Oscillations- 1 Practice Questions - DPP for NEET

 Page 1


DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice questions.
Each question has 4 choices (a), (b), (c) and (d), out of which ONLY
ONE choice is correct.
Q.1 A simple harmonic motion is represented by
() 10sin(20 0.5) Ftt =+ . The amplitude of the S.H.M. is
(a) 30 a = cm (b) 20 a = cm
(c) 10 a = cm (d) 5 a = cm
Q.2 A particle executes a simple harmonic motion of time
period T. Find the time taken by the particle to go directly
from its mean position to half the amplitude
(a)
/2 T
(b) /4 T (c) /8 T (d) /12 T
Q.3 The periodic time of a body executing simple harmonic
motion is 3 sec. After how much time from time 0 t = , its
displacement will be half of its amplitude
(a)
1
sec
8
(b)
1
sec
6
(c)
1
sec
4
(d)
1
sec
3
Q.4 If 
sin
6
xat
p æö
= w+
ç÷
èø
 and ' cos x at =w , then what is the
phase difference between the two waves?
(a) /3 p (b)
/6 p
(c) /2 p (d) p
Q.5 A body is executing S.H.M. when its displacement from
the mean position is 4 cm and 5 cm, the corresponding
velocity of the body is 10 cm/sec and 8 cm/sec. Then the
time period of the body is
(a) 2 sec p (b) / 2sec p
(c) sec p (d) 3 / 2sec p
Page 2


DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice questions.
Each question has 4 choices (a), (b), (c) and (d), out of which ONLY
ONE choice is correct.
Q.1 A simple harmonic motion is represented by
() 10sin(20 0.5) Ftt =+ . The amplitude of the S.H.M. is
(a) 30 a = cm (b) 20 a = cm
(c) 10 a = cm (d) 5 a = cm
Q.2 A particle executes a simple harmonic motion of time
period T. Find the time taken by the particle to go directly
from its mean position to half the amplitude
(a)
/2 T
(b) /4 T (c) /8 T (d) /12 T
Q.3 The periodic time of a body executing simple harmonic
motion is 3 sec. After how much time from time 0 t = , its
displacement will be half of its amplitude
(a)
1
sec
8
(b)
1
sec
6
(c)
1
sec
4
(d)
1
sec
3
Q.4 If 
sin
6
xat
p æö
= w+
ç÷
èø
 and ' cos x at =w , then what is the
phase difference between the two waves?
(a) /3 p (b)
/6 p
(c) /2 p (d) p
Q.5 A body is executing S.H.M. when its displacement from
the mean position is 4 cm and 5 cm, the corresponding
velocity of the body is 10 cm/sec and 8 cm/sec. Then the
time period of the body is
(a) 2 sec p (b) / 2sec p
(c) sec p (d) 3 / 2sec p
2
DPP/ P 27
Q.6 If a simple pendulum oscillates with an amplitude of 50
mm and time period of 2 sec, then its maximum velocity
is
(a) 0.10 m/s (b) 0.15m/s
(c) 0.8 m/s (d) 0.26 m/s
Q.7 The maximum velocity and the maximum acceleration of
a body moving in a simple harmonic oscillator are 2 m/s
and  4 m/s
2
. Then angular velocity will be
(a) 3 rad/sec (b) 0.5 rad/sec
(c) 1 rad/sec (d) 2 rad/sec
Q.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) 1cm (d) 2 cm
Q.9 The amplitude of a particle executing S.H.M. with
frequency of 60 Hz is 0.01 m. The maximum value of the
acceleration of the particle is
(a)
22
144 m/sec p (b)
2
144m/sec
(c)
2
2
144
m/sec
p
(d)
22
288 m/sec p
Q.10A particle executes simple harmonic motion with an
angular velocity and maximum acceleration of 3.5rad/sec
and 7.5 
2
m/s respectively. The amplitude of oscillation
is
(a) 0.28 m (b) 0.36 m (c) 0.53 m (d) 0.61 m
Q.11 What is the maximum acceleration of the particle doing
the SHM 
2sin
2
t
y
péù
= +f
êú
ëû
where y is in cm?
