t −  / 4 ) . If the acceleration of
the particle is written as a = A cos ( t +  ) , then
(a) A = x 0 2 ,  = 3 / 4 (b) A = x0 ,  = − / 4 (c) A = x 0 2 ,  =  / 4 (d) A = x 0 2 ,  = − / 4
6. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple
harmonic motion of time period T, then, which of the following does not change with time?
(a) aT/x (b) aT + 2pv (c) aT/v (d) a2T2 + 4  2 v2
7. A body executing linear simple harmonic motion has a velocity of 3 m/s when its displacement is 4 cm and
a velocity of 4 m/s when its displacement is 3 cm. What is the amplitude of oscillation?
(a) 5 cm (b) 7.5 cm (c) 10 cm (d) 12.5 cm
8. The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16
cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes
8 3 cm/s, will be
(a) 2 3 cm (b) 3 cm (c) 1 cm (d) 2 cm
9. A particle of mass 1 kg is moving in S.H.M. with an amplitude 0.02 and a frequency of 60 Hz. The
maximum force acting on the particle is
(a) 144  2 (b) 188  2 (c) 288  2 (d) None of these
10. If < E > and < U > denote the average kinetic and the average potential energies respectively of mass
describing a simple harmonic motion, over one period, then the correct relation is
(a) < E > = < U > (b) < E > = 2< U > (c) < E > = –2< U > (d) < E > = – < U >
11. Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance h
directly above the tunnel, the motion of the particle is
(a) simple harmonic (b) parabolic (c) oscillatory (d) none of these
2
12. The displacement y(t) = A sin ( t +  ) of a pendulum for  = is correctly represented by
3

(a) (b) (c) (d)

13. The particle executing simple harmonic motion has a kinetic energy K0 cos2 t . The maximum values of
the potential energy and the total energy are respectively
(a) K0/2 and K0 (b) K0 and 2K0 (c) K0 and K0 (d) 0 and 2K0
14. In a simple harmonic oscillator, at the mean position
(a) kinetic energy is minimum, potential energy is maximum
(b) both kinetic and potential energies are maximum
(c) kinetic energy is maximum, potential energy is minimum
(d) both kinetic and potential energies are minimum
15. In S.H.M. the ratio of kinetic energy at mean position to the potential energy when the displacement is half
of the amplitude is
4 2 4 1
(a) (b) (c) (d)
1 3 3 2
16. Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will its
kinetic energy be 75% of the total energy?

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1 1 1 1
(a) s (b) s (c) s (d) s
6 4 3 12
17. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against
its displacement d. Which one of the following represents these correctly? (graphs are schematic and not
drawn to scale)

(a) (b) (c) (d)

18. A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of
10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is
(a) 37.5  2 erg (b) 3.75  2 erg (c) 375  2 erg (d) 0.375  2 erg
19. A particle executes SHM with time period 8 s. Initially, it is at its mean position. The ratio of distance
travelled by it in the 1st second to that in the 2nd second is
(a) 2 :1 ( )
(b) 1: 2 − 1 ( )
(c) 2 + 1 : 2 ( )
(d) 2 − 1 :1
20. A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A =
1 cm). Find the average velocity in the interval in which it moves form equilibrium position to half of its
amplitude.
(a) 4 cm/s (b) 6 cm/s (c) 12 cm/s (d) 16 cm/s
21. A body is moving in a room with a velocity of 20 m/s perpendicular to the two walls separated by 5 m.
There is no friction and the collisions with the walls are elastic. The motion of the body is
(a) not periodic
(b) periodic but not simple harmonic
(c) periodic and simple harmonic
(d) periodic with variable time period
22. Two particles P and Q describe SHM of same amplitude a, same frequency f along the same straight line.
The maximum distance between the two particles is a 2 . The phase difference between the particle is
  
(a) zero (b) (c) (d)
2 6 3
23. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular
frequency  . The amplitude of oscillation is gradually increased. The coin will leave contact with the
platform for the first time
g
(a) at the mean position of the platform (b) for an amplitude of 2

2
g
(c) for an amplitude of 2 (d) at the highest position of the platform

24. A body of mass 5 gram is executing S.H.M. about a fixed point O. With an amplitude of 10 cm, its maximum
velocity is 100 cm/s. Its velocity will be 50 cm s–1 at a distance (in cm)
(a) 5 (b) 5 2 (c) 5 3 (d) 10 2
 t 
25. If y = 2 ( cm ) sin  +  then the maximum acceleration of the particle doing the S.H.M. is
2 
 2 2 2 
(a) cm / s (b) cm / s 2
(c) cm / s 2 (d) cm / s 2
2 2 4 4
26. A body is executing simple harmonic motion. At a displacement x from mean position, its potential energy
is E1 = 2J and at a displacement y from mean position, its potential energy is E2 = 8J. The potential energy
E at a displacement (x + y) from mean position is
(a) 10J (b) 14J (c) 18J (d) 4J
27. The angular velocity and the amplitude of a simple pendulum is  and a respectively. At a displacement x
from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is

