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JEE PHYSICS MCQ: OSCILLATIONS

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A particle moves so that its acceleration a is given by

a = - bx

where

x

is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

View answer and explanation

Correct option: C

\frac{{2\pi }}{{\sqrt b }}

Option C (

\frac{{2\pi }}{{\sqrt b }}

) is correct because it is the correct time period derived from the given acceleration equation and the SHM equation.

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a

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. At a displacement

{y_1}

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{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)