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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A man measures the period of simple pendulum inside a stationary lift and finds it to be

\mathrm{T}

second. If the lift accelerates downwards with acceleration of

\mathrm{g} / 4

, the period of oscillation will be-

View answer and explanation

Correct option: B

\mathrm{T} \times \frac{2}{\sqrt{3}} \mathrm{~s}

Option B (T

\times \frac{2}{\sqrt{3}}

s) is correct because the time period increases when the lift accelerates downwards, reducing the effective gravity.

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