(a)
2
cm/s
2
p
(b)
2
2
cm/s
2
p
(c)
2
cm/s
4
p
(d)
2
cm/s
4
p
Q.12The total energy of a particle executing S.H.M. is
proportional to
(a) Displacement from equilibrium position
(b) Frequency of oscillation
(c) Velocity in equilibrium position
(d) Square of amplitude of motion
Q.13 When the displacement is half the amplitude, the ratio of
potential energy to the total energy is
(a)
1
2
(b)
1
4
(c)1 (d)
1
8
Q.14A particle is executing simple harmonic motion with
frequency f . The frequency at which its kinetic energy
changes into potential energy is
(a) /2 f (b) f (c) 2 f (d) 4 f
Q.15A particle executes simple harmonic motion with a
frequency f . The frequency with which its kinetic energy
oscillates is
(a) /2 f (b) f (c) 2f (d) 4f
Q.16 The kinetic energy of a particle executing S.H.M. is 16 J
when it is in its mean position. If the amplitude of
oscillations is 25 cm and the mass of the particle is 5.12
kg, the time period of its oscillation is
(a)
sec
5
p
(b)
2 sec p
(c)
20 sec p
(d)
5 sec p
Q.17 The displacement x (in metres) of a particle performing
simple harmonic motion is related to time t (in seconds) as
0.05cos 4
4
xt
p æö
= p+
ç÷
èø
. The frequency of the motion will
be
(a) 0.5 Hz (b) 1.0 Hz (c) 1.5 Hz (d) 2.0 Hz
Q.18A particle executes simple harmonic motion
[amplitude = A ] between xA =- and xA =+ . The time
taken for it to go from 0 to /2 A is 
1
T and to go from
/2 A to A is 
2
T . Then
(a)
12
TT < (b)
12
TT > (c)
12
TT = (d)
12
2 TT =
Page 3


DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice questions.
Each question has 4 choices (a), (b), (c) and (d), out of which ONLY
ONE choice is correct.
Q.1 A simple harmonic motion is represented by
() 10sin(20 0.5) Ftt =+ . The amplitude of the S.H.M. is
(a) 30 a = cm (b) 20 a = cm
(c) 10 a = cm (d) 5 a = cm
Q.2 A particle executes a simple harmonic motion of time
period T. Find the time taken by the particle to go directly
from its mean position to half the amplitude
(a)
/2 T
(b) /4 T (c) /8 T (d) /12 T
Q.3 The periodic time of a body executing simple harmonic
motion is 3 sec. After how much time from time 0 t = , its
displacement will be half of its amplitude
(a)
1
sec
8
(b)
1
sec
6
(c)
1
sec
4
(d)
1
sec
3
Q.4 If 
sin
6
xat
p æö
= w+
ç÷
èø
 and ' cos x at =w , then what is the
phase difference between the two waves?
(a) /3 p (b)
/6 p
(c) /2 p (d) p
Q.5 A body is executing S.H.M. when its displacement from
the mean position is 4 cm and 5 cm, the corresponding
velocity of the body is 10 cm/sec and 8 cm/sec. Then the
time period of the body is
(a) 2 sec p (b) / 2sec p
(c) sec p (d) 3 / 2sec p
2
DPP/ P 27
Q.6 If a simple pendulum oscillates with an amplitude of 50
mm and time period of 2 sec, then its maximum velocity
is
(a) 0.10 m/s (b) 0.15m/s
(c) 0.8 m/s (d) 0.26 m/s
Q.7 The maximum velocity and the maximum acceleration of
a body moving in a simple harmonic oscillator are 2 m/s
and  4 m/s
2
. Then angular velocity will be
(a) 3 rad/sec (b) 0.5 rad/sec
(c) 1 rad/sec (d) 2 rad/sec
Q.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) 1cm (d) 2 cm
Q.9 The amplitude of a particle executing S.H.M. with
frequency of 60 Hz is 0.01 m. The maximum value of the
acceleration of the particle is
(a)
22
144 m/sec p (b)
2
144m/sec
(c)
2
2
144
m/sec
p
(d)
22
288 m/sec p
Q.10A particle executes simple harmonic motion with an
angular velocity and maximum acceleration of 3.5rad/sec
and 7.5 
2
m/s respectively. The amplitude of oscillation
is
(a) 0.28 m (b) 0.36 m (c) 0.53 m (d) 0.61 m
Q.11 What is the maximum acceleration of the particle doing
the SHM 
2sin
2
t
y
péù
= +f
êú
ëû
where y is in cm?
