A simple pendulum oscillates with an angular | vedclass

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Question

(A) Let $m$ be the mass of the bob and $l$ be the length of the string.The minimum tension $(T_{min})$ occurs at the extreme position where the veloci

Vedclass · Accepted Answer

$\cos^{-1}(0.75)$

Vedclass · Answer

(A) Let $m$ be the mass of the bob and $l$ be the length of the string.The minimum tension $(T_{min})$ occurs at the extreme position where the velocity is zero: $T_{min} = mg \cos 	heta$.The maximum tension $(T_{max})$ occurs at the lowest point (mean position). Using the conservation of energy, the velocity $v$ at the lowest point is given by $\frac{1}{2}mv^2 = mgl(1 - \cos 	heta)$, so $v^2 = 2gl(1 - \cos 	heta)$.The tension at the lowest point is $T_{max} = mg + \frac{mv^2}{l} = mg + 2mg(1 - \cos 	heta) = mg(3 - 2 \cos 	heta)$.Given $T_{max} = 2T_{min}$, we have $mg(3 - 2 \cos 	heta) = 2mg \cos 	heta$.$3 - 2 \cos 	heta = 2 \cos 	heta$.$3 = 4 \cos 	heta$.$\cos 	heta = 0.75$.Therefore, $	heta = \cos^{-1}(0.75)$.

$A$ simple pendulum oscillates with an angular amplitude $\theta$. If the maximum tension in the string is twice the minimum tension, then $\theta$ is

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