A simple pendulum with length $100\,cm$ and bob of mass $250\,g$ is executing S.H.M. of amplitude $10\,cm$. The maximum tension in the string is found to be $\frac{x}{40}\,N$. The value of $x$ is $..........$.
JEE MAIN 2023, Diffcult
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$\sin \theta_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}$
From conservation of energy
$\frac{1}{2} m v^2=m g l(1-\cos \theta)$
Maximum tension occurs at mean position.
$\therefore T-m g=\frac{m v^2}{l}$
$\Rightarrow T=m g+\frac{m v^2}{l}$
$\therefore T=m g+2 m g(1-\cos \theta)$
$=m g\left[1+2\left(1-\sqrt{1-\sin ^2 \theta}\right)\right]$
$=m g\left[3-2 \sqrt{1-\frac{1}{100}}\right]$
$=\frac{250}{1000} \times 9.8\left[3-2\left(1-\frac{1}{200}\right)\right]=\frac{99}{40}$
$\therefore x=99$
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