(a)
2
cm/s
2
p
(b)
2
2
cm/s
2
p
(c)
2
cm/s
4
p
(d)
2
cm/s
4
p
Q.12The total energy of a particle executing S.H.M. is
proportional to
(a) Displacement from equilibrium position
(b) Frequency of oscillation
(c) Velocity in equilibrium position
(d) Square of amplitude of motion
Q.13 When the displacement is half the amplitude, the ratio of
potential energy to the total energy is
(a)
1
2
(b)
1
4
(c)1 (d)
1
8
Q.14A particle is executing simple harmonic motion with
frequency f . The frequency at which its kinetic energy
changes into potential energy is
(a) /2 f (b) f (c) 2 f (d) 4 f
Q.15A particle executes simple harmonic motion with a
frequency f . The frequency with which its kinetic energy
oscillates is
(a) /2 f (b) f (c) 2f (d) 4f
Q.16 The kinetic energy of a particle executing S.H.M. is 16 J
when it is in its mean position. If the amplitude of
oscillations is 25 cm and the mass of the particle is 5.12
kg, the time period of its oscillation is
(a)
sec
5
p
(b)
2 sec p
(c)
20 sec p
(d)
5 sec p
Q.17 The displacement x (in metres) of a particle performing
simple harmonic motion is related to time t (in seconds) as
0.05cos 4
4
xt
p æö
= p+
ç÷
èø
. The frequency of the motion will
be
(a) 0.5 Hz (b) 1.0 Hz (c) 1.5 Hz (d) 2.0 Hz
Q.18A particle executes simple harmonic motion
[amplitude = A ] between xA =- and xA =+ . The time
taken for it to go from 0 to /2 A is 
1
T and to go from
/2 A to A is 
2
T . Then
(a)
12
TT < (b)
12
TT > (c)
12
TT = (d)
12
2 TT =
DPP/ P 27
3
Q.19A cylindrical piston of mass M slides smoothly inside a
long cylinder closed at one end, enclosing a certain mass
of gas. The cylinder is kept with its axis horizontal. If the
piston is disturbed from its equilibrium position, it
oscillates simple harmonically. The period of oscillation
will be
Gas
P
A
M
h
(a)
2
Mh
T
PA
æö
=p
ç÷
èø
(b)
2
MA
T
Ph
æö
=p
ç÷
èø
(c) 2
M
T
PAh
æö
=p
ç÷
èø
(d) 2 T MPhA =p
Q.20A particle is performing simple harmonic motion along
x-axis with amplitude 4 cm and time period 1.2 sec. The
minimum time taken by the particle to move from x = 2
cm to x = + 4 cm and back again is given by
(a) 0.6 sec (b) 0.4 sec
(c) 0.3 sec (d) 0.2 sec
Q.21A spring of force constant k is cut into two pieces such
that one piece is double the length of the other. Then the
long piece will have a force constant of
(a) (2/3)k (b) (3/2)k
(c) 3k (d) 6k
Q.22A simple pendulum has time period T
1
. The point of
suspension is now moved upward according to equation
y = kt
2
 where k =1m/sec
2
. If new time period is T
2
 then
ratio 
2
1
2
2
T
T
 will be
(a) 2/3 (b) 5/6
(c) 6/5 (d) 3/2
DIRECTIONS (Q.23-Q.25) : In the following questions, more than
one of the answers  given are correct. Select the correct
answers and mark it according to the following codes:
Codes : (a) 1, 2 and 3 are correct
(b) 1 and 2 are correct
(c) 2 and 4 are correct
(d) 1 and 3 are correct
Q.23A particle constrained to move along the x-axis in a
potential V = kx
2
, is subjected to an external time dependent
force f(t)
r
,  here k is a constant, x the distance from the
origin, and t the time. At some time T, when the particle
has zero velocity at x = 0, the external force is removed.
Choose the incorrect options –
(1) Particle executes SHM
(2) Particle moves along +x direction
(3) Particle moves along – x direction
(4) Particle remains at rest
Q.24Three simple harmonic motions in the same direction
having the same amplitude a and same period are
superposed. If each differs in phase from the next by 45°,
then –
(1) The resultant amplitude is (1 2) + a
(2) The phase of the resultant motion relative to the first is
90°
(3) The energy associated with the resulting motion is
(3 2 2) + times the energy associated with any single
motion
(4) The resulting motion is not simple harmonic
Q.25For a particle executing simple harmonic motion, which
of the following statements is correct?
(1) The total energy of the particle always remains the same
(2) The restoring force always directed towards a fixed
point
(3) The restoring force is maximum at the extreme
positions
(4) The acceleration of the particle is maximum at the
equilibrium position
Page 4


DIRECTIONS (Q.1-Q.22) : There are 22 multiple choice questions.
Each question has 4 choices (a), (b), (c) and (d), out of which ONLY
ONE choice is correct.