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(a)
(a 2
− x 2 2 )
(b) 2
x 2 2
(c)
(a 2
− x2 )
(d) 2
x2
x 2 2 ( a − x2 2 ) x2 ( a − x2 )
TOPIC 2: Time Period, Frequency, Simple Pendulum and Spring Pendulum
28. The necessary condition for the bob of a pendulum to execute SHM is
(a) the length of pendulum should be small (b) the mass of bob should be small
(c) amplitude of oscillation should be small (d) the velocity of bob should be small
29. A child swinging on swing in sitting position stands up. The time period of the swing will
(a) increase (b) decrease (c) remain same(d) increase if the child is tall and decrease if the child is short.
30. A simple pendulum oscillates in air with time period T and amplitude A. As the time passes
(a) T and A both decrease (b) T increases and A is constant
(c) T remains same and A decreases (d) T decreases and A is constant
31. A simple pendulum is made of a body which is a hollow sphere containing mercury suspended by means
of a wire. If a little mercury is drained off, the period of pendulum will

(a) remain unchanged (b) increase (c) decrease (d) become erratic
32. If the magnitude of displacement is numerically equal to that of acceleration, then the time period is
(a) 1 second (b) p second (c) 2p second (d) 4p second
33. A simple pendulum has a metal bob, which is negatively charged. If it is allowed to oscillate above a
positively charged metallic plate, then its time period will
(a) increase (b) decrease (c) become zero (d) remain the same

34. Identify the wrong statement from the following

1
(a) If the length of a spring is halved, the time period of each part becomes times the original
2
1 1 1
(b) The effective spring constant K of springs in parallel is given by = + + ......
K K1 K 2
(c) The time period of a stiffer spring is less than that of a soft spring
(d) The spring constant is inversely proportional to the spring length
35. A vertical mass-spring system executes simple harmonic oscillations with a period of 2s. A quantity of this
system which exhibits simple harmonic variation with a period of 1 s is
(a) velocity (b) potential energy
(c) phase difference between acceleration and displacement
(d) difference between kinetic energy and potential energy
36. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4
m/s. The period of oscillation is
(a) 0.01 s (b) 10 s (c) 0.1 s (d) 100 s
37. The spring constant from the adjoining combination of springs is

(a) K (b) 2 K (c) 4 K (d) 5 K/2

38. A particle at the end of a spring executes S.H.M with a period t1. While the corresponding period for another
spring is t2. If the period of oscillation with the two springs in series is T then

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(a) T −1 = t1−1 + t −21 (b) T 2 = t11 + t 22 (c) T = t1 + t 2 (d) T −2 = t1−2 + t 2−2

39. Two particles A and B of equal masses are suspended from two massless springs of spring constants k1 and
k2, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of A and B
is
k1 k k2 k
(a) (b) 2 (c) (d) 1
k2 k1 k1 k2

40. In an experiment for determining the gravitational acceleration g of a place with the help of a simple
pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.

What is the value of g at the place?

(a) 9.81 m/s2 (b) 9.87 m/s2 (c) 9.91 m/s2 (d) 10.0 m/s2
41. A ring is suspended from a point S on its rim as shown in the figure. When displaced from equilibrium, it
oscillates with time period of 1 second. The radius of the ring is (take g =  2 )

(a) 0.15 m (b) 1.5 m (c) 1.0 m (d) 0.5 m

42. A particle moves such that its acceleration ‘a’ is given by a = – bx where x is the displacement from
equilibrium position and b is constant. The period of oscillation is
(a) 2  /b (b) 2 π / b (c) 2 / b (d) 2  / b
1
43. The total mechanical energy of a spring-mass system in simple harmonic motion is E = m2 A 2 . Suppose
2
the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the
same. The new mechanical energy spring pendulum will

(a) become 2E (b) become E/2 (c) become 2E (d) remain E

44. Two springs, of force constants k1 and k2 are connected to a mass m as shown. The frequency of oscillation
of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation
becomes
(a) 2 f (b) f /2 (c) f /4 (d) 4 f
45. If T1 and T2 are the time-periods of oscillation of a simple pendulum on the surface of earth (of radius R)
and at a depth d, then d is equal to
 T12   T22   T   T 
(a) 1 − 2  R (b) 1 − 2  R (c) 1 − 1  R (d) 1 − 2  R
 T2   T1   T2   T1 
46. A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme
position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken
to the poles it will

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(a) run slow (b) run fast (c) stop working (d) give correct time
47. A rectangular block of mass m and area of cross-section A floats in a liquid of density  . If it is given a
small vertical displacement from equilibrium. It undergoes oscillations with a time period T then
(a) T  m (b) T   (c) T  1 / A (d) T  1 / 
48. A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude 'a'. If
the coefficient of friction is '  ', then the block just starts to slip when the frequency of oscillation is
1 g g a a
(a) (b) (c) 2 (d)
2 a a g g

