A simple pendulum of length ' ell ' has a bob | vedclass

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Question

(D) For a simple pendulum executing $S$.$H$.$M$. with small amplitude $A$,the tension $T$ in the string at any angle $	heta$ is given by $T = mg \cos

Vedclass · Accepted Answer

$mg\left(\frac{A^{2}}{\ell^{2}}+1\right)$

Vedclass · Answer

(D) For a simple pendulum executing $S$.$H$.$M$. with small amplitude $A$,the tension $T$ in the string at any angle $	heta$ is given by $T = mg \cos 	heta + \frac{mv^2}{\ell}$.At the mean position,the velocity $v$ is maximum,and $	heta = 0$,so $\cos 	heta = 1$. Thus,the maximum tension is $T_{\max} = mg + \frac{mv_{\max}^2}{\ell}$.In $S$.$H$.$M$.,the velocity is given by $v = A\omega \cos(\omega t)$,where $\omega = \sqrt{\frac{g}{\ell}}$.The maximum velocity is $v_{\max} = A\omega = A\sqrt{\frac{g}{\ell}}$.Therefore,$v_{\max}^2 = A^2 \frac{g}{\ell}$.Substituting this into the expression for $T_{\max}$:$T_{\max} = mg + \frac{m}{\ell} \left( A^2 \frac{g}{\ell} ight) = mg + mg \frac{A^2}{\ell^2} = mg \left( 1 + \frac{A^2}{\ell^2} ight)$.

$A$ simple pendulum of length ' $\ell$ ' has a bob of mass 'm'. It executes $S$.$H$.$M$. of small amplitude '$A$'. The maximum tension in the string is ($g=$ acceleration due to gravity).

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