Q.1 A simple harmonic motion is represented by
() 10sin(20 0.5) Ftt =+ . The amplitude of the S.H.M. is
(a) 30 a = cm (b) 20 a = cm
(c) 10 a = cm (d) 5 a = cm
Q.2 A particle executes a simple harmonic motion of time
period T. Find the time taken by the particle to go directly
from its mean position to half the amplitude
(a)
/2 T
(b) /4 T (c) /8 T (d) /12 T
Q.3 The periodic time of a body executing simple harmonic
motion is 3 sec. After how much time from time 0 t = , its
displacement will be half of its amplitude
(a)
1
sec
8
(b)
1
sec
6
(c)
1
sec
4
(d)
1
sec
3
Q.4 If 
sin
6
xat
p æö
= w+
ç÷
èø
 and ' cos x at =w , then what is the
phase difference between the two waves?
(a) /3 p (b)
/6 p
(c) /2 p (d) p
Q.5 A body is executing S.H.M. when its displacement from
the mean position is 4 cm and 5 cm, the corresponding
velocity of the body is 10 cm/sec and 8 cm/sec. Then the
time period of the body is
(a) 2 sec p (b) / 2sec p
(c) sec p (d) 3 / 2sec p
2
DPP/ P 27
Q.6 If a simple pendulum oscillates with an amplitude of 50
mm and time period of 2 sec, then its maximum velocity
is
(a) 0.10 m/s (b) 0.15m/s
(c) 0.8 m/s (d) 0.26 m/s
Q.7 The maximum velocity and the maximum acceleration of
a body moving in a simple harmonic oscillator are 2 m/s
and  4 m/s
2
. Then angular velocity will be
(a) 3 rad/sec (b) 0.5 rad/sec
(c) 1 rad/sec (d) 2 rad/sec
Q.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) 1cm (d) 2 cm
Q.9 The amplitude of a particle executing S.H.M. with
frequency of 60 Hz is 0.01 m. The maximum value of the
acceleration of the particle is
(a)
22
144 m/sec p (b)
2
144m/sec
(c)
2
2
144
m/sec
p
(d)
22
288 m/sec p
Q.10A particle executes simple harmonic motion with an
angular velocity and maximum acceleration of 3.5rad/sec
and 7.5 
2
m/s respectively. The amplitude of oscillation
is
(a) 0.28 m (b) 0.36 m (c) 0.53 m (d) 0.61 m
Q.11 What is the maximum acceleration of the particle doing
the SHM 
2sin
2
t
y
péù
= +f
êú
ëû
where y is in cm?
(a)
2
cm/s
2
p
(b)
2
2
cm/s
2
p
(c)
2
cm/s
4
p
(d)
2
cm/s
4
p
Q.12The total energy of a particle executing S.H.M. is
proportional to
(a) Displacement from equilibrium position
(b) Frequency of oscillation
(c) Velocity in equilibrium position
(d) Square of amplitude of motion
Q.13 When the displacement is half the amplitude, the ratio of
potential energy to the total energy is
(a)
1
2
(b)
1
4
(c)1 (d)
1
8
Q.14A particle is executing simple harmonic motion with
frequency f . The frequency at which its kinetic energy
changes into potential energy is
(a) /2 f (b) f (c) 2 f (d) 4 f
Q.15A particle executes simple harmonic motion with a
frequency f . The frequency with which its kinetic energy
oscillates is
(a) /2 f (b) f (c) 2f (d) 4f
Q.16 The kinetic energy of a particle executing S.H.M. is 16 J
when it is in its mean position. If the amplitude of
oscillations is 25 cm and the mass of the particle is 5.12
kg, the time period of its oscillation is
(a)
sec
5
p
(b)
2 sec p
(c)
20 sec p
(d)
5 sec p
Q.17 The displacement x (in metres) of a particle performing
simple harmonic motion is related to time t (in seconds) as
0.05cos 4
4
xt
p æö
= p+
ç÷
èø
. The frequency of the motion will
be
(a) 0.5 Hz (b) 1.0 Hz (c) 1.5 Hz (d) 2.0 Hz
Q.18A particle executes simple harmonic motion
[amplitude = A ] between xA =- and xA =+ . The time
taken for it to go from 0 to /2 A is 
1
T and to go from
/2 A to A is 
2
T . Then
(a)
12
TT < (b)
12
TT > (c)
12
TT = (d)
12
2 TT =
DPP/ P 27
3
Q.19A cylindrical piston of mass M slides smoothly inside a
long cylinder closed at one end, enclosing a certain mass
of gas. The cylinder is kept with its axis horizontal. If the
piston is disturbed from its equilibrium position, it
oscillates simple harmonically. The period of oscillation
will be
Gas
P
A
M
h
(a)
2
Mh
T
PA
æö
=p
ç÷
èø
(b)
2
MA
T
Ph
æö
=p
ç÷
èø
(c) 2
M
T
PAh
æö
=p
ç÷
èø