49. A particle of mass is executing oscillations about the origin on the x-axis. Its potential energy is V(x) =
k| x |3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is
3
1
(a) proportional to (b) proportional to a (c) independent a 2 (d) None of these
a
50. A pendulum is undergoing S.H.M. The velocity of the bob in the mean position is v. If now its amplitude
is doubled, keeping the length same, its velocity in the mean position will be
(a) v/2 (b) v (c) 2 v (d) 4 v
51. A circular hoop of radius R is hung over a knife edge. The period of oscillation is equal to that of a simple
pendulum of length
(a) R (b) 2R (c) 3R (d)3R/2
52. On Earth, a body suspended on a spring of negligible mass causes extension L and undergoes oscillations
along length of the spring with frequency f. On the Moon, the same quantities are L/n and f ' respectively.
The ratio f '/f is
1
(a) n (b) (c) n–1/2 (d) 1
n
53. A mass m = 1.0 kg is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the
spring and the pan is negligible. When pressed slightly and released, the mass executes simple harmonic
motion. The spring constant is 500 N/m. What is the amplitude A of the motion, so that the mass m tends
to get detached from the pan ? (Take g = 10 m/s2).
The spring is stiff enough so that it does not get distorted during the motion.

(a) A > 2.0 cm (b) A = 2.0 cm (c) A < 2.0 cm (d) A = 1.5 cm
54. A simple pendulum attached to the roof of a lift has a time period of 2s in a stationary lift. If the lift is
allowed to fall freely the frequency of oscillations of pendulum will be
(a) zero (b) 2Hz (c) 0.5 Hz (d) infinity
55. A Second’s pendulum is placed in a space laboratory orbiting around the earth at a height 3 R from the
earth’s surface where R is earth’s radius. The time period of the pendulum will be
(a) zero (b) 2 3 (c) 4 sec (d) infinite
56. A simple spring has length l and force constant K. It is cut into two springs of lengths l1 and l2 such that
l1 = n l2 (n = an integer). The force constant of spring of length l1 is
(a) K (1 + n) (b) (K/n) (1 + n) (c) K (d) K/(n + 1)
57. The length of a second’s pendulum at the surface of earth is 1 m. The length of second’s pendulum at the
surface of moon where g is 1/6th that at earth’s surface is
(a) 1/6 m (b) 6 m (c) 1/36 m (d) 36 m

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58. A small ball of density 4  0 is released from rest just below the surface of a liquid. The density of liquid
increases with depth as  = 0 (1 + ay) where a = 2m–1 is a constant. Find the time period of its oscillation.
(Neglect the viscosity effects).
2   3
(a) sec (b) sec (c) sec (d) sec
5 5 2 5 2 5
d 2 x2
59. If a simple harmonic motion is represented by 2 +  x = 0 , its time period is
dt
2 2
(a) (b) (c) 2  (d) 2
 

TOPIC 3: Damped & Forced Oscillations and Resonance

60. On account of damping, the frequency of a vibrating body
(a) remains unaffected (b) increases (c) decreases (d) changes erratically
61. A block connected to a spring oscillates vertically. A damping force Fd, acts on the block by the surrounding
medium. Given as Fd = – bV, b is a positive constant which depends on:
(a) viscosity of the medium (b) size of the block (c) shape of the block (d) All of these
62. Resonance is an example of
(a) tuning fork (b) forced vibration (c) free vibration (d) damped vibration
63. If a body oscillates at the angular frequency d of the driving force, then the oscillations are called
(a) free oscillations (b) coupled oscillations (c) forced oscillations (d) maintained oscillations
64. In case of a forced vibration, the resonance wave becomes very sharp when the
(a) restoring force is small (b) applied periodic force is small
(c) quality factor is small (d) damping force is small
65. In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value a0 at the end
of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be
(a) a0/2 (b) a0/4 (c) a0/6 (d) a0/9
66. The amplitude of velocity of a particle is given by, Vm = V0/ ( a − b + c ) where V0, a, b and c are positive:
2

The condition for a single resonant frequency is

(a) b2 < 4ac (b) b2 = 4a c (c) b2 = 5ac (d) b2 = 7ac
67. When an oscillator completes 100 oscillations its amplitude reduces to1/3 of its initial value. What will be
its amplitude, when it completes 200 oscillations?
1 2 1 1
(a) (b) (c) (d)
8 3 6 9
68. A forced oscillator is acted upon by a force F = F0 sin t . The amplitude of oscillation is given by
55
. The resonant angular frequency is
22 − 36 + 9
(a) 2 unit (b) 9 unit (c) 18 unit (d) 36 unit
69. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it
will decrease to a times its original magnitude, where a equals
(a) 0.7 (b) 0.81 (c) 0.729 (d) 0.6
70. Bob of a simple pendulum of length l is made of iron. The pendulum is oscillating over a horizontal coil
carrying direct current. If the time period of the pendulum is T then :
l
(a) T  2 and damping is smaller than in air alone.
g
l
(b) T = 2 and damping is larger than in air alone.
g
l
(c) T  2 and damping is smaller than in air alone.
g

