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  4. A man measures the period of a simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates upwards with an acceleration g / 4, then the period of pendulum will be

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Q. A man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ second. If the lift accelerates upwards with an acceleration $g / 4$, then the period of pendulum will be

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Solution:

As, $T=2 \pi \sqrt{\frac{l}{g}}$. When lift is accelerated upwards with acceleration a $(=g / 4)$, then effective acceleration due to gravity inside the lift
$ g_{1}=g+a=g+\frac{g}{4}=\frac{5 g}{4} $
$\therefore T_{1}=2 \pi \sqrt{\frac{l}{5 g / 4}}=2 \pi \sqrt{\frac{l}{g}} \times \frac{2}{\sqrt{5}}=\frac{2 T}{\sqrt{5}}$