Question

A simple pendulum has time period '$t$'. Its time period in a lift which is moving upwards with acceleration $3\,ms^{-2}$ is

• A. $t \sqrt{\frac{9.8}{12.8}}$
• B. $t \sqrt{\frac{12.8}{9.8}}$
• C. $t \sqrt{\frac{9.8}{6.8}}$
• D. $t \sqrt{\frac{6.8}{9.8}}$

Answer 1

Answer: (A)

$t \propto \frac{1}{\sqrt{9.8}}, t^{\prime} \propto \frac{1}{\sqrt{12.8}}$
$\left(\therefore g^{\prime}=9.8+3=12.8\right)$
$\therefore \frac{t^{\prime}}{t}=\sqrt{\frac{9.8}{12.8}}$
$\Rightarrow t^{\prime}=\sqrt{\frac{9.8}{12.8}} t$