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l
(d) T  2 and damping is larger than in air alone.
g

NEET PREVIOUS YEARS QUESTIONS

1. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a
simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m
from the mean position. The time period of oscillation is [2018]
(a) 2  s (b)  s (c) 1 s (d) 2 s
2. A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new
force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is [2017]
(a) 1 : 9 (b) 1 : 11 (c) 1 : 14 (d) 1 : 6
3. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm
from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period
in seconds is [2017]
5 4 2 5
(a) (b) (c) (d)
2 5 3 
4. When two displacements represented by y1 = a sin( t ) and y2 = b cos( t ) are superimposed the motion
is: [2015]
a
(a) simple harmonic with amplitude (b) simple harmonic with amplitude a 2 + b 2
b
(a + b)
(c) simple harmonic with amplitude (d) not a simple harmonic
2
5. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position
are V1 and V2, respectively. Its time period is [2015]
x 2 − x1
2 2
V1 + V2
2 2
V1 − V2
2 2
x1 − x 22
2
(a) 2 (b) 2  (c) 2  (d) 2 
V12 − V22 x12 + x 22 x12 − x 22 V12 − V22
6. A particle is executing a simple harmonic motion. Its maximum acceleration is  and maximum velocity is
 . Then its time period of vibration will be : [2015]
 2 2 2
(a) (b) (c) (d) 2
   
7. The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos ( t )
where X = displacement at time t,  = frequency of oscillation Which one of the following graphs shows
correctly the variation of ‘a’ with ‘t’? [2014]

(a) (b) (c) (d)

8. The displacement of a particle executing simple harmonic motion is given by y = A0 + A sin t + B cos t .
Then the amplitude of its oscillation is given by : [NEET – 2019]
A02 + ( A + B )
2
1) A0 + A2 + B 2 2) A2 + B 2 3) 4) A + B
9. Average velocity of a particle executing SHM in one complete vibration is : [NEET – 2019]
A A 2
1) 2) A 3) 4) zero
2 2
10. The radius of circle the period of revolution initial position and sense of revolution are indicated in the
fig. [NEET – 2019]

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y-projection of the radius vector of rotating particle P is :

 t 
1) y ( t ) = −3cos 2 t , where y in m 2) y ( t ) = 4sin   , where y in m
 2
 3 t   t 
3) y ( t ) = 3cos   , where y in m 4) y ( t ) = 3cos   , where y in m
 2   2
11. A mass falls from a height 'h' and its time of fall 't' is recorded in terms of time period T of a simple
pendulum. On the surface of earth it is found that t = 2T. The entire set up is taken on the surface of
another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and
corresponding times noted as t' and T', then [NEET – 2019(ODISSA)]
1) t ' = 2T ' 2) t '  2T ' 3) t '  2T ' 4) t ' = 2T '
12. Identify the function which represents a periodic motion [NEET – 2020(COVID-19)]

(1) et (2) log e ( t ) (3) sin t + cos t (4) e −t

13. The phase difference between displacement and acceleration of a particle in a simple harmonic motion
is [NEET – 2020]
3 
1) zero 2)  rad 3) rad 4) rad
2 2
14. A body is execut simple harmonic motion with frequency ‘n’, the frequency of its potential energy is
[NEET-2021]
1) 2n 2) 3n 3) 4n 4) n
15. A spring is stretched by 5cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is
suspended by it is : [NEET-2021]
1) 6.28 s 2) 3.14 s 3) 0.628 s 4) 0.0628 s
16. Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their
mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which
the two are again in phase at the mean position is : [NEET-2022]
1) 11 2) 9 3) 10 4) 8

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NCERT LINE BY LINE QUESTIONS – ANSWERS

1. (b) 2. (c) 3. (d) 4. (d) 5. (d) 6. (c) 7. (a) 8. (a) 9. (b) 10. (b)
11. (d) 12. (a) 13. (a) 14. (a) 15. (b) 16. (d) 17. (b) 18. (a) 19. (b) 20. (b)

NCERT BASED PRACTICE QUESTIONS-ANSWERS

1) 2 2) 3 3) 3 4) 2 5) 3 6) 2 7) 3 8) 3 9) 4 10) 1
11) 3 12) 2 13) 3 14) 2 15) 2 16) 4 17) 2 18) 3 19) 4 20) 3
21) 2 22) 3 23) 4 24) 2 25) 2 26) 4 27) 3 28) 3 29) 1 30) 3
31) 1 32) 1 33) 3 34) 1 35) 1 36) 3

