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NEET Exam  >  NEET Test  >  Topic-wise MCQ Tests  >  31 Year NEET Previous Year Questions: Oscillations -2 - NEET MCQ

31 Year NEET Previous Questions: Oscillations -2 Free MCQ Practice Test


MCQ Practice Test & Solutions: 31 Year NEET Previous Year Questions: Oscillations -2 (30 Questions)

You can prepare effectively for NEET Topic-wise MCQ Tests for NEET with this dedicated MCQ Practice Test (available with solutions) on the important topic of "31 Year NEET Previous Year Questions: Oscillations -2". These 30 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Number of Questions: 30

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31 Year NEET Previous Year Questions: Oscillations -2 - Question 1
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A particle moving along the X-axis, executes simple harmonic motion then the force acting on it is given by [1988]

Detailed Solution: Question 1

For simple harmonic motion, F = – kx.
Here,  k = Ak.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 2
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A particle, with restoring force proportional to displacement and resistve force proportional to velocity is subjected to a force F sin ω0. If the amplitude of the particle is maximum for ω = ω1 and the energy of the particle is maximum for ω = ω2, then [1989,1998]

Detailed Solution: Question 2

At maximum energy of the particle, velocity resonance takes place, which occurs when frequency of external periodic force is equal to natural frequency of undamped vibrations, i.e. ω2 = ω0.
Further, amplitude resonance takes place at a frequency of external force which is less than the frequency of undamped natural vibrations, i.e. ω1 ≠  ω0 .

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31 Year NEET Previous Year Questions: Oscillations -2 - Question 3
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A mass m is suspen ded from a two coupled springs, connected in series. The force constant for springs are k1 and k2. The time period of the suspended mass will be [1990]

Detailed Solution: Question 3

The effective spring constant of two springs in series;

Time period,

31 Year NEET Previous Year Questions: Oscillations -2 - Question 4
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The composition of two simple h ar monic motions of equal periods at right angle to each other and with a phase difference of π results in the displacement of the particle along [1990]

Detailed Solution: Question 4

x = a sinωt and y = b sin (ωt + π) = – b sinωt

It is an equatation of a straight line.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 5
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The angular velocity and the amplitude of a simple pendulum is ω 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 [1991]

Detailed Solution: Question 5

PE ,

and K.E.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 6
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A body is executing S .H. M . When the displacements from the mean position are 4cm and 5 cm, the corresponding velocities of the body are 10 cm per sec and 8 cm per sec. Then the time period of the body is [1991]

Detailed Solution: Question 6

For S.H.M., Velocity,

at displacement x.

Putting this value in equation (2) mentioned above,

 radian / sec .

Time period

31 Year NEET Previous Year Questions: Oscillations -2 - Question 7
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A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by , where g is equal to [1991]

Detailed Solution: Question 7

The effective value of acceleration due to gravity is

31 Year NEET Previous Year Questions: Oscillations -2 - Question 8
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If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2.0 m/s2 at any time, the angular frequency of the oscillator is equal to [1992]

Detailed Solution: Question 8

ω2= 100 or ω = 10 rad/s

31 Year NEET Previous Year Questions: Oscillations -2 - Question 9
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A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from x = A to x = A/2 is [1992]

Detailed Solution: Question 9

For S.H.M.,

When x = A, 

When

or, 

Now, time taken to travel from x = A to x = A/2 is  (T/4 – T/12) = T/6

31 Year NEET Previous Year Questions: Oscillations -2 - Question 10
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A wave has S.H.M (Simple Harmonic Motion) whose period is 4 seconds while another wave which also possess SHM has its period 3 seconds. If both are combined, then the resultant wave will have the period equal to [1993]

Detailed Solution: Question 10

Beats are produced. Frequency of beats will be

 Hence time period = 12 s.

[Alt : Frequency of combined S.H.M

31 Year NEET Previous Year Questions: Oscillations -2 - Question 11
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A body executes S.H.M with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ? [1993]

Detailed Solution: Question 11

Total energy 

[Alt : The displacement at which the P.E. becomes n times the total energy is given by

where A is the amplitude.]

31 Year NEET Previous Year Questions: Oscillations -2 - Question 12
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Which one of the following is a simple harmonic motion? [1994]

Detailed Solution: Question 12

A wave moving through a string fixed at both ends, is a transverse wave formed as a result of simple harmonic motion of particles of the string.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 13
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A par ticle is subjected to two mutually perpendicular simple harmonic motions such that its x and y coordinates are given by   

 

x = 2 sin ωt;  

 The path of the particle will be [1994]

Detailed Solution: Question 13

As phase difference  the resultant path of particle is an ellipse.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 14
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In a simple harmonicmotion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic? [1996]

Detailed Solution: Question 14

Total energy of particle executing S.H.M. of amplitude (A).

