A simple pendulum is placed inside a lift, which is moving with a uniform acceleration. If the time periods of the pendulum while the lift is moving upwards and downwards are in the ratio $1: 2$, then the acceleration of the lift is (Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ )

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Aicharya · Accepted Answer

Given, time period of the pendulum while the lift is moving upwards and downward are in the ratio, $ T_1: T_2=1: 2 $ Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ We know that, If the lift is moving upward, then total time-period,

when the lift is moving downwards, then the total time period,

By dividing Eq. (i) to (ii), we get $ \begin{array}{rlrl} \therefore & \frac{T_1}{T_2} & =\sqrt{\frac{g-a}{g+a}} \\ & \text { Now, } & \frac{1}{2} & =\sqrt{\frac{g-a}{g+a}} \end{array} $ Square on the both sides, we get $ \begin{aligned} & \text { or } & \left(\frac{1}{2}\right)^2 & =\frac{g-a}{g+a} \text { or } \frac{g-a}{g+a}=\frac{1}{4} \\ & \text { or } & 4 g-4 a & =g+a \text { or } 3 g=5 a \\ & \text { or } & a & =\frac{3 g}{5} \\ \Rightarrow & & a & =\frac{30}{5}=6 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $ So, the acceleration of the lift is $6 \mathrm{~m} / \mathrm{s}^2$

A simple pendulum is placed inside a lift, which is moving with a uniform acceleration. If the time periods…

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