The time period of a simple pendulum on the surface of the earth is T. The height above the surface of the earth at which the time period of the pendulum becomes $2 \mathrm{~T}$ is (Radius of the earth $=6400 \mathrm{~km}$ )

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Aicharya · Accepted Answer

Time period of a simple pendulum is given by, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

$2 \mathrm{~T}=\frac{\mathrm{T}}{1+\frac{\mathrm{h}}{\mathrm{R}}} \Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=\frac{1}{2}$ $\begin{aligned} & \frac{\mathrm{h}}{\mathrm{R}}=-\frac{1}{2} \Rightarrow \mathrm{h}=-\frac{\mathrm{R}}{2}=\frac{6400}{2} \\ & \mathrm{~h}=3200 \mathrm{~km}\end{aligned}$

The time period of a simple pendulum on the surface of the earth is T. The height above the surface of the…

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