TOPIC WISE PRACTICE QUESTIONS - ANSWERS

1) 4 2) 4 3) 1 4) 3 5) 1 6) 1 7) 1 8) 4 9) 3 10) 1
11) 3 12) 1 13) 3 14) 3 15) 1 16) 1 17) 4 18) 3 19) 4 20) 3
21) 2 22) 2 23) 2 24) 3 25) 2 26) 3 27) 3 28) 3 29) 2 30) 3
31) 2 32) 3 33) 2 34) 2 35) 2 36) 1 37) 1 38) 2 39) 3 40) 2
41) 4 42) 2 43) 4 44) 1 45) 1 46) 4 47) 1 48) 1 49) 1 50) 3
51) 2 52) 4 53) 3 54) 1 55) 4 56) 1 57) 1 58) 1 59) 1 60) 3
61) 4 62) 2 63) 3 64) 4 65) 4 66) 2 67) 4 68) 2 69) 3 70) 4

NEET PREVIOUS YEARS QUESTIONS-ANSWERS

1) 2 2) 2 3) 2 4) 2 5) 1 6) 3 7) 3 8) 2 9) 4 10) 4
11) 4 12) 3 13) 2 14) 1 15) 3 16) 1

NCERT BASED PRACTICE QUESTIONS - SOLUTIONS

2. A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of
individual SHMs.
3. It will be a SHM as projection of uniform circular motion on axis in SHM.

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l
4. T= , LA = LC so TA = TC  nA = nC
g
So C will be in resonance, hence C will vibrate with maximum amplitude.
5. Term –kx represents restoring force but −bv represents damping force.
6. ndisp = nvelocity = aaccn = n
but nPE = 2n
so TPE=T/2 = 1sec.
7. x = acos(t)2
It is a cosine function which varies between –a and +a so it is oscillatory. Now for periodic motion.
x(t + T) = x(t) but
x(t + T) = acos( (t + T))2= acos[2t2 + 2tT+ 2T2]
 x(t) so not periodic
8. TEav = KEmax. = PEmax. = ½ KA2
Vav = AW < cos t > = 0
V
but Vrms = max
2
9. a sint + bcost = a 2 + b 2 sin(t +  )
b
Where  = tan −1   so it is SHM with amplitude a 2 + b 2
a
10. At x = ±A, v = 0 and a = amax
11. Velocity is ahead by /2 phase angle from displacement
1
12. E = KA2e − bt / m (exponentially)
2
13. b = 0, A = 
14. x = sin2wt

15. at t = 2sec. and t = 6 sec oscillater has same consecutime extremes means phased difference is zero.
16. Motion of arrow released from a bow is not periodic (Not repeative motion)
17. In damped oscillations

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so graph between disp. time is exponentially decreasing.

18. F = F0cos(dt)
19. In forced oscillations -amplitude

If damping is minimum then peak amplitude maximum.

20. Natural frequency of the body depends on shape and size of body and it material.
21. No. of oscillations per unit time is different means frequencies of both curve are not same.
23. For SHM inertia and elasticity are essential properties here total energy of system remains conserved.
24:

25. Given

As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine
and cosine functions, hence it is periodic motion
26: (d) For motion to be SHM acceleration of the particle must be opposite of restoring force and proportional
to negative of displacement. So, F = ma = m(–2x)

Hence, ax = –2x
h
27: Motion of an oscillating liquid column in a U tube is SHM with period, T = 2 , where h is the height
g
of liquid column in one arm of U tube in equilibrium position of liquid. Therefore, T is independent of
density of liquid.
28 Given
x = a cos ω t . . . (i)
y = a sin ω t
Squaring and adding (i) and (ii), we get
x2+y2=a2cos2ωt+a2sin2ωt=a2
It is an equation of circle. Thus, trajectory of motion will be circle.
29.

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30.

31

TOPIC WISE PRACTICE QUESTIONS - SOLUTIONS

1. (d) In one complete oscillation particle comes to its starting position again to displacement is zero.
2. (d)
3. (a) For simple harmonic motion, F = – kx. Here, k = Ak.

4. (c) As phase difference = , the resultant path of particle is an ellipse.
4
5. (a) Here, displacement x = x 0 cos ( t −  / 4 )

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a
6. (a) For an SHM, the acceleration a = − 2 x where  2 is a constant. Therefore, is a constant. The time
x
aT
period T is also constant. Therefore, is a constant.
x
7. (a) Velocity in SHM is given by v =  a 2 − y 2
At y = 4 cm = 0.04 m, v = 3m/ s
 3 =  a 2 − ( 0.04 ) ---------(1)
2

At y = 3 cm = 0.03 m, v = 4 m/s
 4 =  a 2 − ( 0.03) ---------(2)
2

Dividing (2) by (1), we get a = 0.05 = 5 cm

8. (d) At mean position velocity is maximum

9. (c) Max. force = mass × max. acceleration

= m 42  2a = 1 4  2  ( 60 )  0.02 = 2882
2

1
E av = U av = m2 A 2
10. (a) 2
11. (c) When a particle is dropped from a height h above the centre of tunnel.
(i) It will oscillate, through the earth to a height h on both sides
(ii) The motion of particle is periodic
(iii) The motion of particle will not be SHM.
12. (a) Displacement y ( t ) = Asub ( t +  ) [Given]
2 2
For  = , at t = 0; y = A sin  = A sin
3 3
= A sin 120° = 0.87 A [ sin 120° = 0.866]
Graph (a) depicts y = 0.87A at t = 0
13. (c) The kinetic energy of particle executing S.H.M is K.E=K0cos2ωt
Maximum value K.E is K.Emean =K0
Now, the total energy of S.H.M is same as the maximum value of K.E, which is K0
Similarly the maximum value of potential energy of S.H.M is same as the maximum value of K.E, which is K0

