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A pendulum is undergoing SHM with frequency $f$. What is the frequency of its kinetic energy ?
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$2 f$
Let, $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}$ $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}$ Kinetic energy, $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2$
$\begin{aligned} & \Rightarrow \mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 \cos ^2 \omega \mathrm{t} \\ & \Rightarrow \mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2\left(\frac{1+\cos 2 \omega \mathrm{t}}{2}\right) \\ & \Rightarrow \mathrm{K}=\frac{1}{4} \mathrm{~m} \omega^2 \mathrm{~A}^2\left(1+\cos ^2 \omega \mathrm{t}\right) \\ & \therefore \omega_{\mathrm{K}}=2 \omega\end{aligned}$
$\therefore$ Frequency of oscillation of $\mathrm{K} . \mathrm{E}=2 \mathrm{f} \quad\left[\mathrm{f}=\frac{\omega}{2 \pi}\right]$
$\begin{aligned} & \Rightarrow \mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 \cos ^2 \omega \mathrm{t} \\ & \Rightarrow \mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2\left(\frac{1+\cos 2 \omega \mathrm{t}}{2}\right) \\ & \Rightarrow \mathrm{K}=\frac{1}{4} \mathrm{~m} \omega^2 \mathrm{~A}^2\left(1+\cos ^2 \omega \mathrm{t}\right) \\ & \therefore \omega_{\mathrm{K}}=2 \omega\end{aligned}$
$\therefore$ Frequency of oscillation of $\mathrm{K} . \mathrm{E}=2 \mathrm{f} \quad\left[\mathrm{f}=\frac{\omega}{2 \pi}\right]$
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