In This Post we are providing Chapter-14 OSCILLATIONS NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.
NCERT MOST IMPORTANT QUESTIONS ON OSCILLATIONS
1.The springs of spring factor k, 2k, k respectively are connected in parallel to a mass m. If the mass = 0.08kg m and k = 2N|m, then find the new time period?
Ans. Total spring constant, K1 = K1 + K2 + K3 (In parallel)
= K + 2K + K
= 4K
= 4 × 2 (k = 2 N | m)
= 8 N | m
Time period,
2.The bob of a vibrating simple pendulum is made of ice. How will the period of swing will change when the ice starts melting?
Ans .The period of swing of simple pendulum will remain unchanged till the location of centre of gravity of the bob left after melting of the ice remains at the fixed position from the point of suspension. If centre of gravity of ice bob after melting is raised upwards, then effective length of pendulum decreases and hence time period of swing decreases. Similarly, if centre of gravity shifts downward, time period increases.
3.An 8 kg body performs S.H.M. of amplitude 30 cm. The restoring force is 60N, when the displacement is 30cm. Find: – a) Time period b) the acceleration c) potential and kinetic energy when the displacement is 12cm?
Ans.Here m = 8 kg
m = Mass, a = amplitude
a = 30cm = 0.30m
a) f = 60 N, Y = displacement = 0.30m
K = spring constant
Since, F = Ky
K = 
= 
As, Angular velocity = w = 
Time period, T = 
b) Y = displacement = 0.12m
Acceleration, A = w2 y
A = (5)2 × 0.12
A = 3.0m |s2
P.E. = Potential energy = 
Kinetic energy = K.E = 
= 
Kinetic energy = K. E. = 7.56J
4.A particle executing SH.M has a maximum displacement of 4 cm and its acceleration at a distance of 1 cm from its mean position is 3 cm/s2. What will be its velocity when it is at a distance of 2cm from its mean position?
Ans.The acceleration of a particle executing S.H.M is –
A = w2 Y
w = Angular frequency ; Y = Displacement
A = Acceleration
Given A = 3cm / s2 ; Y = 1cm
So, 3 = w2 × 1
w = 
The velocity of a particle executing S.H.M is :-
a = amplitude
5.What is ratio of frequencies of the vertical oscillations when two springs of spring constant K are connected in series and then in parallel?
Ans .If two spring of spring constant K are connected in parallel, then effective resistance in parallel = KP = K + K = 2K
Let fP = frequency in parallel combination.
In Series combination, effective spring constant for 2 sprigs of spring constant K is :-
Let fS = frequency in series combination
Divide equation 2) by 1)
6. A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -α ¸, where J is the restoring couple and ¸ the angle of twist).
Ans. Mass of the circular disc, m = 10 kg
Radius of the disc, r = 15 cm = 0.15 m
The torsional oscillations of the disc has a time period, T = 1.5 s
The moment of inertia of the disc is:
I 
= 
= 0.1125 kg 
Time period, 
α is the torsional constant.
= 1.972 Nm/rad
Hence, the torsional spring constant of the wire is 1.972 Nm rad–1.
7. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
Ans.The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = g
Centripetal acceleration 
Where,
v is the uniform speed of the car
R is the radius of the track
Effective acceleration (
) is given as:
Time period, 
Where,l is the length of the pendulum
∴Time period, T 
9. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Ans.(b) and (c) are SHMs
(a) and (d) are periodic, but not SHMs
(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.
(b) An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.
(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.
(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.
10. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ( is any positive constant):
(a)
(b)
(c) 3 cos (π/4 – 2t)
(d) cos t + cos 3t + cos 5t
(e) exp
(f) 1 + t +
Ans.(a) SHM
The given function is:
This function represents SHM as it can be written in the form:
Its period is: 
(b) Periodic, but not SHM
The given function is:
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