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JEE PHYSICS MCQ: OSCILLATIONS

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A simple harmonic oscillator has an amplitude of

0.5 \mathrm{~m}

and a time period of 2 seconds. What is the magnitude of acceleration when it is displaced from the mean position by

0.25 \mathrm{~m}

?

View answer and explanation

Correct option: C

\frac{\pi^2}{4} \mathrm{~m} / \mathrm{s}^2

Option C (

\frac{\pi^2}{4} \mathrm{~m} / \mathrm{s}^2

) is correct because the acceleration in SHM is proportional to the displacement and can be calculated using the angular frequency and displacement.

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Assertion: A particle executing simple harmonic motion comes to rest at the extreme positions.

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x

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y

-axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute A particle is executing SHM with amplitude

a

. When the PE of a particle is one-fourth of its maximum value during the oscillation, its displacement from the equilibrium position will be The displacement, velocity and acceleration in a simple harmonic motion are related as the Assertion: Angular velocity of earth around sun is lesser than the angular velocity of moon about earth.
Reason: Time taken by moon revolve around earth is less than time taken by earth to revolve around sun. A particle is executing

SHM

. At a displacement

{y_1}

its potential energy is

{U_1}

and at a displacement

{y_2}

its potential energy is

{U_2}

. The potential energy of the particle at displacement

\left( {{y_1} + {y_2}} \right)