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JEE PHYSICS MCQ: OSCILLATIONS

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The acceleration

d^2 x / d t^2

of a particle varies with displacement

x

as

\frac{d^2 x}{d t^2}=-k x

where

k

is a constant of the motion. The time period

T

of the motion is equal to

View answer and explanation

Correct option: C

2 \pi / \sqrt{k}

Option C is correct because the time period T is calculated using

T = 2 \pi / \sqrt{k}

based on the given acceleration equation.

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