Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :

Simple Harmonic Motion PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q48 with Solution

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Correct answer: B

For a pendulum of length $l$ , when the lift is stationary, effective acceleration is just $g$ .

So,

T = 2\pi \sqrt{\frac{l}{g}}

If the lift moves upward with acceleration $a = \dfrac{g}{2}$ , the effective gravitational acceleration becomes

g_{\text{eff}} = g + a = g + \frac{g}{2} = \frac{3g}{2}

The time period in an effective gravity $g_{\text{eff}}$ is

T' = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{3g/2}}

Rewrite:

T' = 2\pi \sqrt{\frac{2l}{3g}}

Now use

T = 2\pi \sqrt{\frac{l}{g}}

Hence,

T' = \sqrt{\frac{2}{3}}\,T

T' = \sqrt{\frac{2}{3}}\,T

This matches Option B.