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14. (c) So, at mean position, x = 0
1
K.E. = m2a 2 (maximum) ; P.E. = 0
2
1 1
15.
2
( )
(a) K.E. = m2 a 2 − y 2 and P.E. = m2 y 2
2
1
At mean position, y = 0;  K.E. = m2a 2
2
2
a 1 a 1
At, y = ; P.E. = m2 = m2a 2
2 2 4 8

 Ratio =
(1/ 2 ) m a = K.E.at mean position = 4
2 2

(1/ 8) m2a 2 PE at a / 2 1
16. (a) K.E. of a body undergoing SHM is given by,
1 1
K.E. = ma 22 cos 2 t and T.E. = ma 2 2
2 2

Given K.E. = 0.75 T.E.  0.75 = cos 2 t  t =
6
  2 1
t= t= t= s
6  6  2 6
1 1
17.
2
( )
(d) K.E. = k A 2 − d 2 and P.E. = kd 2
2
At mean position d = 0. At extrement positions d = A
1
18. (
(c) Kinetic energy K = m2 a 2 − y 2
2
)
2
1  2 
= 10    102 − 52  = 375 r 2 erg
2  2 
19. (b)

20. (c) Average velocity is simply the displacement of the particle divided by time taken.
Simple harmonic motion can be described by x=A sin ωt where ω=2π/T if the particle starts at the mean position
at t=0
1
t = sin −1
time when particle reaches x=A/2 is when 2 which is t =1/24 seconds
Total displacement is A/2
1/ 2 cm
= = 12cm / s
1
s
Hence average velocity 24
21. (b) Since there is no friction and collision is elastic, no loss of energy takes place and the body strikes again
and again with two perpendicular walls. So the motion of the ball is periodic. But here, there is no restoring
force. So the characteristics of SHM will not be satisfied.
22. (b)

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23. (b) The coin will leave contact at any point when the acceleration of platform is more than 'g' and in the same direction of 'g'.
At the highest point, both of the conditions are satisfied. Acceleration at the highest point is equal to ω2A, so we have
g
2 A  g  A 
2
24. (c) v max = 100 = a ;  = 100/a = 100/10 = 10 rad/s
v2 = 2 (a2 – y2) or 502 = 102 (102 – y2) or 25 = 100 – y2
or y = 75 = 5 3 cm.
 t  dy   t 
25. (b) ay = 2sin  +   v = = 2  cos  +  
2  dt 2 2 
d2 y 2  t  2
a= = − sin  +    a max =
dt 2 2  2
1 1
26. (c) E1 = kx 2 , E 2 = ky 2
2 2
1 1 1
E = k ( x + y ) = kx 2 + ky 2 + kxy = E1 + E 2 + 2 E1E 2
2

2 2 2
= 2 + 8 + 2 16 = 18 J
1 1
27. (c) P.E., V = m2 x 2 and K.E., T = m2 a 2 − x 2
2 2
( )
T a −x 2 2
 =
V x2
28. (c) the mass of bob should be small
29. (b) When the child stands up, the position of centre of mass of the system changes. Hence the effective length of the
pendulum decreases. As T  1 ,so time period will decrease.
30. (c) Air friction damps the oscillation so, temperature increases and Amplitude
31. (b) When some mercury is drained off, the centre of gravity of the bob moves down and so length of the
pendulum increases, which result increase in time period.
4 2
32. (c) y = 2 y, 2 = 1, 2 = 1 or T 2 = 4 2
T
or T = 2  second
33. (b) A simple pendulum has a metal bob, which is negatively charged. If it is allowed to oscillate above a
positively charged metallic plate, then its time period will decrease
34. (b) In parallel combination, the effective spring constant
(K) of springs
K = K1 + K2 + ... Hence, option (b) is wrong.