(d) 2 T MPhA =p
Q.20A particle is performing simple harmonic motion along
x-axis with amplitude 4 cm and time period 1.2 sec. The
minimum time taken by the particle to move from x = 2
cm to x = + 4 cm and back again is given by
(a) 0.6 sec (b) 0.4 sec
(c) 0.3 sec (d) 0.2 sec
Q.21A spring of force constant k is cut into two pieces such
that one piece is double the length of the other. Then the
long piece will have a force constant of
(a) (2/3)k (b) (3/2)k
(c) 3k (d) 6k
Q.22A simple pendulum has time period T
1
. The point of
suspension is now moved upward according to equation
y = kt
2
 where k =1m/sec
2
. If new time period is T
2
 then
ratio 
2
1
2
2
T
T
 will be
(a) 2/3 (b) 5/6
(c) 6/5 (d) 3/2
DIRECTIONS (Q.23-Q.25) : In the following questions, more than
one of the answers  given are correct. Select the correct
answers and mark it according to the following codes:
Codes : (a) 1, 2 and 3 are correct
(b) 1 and 2 are correct
(c) 2 and 4 are correct
(d) 1 and 3 are correct
Q.23A particle constrained to move along the x-axis in a
potential V = kx
2
, is subjected to an external time dependent
force f(t)
r
,  here k is a constant, x the distance from the
origin, and t the time. At some time T, when the particle
has zero velocity at x = 0, the external force is removed.
Choose the incorrect options –
(1) Particle executes SHM
(2) Particle moves along +x direction
(3) Particle moves along – x direction
(4) Particle remains at rest
Q.24Three simple harmonic motions in the same direction
having the same amplitude a and same period are
superposed. If each differs in phase from the next by 45°,
then –
(1) The resultant amplitude is (1 2) + a
(2) The phase of the resultant motion relative to the first is
90°
(3) The energy associated with the resulting motion is
(3 2 2) + times the energy associated with any single
motion
(4) The resulting motion is not simple harmonic
Q.25For a particle executing simple harmonic motion, which
of the following statements is correct?
(1) The total energy of the particle always remains the same
(2) The restoring force always directed towards a fixed
point
(3) The restoring force is maximum at the extreme
positions
(4) The acceleration of the particle is maximum at the
equilibrium position
4
DPP/ P 27
DIRECTIONS (Q.26-Q.27) : Read the passage given below and
answer the questions that follows :
The differential equation of a particle undergoing SHM is given
by a
2
2
dx
bx0
dt
+= . The particle starts from the extreme position.
Q.26 The ratio of the maximum acceleration to the maximum
velocity of the particle is –
(a)
b
a
(b)
a
b
(c)
a
b
(d)
b
a
Q.27 The equation of motion may be given by :
(a) x = Asin 
b
t
a
æö
ç÷
èø
(b) x = Acos 
b
t
a
æö
ç÷
èø
(c) x = A sin 
b
t
a
æö
+q
ç÷
èø
 where q ¹ p/2
(d) None of these
DIRECTIONS (Q.28-Q.30) : Each of these questions contains two
statements: Statement-1 (Assertion) and Statement-2 (Reason). Each
of these questions has four alternative choices, only one of which is
the correct answer. You have to select the correct choice.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for  Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is
NOT a correct explanation for Statement-1.
(c) Statement -1 is False, Statement-2 is True.
(d) Statement -1 is True, Statement-2 is False.
Q.28Statement-1 : In S.H.M., the motion is ‘to and fro’ and
periodic.
Statement-2 : Velocity of the particle
22
()=w- v kx (where x is the displacement and k is
amplitude)
Q.29 Statement-1 : In simple harmonic motion, the velocity is
maximum when acceleration is minimum.
Statement-2 : Displacement and velocity of S.H.M. differ
in phase by 
/2 p
Q.30Statement-1 : The graph of total energy of a particle in
SHM w.r.t., position is a straight line with zero slope.
Statement-2 : Total energy of particle in SHM remains
constant throughout its motion.