The period of a simple pendulum measured inside a stationary lift is ‘T’. If the lift starts moving upwards with a acceleration g / 3. What will be the time period

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T / 3

3 T

$\frac{\sqrt{3} T}{2}$

$\sqrt{\frac{3}{2}} T$

answer is C.

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Detailed Solution

$\begin{array}{l} T \propto \frac{1}{\sqrt{g}} \\ T_{1} \sqrt{g_{1}} = T_{2} \sqrt{g_{2}} \Rightarrow T \sqrt{g} = T_{2} \sqrt{g + \frac{g}{3}} \\ T \sqrt{g} = T_{2} \sqrt{\frac{4}{3}} g \Rightarrow T = T_{2} \sqrt{\frac{4}{3}} T_{2} = \frac{\sqrt{3} T}{2} \end{array}$

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