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35. (b) Potential energy because the time period of potential energy and kinetic energy is 1/2 that of the simple
harmonic motion
2
36. (a) Maximum velocity, v max = a  v max = a 
T
−3
2a 2  3.14  7 10
T= =  0.01s
v max 4.4
37. (c) Here all the three springs are connected in parallel to mass m. Hence equivalent spring constant
k = K + K + 2 K = 4 K.
38. (b) Let mass of the particle be m and k represents the spring constant.
m
t = 2
Time period of oscillation of a particle executing SHM k
m 4 2 m
t1 = 2  k1 = 2
k1 t1
Thus for 1st spring
4 2 m
k2 = 2
Similarly t2
1 1 1 kk
= +  kc = 1 2
Now spring constant of the combination of two springs in series k c k1 k 2 k1 + k 2
m
T = 2
kc
Time period of the oscillation of combined springs

39. (c) Maximum velocity during SHM = A

k
But = m2 ;  =
m
k
 Maximum velocity = A
m
Here the maximum velocity is same and m is also same
A1 k2
 A1 k1 = A 2 k 2  =
A2 k1
40. (b) Using, The expression for the time period of a pendulum,

45
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0.5
1 1
T = 2   ; T 2 = 4 2
g g
4 2
Thus, the slope of the graph gives, the value of
g
Hence, slope is 4 for the graph given.
Thus, the value of g is 9.87
Isup port
41. (d) We know that the time period of a physical pendulum is given by T = 2
mg cm

Isup port = mR + mR = 2mR ;

2 2 2
cm =R

T = 2
( 2mR ) = 2
2
2R T 2g
R= 2
mgR g 8
A seconds pendulum is a pendulum whose period is precisely two seconds; one second for a swing in one
22  2
= 0.5m
direction and one second for the return swing. i.e. T=2s, so we have R = 8
2

displacement x
42. (b) T = 2 = 2 = 2 / b
acceleration bx
1 k 1
43. (d) E = m A 2  E = kA 2
2 m 2
 E does not depend on m
44. (a) The two springs are in parallel.
 Effective spring constant,
k = k1 + k2
Now, frequency of oscillation is given by
When both k1 and k2 are made four times their original values, the new frequency is given by
1 k 1 k1 + k 2
f= or f = --------(i)
2 m 2 m
When both k1 and k2 are made four times their original values, the new frequency is given by
1 4k1 + 4k 2
f| =
2 m
1 4 ( k1 + k 2 )  1 k1 + k 2 
= = 2   = 2f
2 m  2 m 
k k
45. (a) T1 = and T2 =
g  d
g 1 − 
 R
2
T d T  d
So, 1 = 1 − =  1  = 1 −
T2 R  T2  R
 T   2

d = 1 −  1   R
  T2  
46. (d) If it gives correct time at equator, it will give correct time at poles also because the time period of spring
mass system is independent of g.
47. (a) A rectangular block of mass m and area of cross-section. A floats in a liquid of density r . If it is given
a small vertical displacement from equilibrium. It undergoes oscillations with a time period T then: T  m
48. (a) For the block is about to slip, g = 2 a

46
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g 1 g
= v=
a 2 a
49. (a) V(x)=K∣x∣3
 V   ML T 
2 −2

x  L
3 3

∴[K]= = =[ML−1T−2]
Now time period may depends on
T ∝(mass)x(amplitude)y(K)z ; [M0L0T]=[M]x[L]y[ML−1T−2]z
−1 −1
Equating powers we get, −2z = 1, z = , y − z = 0, y =
2 2
As T  ( amplitude )
z

−1
1
T  (a ) 2 ; T 
a
50. (c) vmax = aω ; v|max = 2a  = 2 vmax = 2v (since vmax = v)
2R
51. (b) T = 2 = 2
g g
52. (d) Oscillations along spring length are independent of gravitation.
53. (c) As F = -kx
54. (a) When lift is falling freely, the effective acceleration due to gravity inside the lift is zero i.e. g' = g – g =
0. Therefore time period will be infinity and frequency is zero.
55. (d) The second pendulum placed in a space laboratory orbiting around the earth is in a weightlessness state.
Hence g = 0 so T = 
56. (2) For a spring kℓ = constant
k1ℓ1 = k2ℓ2 = kℓ (1)
Spring is divided in two parts having lengths ℓ1 & ℓ2 (i.e. ℓ1 + ℓ2 = ℓ)
with spring constants k1 & k2 respectively.
as ℓ1 = nℓ2 & ℓ1 + ℓ2 = ℓ
hence nℓ2 + ℓ2 = ℓ
∴ℓ2(n + 1) = ℓ ∴ (ℓ / ℓ2) = n + 1 (2)
from (1) k2 = {kℓ/ℓ2} = k ∙ (ℓ/ℓ2)
∴ from (2), k2 = k(1 + n)
|
57. (a) T = 2 ; 2 = 2 = 2
g g ( g / 6)
Time period will remain constant if on moon,
|
= / 6 = 1/ 6m
58. (a)  = 0 (1 + ay ) , a = 2m −1 , T = ?
0 = 40 ; 0 = 0 (1 + ay )
40 = 0 (1 + 2y ) ; 40 = 0 + 20 y
F = −gV ; ma = −gV
V= volume of ball
40 .V.a = −20 Vgy
y 2
2a = −gy = − , g = 10
a g
y 2 1
=− =−
a 10 5

47
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displacement 1
T = 2  T = 2
acceleration 5 ; T = 2 sec
5
2
d x 2 2
59. (a) = −x = −2 x   =  or T = =
dt 2
 
60. (c) Damping is caused by opposing force, which decreases the frequency.

