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Oscillations NEET Questions — Free MCQ Practice

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A block having mass M is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant k. The other end of the spring is attached to a rigid wall. This system consisting of spring and mass M is executing SHM with amplitude A and frequency f. When the block is passing through the mid-point of its path of motion, a body of mass mis placed on top of it, as a result of which its amplitude and frequency changes to Aâ€™ and fâ€™. If the velocity before putting the mass and after putting it is $ \nu $ and $ \nu ' $ respectively, then $ { \nu' \over \nu } $=...................................

A. $ \left ( { M \over m+M } \right) $ B. $ \left ( { M + m \over M } \right) $ C. $ \left( { M+m \over M - m } \right) { A' \over A } $ D. $ \left( { M- m \over M + m } \right) { A' \over A } $

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Explanation

According to the law of conservation of momentum, $ MV = (M + m) \nu ' $

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A block having mass M is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant k. The other end of the spring is attached to a rigid wall. This system consisting of spring and mass M is executing SHM with amplitude A and frequency f. When the block is passing through the mid-point of its path of motion, a body of mass mis placed on top of it, as a result of which its amplitude and frequency changes to Aâ€™ and fâ€™. The ratio of amplitudes $ { A^1 \over A } $ = ........

A. $ \sqrt { \left( M + m \over m \right) } $ B. $ \sqrt { \left( m \over M + m \right) } $ C. $ \sqrt { \left( M \over M+m \right) } $ D. $ \sqrt { \left( M+m \over M \right) } $

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Explanation

According to the law of conservation of energy, Kinetic Energy at mid point = potential Energy at the end points $ \therefore { 1 \over 2 } M \nu^2 = { 1 \over 2 } k A^2 $ $ And { 1 \over 2} ( M + m ) \nu ^{1 ^2 } = {1 \over 2 } k A^{1 ^2 } $

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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The motion of the particle is â€¦â€¦â€¦â€¦..

A. Damped motion B. Periodic motion C. Rotational motion D. S.H.M.

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Explanation

$ y = 3 cos 2t + 4 sin 2 t $ $ \therefore A sin \phi = 3 and A cos \phi = 4 $ $ \therefore y = A sin \phi cos 2 t + A cos \phi sin 2 t $ $ \therefore y = A sin ( \omega t + \phi ) $ Which shows that the motion is simple harmonic motion

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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t The periodic time of oscillation is â€¦â€¦â€¦

A. 2s B. ps C. (/2)s D. 2 ps

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Explanation

$ T = { 2 \pi \over \omega } = {2 \pi \over 2 } = \pi s $

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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The amplitude of oscillation is â€¦â€¦cm

A. 1 B. 3 C. 5 D. 7

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Explanation

$ amplitude A = \sqrt { 3^2 + 4^2 } = 5 cm $

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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The maximum acceleration of the particle isâ€¦â€¦â€¦â€¦cm / s2.

A. 4 B. 12 C. 20 D. 28

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Explanation

Maximum Accelaration of $ A particle = A \omega^2 = 5 (2)^2 = 20 cms^{-2} $

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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . If the mass of the particle is 5 gm, then the total energy of the particle is â€¦â€¦erg.

A. 250 B. 125 C. 500 D. 375

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Explanation

$ Mechanical energy = {1 \over 2 } m \omega ^2 A^2 = 250 erg $

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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The frequency of the particle is â€¦â€¦â€¦s- 1 .

A. (1/ p) B. p C. (1 /2p ) D. ( p /2)

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Explanation

$ Frequenct of the particle f = {1 \over T } = { 1 \over \pi } s^{-1} $

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The function $Sin^2(Ã¹t )$ representsâ€¦â€¦

A. ASHM with periodic time p/Ã¹. B. ASHM with a periodic time 2p/Ã¹. C. A periodic motion with periodic time p/Ã¹ D. A periodic motion with period 2p/Ã¹.

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Explanation

$ y = sin^2 \omega t = {1 - cos 2 \omega t \over 2 } = {1 \over 2 } - {1 \over 2 } cos 2 \omega t .....(1) $ $ \therefore \nu = { 1 \over 2 } 2 \omega sin ( 2 \omega t ) = \omega sin 2 \omega t $ $ \therefore a = 2 \omega^2 cos 2 \omega t $ $ = 2 \times 2 \omega^2 \left( {1 \over 2} - Y \right) \{ From eqn (1) \} $ $ = - 4 \omega^2 \left( { 1 \over 2} - y \right) $ $ \therefore a \alpha - y \{ \therefore SHM \} $ $ Now , { 2 \pi \over T} = 2 \omega \Rightarrow T = { \pi \over \omega } $

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A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them is:

A.

$\frac{π}{3}$

B.

$\frac{2 π}{3}$

C.

$\frac{π}{6}$

D.

$\frac{π}{2}$

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Explanation

2.

Resultant amplitude

$\begin{aligned} R = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos ϕ} \\ a^{2} = a^{2} + a^{2} + 2 a^{2} \cos ϕ \\ \cos ϕ = - \frac{1}{2} \Rightarrow ϕ = \frac{2 π}{3} \end{aligned}$

Oscillations MCQs for NEET — Physics Questions with Answers

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