A simple pendulum, consisting of a small ball | vedclass

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Question

(A) Let $l$ be the length of the string. At the extreme position, the velocity is zero, so the tension is $T_{\min } = mg \cos 	heta$, where $	heta$

Vedclass · Accepted Answer

$\frac{3 mg}{2}$

Vedclass · Answer

(A) Let $l$ be the length of the string. At the extreme position, the velocity is zero, so the tension is $T_{\min } = mg \cos 	heta$, where $	heta$ is the maximum angular displacement.At the lowest point (mean position), the tension is maximum, given by $T_{\max } = mg + \frac{mv^2}{l}$.By the law of conservation of energy between the extreme position and the mean position: $\frac{1}{2} mv^2 = mgl(1 - \cos 	heta)$, which implies $v^2 = 2gl(1 - \cos 	heta)$.Substituting $v^2$ into the expression for $T_{\max }$: $T_{\max } = mg + \frac{m(2gl(1 - \cos 	heta))}{l} = mg + 2mg(1 - \cos 	heta) = mg(1 + 2 - 2 \cos 	heta) = mg(3 - 2 \cos 	heta)$.Given $T_{\max } = 2 T_{\min }$, we have $mg(3 - 2 \cos 	heta) = 2(mg \cos 	heta)$.$3 - 2 \cos 	heta = 2 \cos 	heta \implies 4 \cos 	heta = 3 \implies \cos 	heta = \frac{3}{4}$.Substituting $\cos 	heta = \frac{3}{4}$ into the expression for $T_{\max }$: $T_{\max } = 2mg \cos 	heta = 2mg \left(\frac{3}{4}ight) = \frac{3mg}{2}$.

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