61. (d) F = – bV, b depends on all the three i.e., shape and size of the block and viscosity of the medium.
62. (b) Resonance can occur only if external force is applied at same frequency as that of natural frequency, hence it is forced
vibration.
63. (c) forced oscillations
64. (d) In case of a forced vibration, the resonance wave becames very sharp when the damping force is small.
65. (d) In damped oscillation, amplitude goes on decaying exponentially

66. (b) Vm = V0 / ( a2 − b + c )

If there is a single resonant frequency, then this equation should be satisfied for only one that particular
resonant frequency, hence a2 − b + c = 0 will have equal roots therefore;
D = 0  (–b)2 – 4ac = 0  b2 = 4ac
67. (d) This is a case of damped vibration as the amplitude of vibration is decreasing with time. Amplitude of
vibrations at any instant t is given by a = a0e-bt, where a0 is the initial amplitude of vibrations and b is the
damping constant. Now, when t = 100 T, a = a0/3 [T is time period] Let the amplitude be a' at t =200 T. i.e.
after completing 200 oscillations.

∴ The amplitude will be reduced to 1/9 of initial value.

68. (b) At resonance, amplitude of oscillation is maximum
 22 − 36 + 9 is minimum
 4 − 36 = 0 (derivative is zero)   = 9

48
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bt
69. (c) A = A 0 e (where, A0 = maximum amplitude)
2m

According to the questions, after 5 second,

b( 5)

0.9A 0 = A 0 e 2m -------------(i)
After 10 more second,
b(15)
−
A = A 0 e 2m --------------(ii)
From eqns (i) and (ii)
A = 0.729 A0  = 0.729
70. (d) When the pendulum is oscillating over a current carrying coil, and when the direction of oscillating
pendulum bob is opposite to the direction of current. Its instantaneous acceleration increases.

Hence time period T  2 and damping is larger than in air alone due energy dissipation.
g
NEET PREVIOUS YEARS QUESTIONS-EXPLANATIONS
1. (b) From question, acceleration, a = 20 m/s2, and displacement, y = 5m
a = 2 y  20 = 2 ( 5 )   = 2 rad / s
2 2
Time period of pendulum, T = = = s
 2
2. (b) Let be the complete length of the spring.
Length when cut in ratio, 1 : 2 : 3 are , , and
6 3 2
1
Spring constant (k) 
length ( )
Spring constant for given segments
k1 = 6k, k2 = 3k and k3 = 2k
When they are connected in series
1 1 1 1 1 6
|
= + +  |=
k 6k 3k 2k k 6k
 Force constant k' = k
And when they are connected in parallel ; k || = 6k + 3k + 2k  k || = 11k
k| 1
Then the ratios ; || = i.e., k | : k || = 1:11
k 11
3. (b) Given, Amplitude A = 3 cm
Which particle is at x = 2 cm
According to question, magnitude of velocity = acceleration
2 4 4
( 3) − ( 2 ) = 2   ;
2 2
 A 2 − x 2 = x2 ; 5= T=
 T  T 5
4. (b) The two displacements equations are y1 = a sin ( t )
 
and y2 = b cos ( t ) = b sin  t + 
 2

y eq = y1 + y 2 = a sin t + b cos t = a sin t + b sin  t + 
 2
Since the frequencies for both SHMs are same, resultant motion will be SHM.

Now A eq = a 2 + b 2 + 2ab cos
2

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5. (a) As we know, for particle undergoing SHM,

V =  A 2 − X 2 ; V12 = 2 ( A 2 − x12 ) , V22 = 2 ( A 2 − x 22 )
Subtracting we get,

6. (c) As, we know, in SHM

Maximum acceleration of the particle,  = A2
Maximum velocity,  = A
 2 2  2 
= ; T= =  = 
    T
7. (c) Displacement, x = A cos ( t ) (given)
dx
Velocity, v = = −A sin ( t )
dt
dv
Acceleration, a = = −A2 cos t
dt
Hence graph (c) correctly dipicts the variation of a with t.
8.

9. Displacement = zero in one complete oscillation

10.

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2h 1
= 
g g
11. Time of flight
1 1
= 2 
g g
Time period of pendulum
Ratio of time of flight & time period of pendulum is independent of g. Hence t = 2T
|

12. Option (3) is a combination of SHM of same  and same axis so its resultant is also a SHM which is
periodic.
13. Displacement (x) equation of SHM
y = A sin t
y
A

a = − 2 y = − 2 A sin t
y
2 A

− 2 A

Phase difference between displacement and acceleration is  radians

14. In each 1 oscillation P.E varies from min to maximum 2 times
15.
F=Kx
10 = K  5  510−2
K = 200
m 2 2 6.28
T = 2 = 2 = = = 0.628sec
K 200 10 10
1 g
16. Frequency n =
2 l
n l = constant
Shorter pendulum completes one oscillation more than longer pendulum
( n + 1) 100 = n 121
( n + 1)10 = n (11)
On solving, n = 10
 Shorter pendulum makes 11 oscillations

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