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(B) The tension $T$ in the string of a simple pendulum at any angle $	heta$ is given by the radial force balance equation:$T = mg \cos 	heta + \frac

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$mg\left[1+\left(\frac{A}{l}\right)^2\right]$

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(B) The tension $T$ in the string of a simple pendulum at any angle $	heta$ is given by the radial force balance equation:$T = mg \cos 	heta + \frac{mv^2}{l}$For small oscillations,$\cos 	heta \approx 1$ and the velocity $v$ is maximum at the mean position $(	heta = 0)$.Thus,the maximum tension $T_{\max}$ occurs at the mean position:$T_{\max} = mg + \frac{mv_{\max}^2}{l}$For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{l}}$.The maximum velocity in $S$.$H$.$M$. is $v_{\max} = A\omega$.Substituting $v_{\max}$ into the tension equation:$T_{\max} = mg + \frac{m(A\omega)^2}{l}$$T_{\max} = mg + \frac{m A^2 \omega^2}{l}$Since $\omega^2 = \frac{g}{l}$,we get:$T_{\max} = mg + \frac{m A^2 (g/l)}{l}$$T_{\max} = mg + \frac{m A^2 g}{l^2}$$T_{\max} = mg \left[1 + \left(\frac{A}{l}ight)^2ight]$

$A$ simple pendulum of length '$l$' and a bob of mass '$m$' is executing $S$.$H$.$M$. of small amplitude '$A$'. The maximum tension in the string will be ($g=$ acceleration due to gravity).

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