K.E.of the particle

Clearly , 

[Alt : If the displacement be n times the amplitude, then 

31 Year NEET Previous Year Questions: Oscillations -2 - Question 15
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A linear harmonic oscillator of force constant 2 × 106 N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its [1996]

Detailed Solution: Question 15

 k = 2 × 106 N/m
Amplitude (A) = 0.01 m
Total mechanical energy = 160J
As we know that total mechanical energy is equal to total maximum potential energy.
Maximum K. E.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 16
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A particle starts simple harmonic motion from the mean position. Its amplitude is A and time period is T.What is its displacement when its speed is half of its maximum speed [1996]

Detailed Solution: Question 16

[Alt : The displacement at which the speed is n times the maximum speed is given by

31 Year NEET Previous Year Questions: Oscillations -2 - Question 17
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If the length of a simple pendulum is increased by 2%, then the time period [1997]

Detailed Solution: Question 17

We know that 

If length is increased by 2%., time period increases by 1%.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 18
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Two  simple harmonic motions with the same frequency act on a particle at right angles i.e., along x and y axis. If the two amplitudes are equal and the phase difference is ω/2, the resultant motion will be [1997]

Detailed Solution: Question 18

Equation of two simple harmonic motions

....(1)

....(2)

On squaring and adding equations  (1) and (2),
x2 + y2 = A2

This is an equation of a circle. Hence, resulting motion will be a circular motion.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 19
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A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will [1997]

Detailed Solution: Question 19

Time period of simple pendulum

where l is effective
length. [i.e distance between centre of suspension and centre of gravity of bob]

Initially, centre of gravity is at the centre of sphere. When water leaks the centre of gravity goes down until it is half filled; then it begins to go up and finally it again goes at the centre. That is effective length first increases and then decreases. As  , so time period first increases and then decreases.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 20
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Two simple pendulums of length 5m and 20m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed ....... oscillations [1998]

Detailed Solution: Question 20

Let the pendulums be in phase after t sec of start. Within this time, if the bigger pendulum executes n oscillations, the smaller one will have executed (n + 1) oscillations.
Now, the time of n oscillation =& the time of (n + 1) oscillation

To be in phase

or,      2n = n + 1
or,      n = 1
Hence, the no. of oscillations executed by shorter pendulum = n + 1 = 1 + 1 = 2

31 Year NEET Previous Year Questions: Oscillations -2 - Question 21
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A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4 m is suspended from the same spring? [1998]

Detailed Solution: Question 21

On putting the value of n  we get

31 Year NEET Previous Year Questions: Oscillations -2 - Question 22
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The time period of a simple pendulum is 2 seconds. If its length is increased by 4-times, then its period becomes [1999]

Detailed Solution: Question 22

If  is increased by 4 times, time period will increase by two times.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 23
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A particle executing S.H.M. has amplitude 0.01m and frequency 60 Hz. The maximum acceleration of the particle is [1999]

Detailed Solution: Question 23

Amplitude (A) = 0.01 m, Frequency = 60 Hz
Maximum acceleration = Aω2 = 0.01 x (2πn)2
= 0.01 x 4π2 x 60 x 60 = 144π2 m/sec2

31 Year NEET Previous Year Questions: Oscillations -2 - Question 24
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Masses MA and MB hanging from the ends of strings of lengths LA and LB are executing simple harmonic motions. If their frequencies are fA = 2fB, then [2000]

Detailed Solution: Question 24

regardless of mass.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 25
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Two simple harmonic motions act on a particle.
These harmonic motions are x = A cos (ωt + δ), y  the resulting motion is [2000]

Detailed Solution: Question 25

....(1)

When 

x = -A sin (ωt + α)            ....(2)
Squaring (1) and (2)
and then adding x2 + y2 = A2 [cos2 (ωt +α) + sin2 (ωt +α)] or x2 + y2 = A2, which is the equation of a circle.
The present motion is anticlockwise.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 26
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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 [2001]

Detailed Solution: Question 26

We know that the time period (T)

Since the negatively charged bob is attracted by the positively charged plate, therefore acceleration due to gravity will increase and time period will decrease.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 27
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There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force F = –kx. The total energy of body depends upon[2001]

Detailed Solution: Question 27

Total Energy of body perfor ming simple harmonic motion = where k = mω2

Hence energy depends upon amplitude and k (spring constant).

31 Year NEET Previous Year Questions: Oscillations -2 - Question 28
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A body of mass M, executes vertical SHM with periods t1 and t2, when separately attached to spring A and spring B respectively. The period of SHM, when the body executes SHM, as shown in the figure is t0. Then [2002] 

Detailed Solution: Question 28

⇒  k = Const.t-2

Here the springs are joined in parallel. So k0 = k1+k2 where k0 is resultant force constant

31 Year NEET Previous Year Questions: Oscillations -2 - Question 29
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The amplitude of a pendulum executing damped simple harmonic motion falls to 1/3 the original value after 100 oscillations. The amplitude falls to S times the original value after 200 oscillations, where S is [2002]

Detailed Solution: Question 29

In damped harmonic oscillator, amplitude falls exponentially.

After 100 oscillations amplitude falls to times.

∴ After next 100 oscillations i.e., after 200 oscillations amplitude falls to  times.

31 Year NEET Previous Year Questions: Oscillations -2 - Question 30
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A particle is executing a simple harmonic motion of amplitude a. Its potential energy is maximum when the displacement from the position of the maximum kinetic energy is [2002]

Detailed Solution: Question 30

P.E. of particle executing S.H.M. = 

At x = a, P.E. is maximum i.e.

At x = 0, K.E. is maximum.

Hence, displacement from position of maximum Kinetic energy = ±a.

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