A simple pendulum oscillates with an angular | vedclass

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(B) Let $m$ be the mass of the bob and $l$ be the length of the string. The tension in the string at any angle $\phi$ is given by $T = mg \cos \phi +

Vedclass · Accepted Answer

$\cos ^{-1}(0.5)$

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(B) Let $m$ be the mass of the bob and $l$ be the length of the string. The tension in the string at any angle $\phi$ is given by $T = mg \cos \phi + \frac{mv^2}{l}$.At the extreme position,$\phi = 	heta$ and $v = 0$,so the minimum tension is $T_{min} = mg \cos 	heta$.At the lowest point,$\phi = 0$. Using conservation of energy,$mg(l - l \cos 	heta) = \frac{1}{2}mv^2$,so $mv^2 = 2mgl(1 - \cos 	heta)$.The maximum tension is $T_{max} = mg + \frac{mv^2}{l} = mg + 2mg(1 - \cos 	heta) = mg(3 - 2 \cos 	heta)$.Given $T_{max} = 4 T_{min}$,we have $mg(3 - 2 \cos 	heta) = 4mg \cos 	heta$.$3 - 2 \cos 	heta = 4 \cos 	heta \implies 6 \cos 	heta = 3 \implies \cos 	heta = 0.5$.Therefore,$	heta = \cos^{-1}(0.5)$.

$A$ simple pendulum oscillates with an angular amplitude $\theta$. If the maximum tension in the string is $4$ times the minimum tension,then the value of $\theta$ is

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