In a simple pendulum, the tension at the lowest point of the string is double the weight of the bob. The angular amplitude of the oscillation of the bob is

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$\cos^{- 1} (\frac{1}{3})$

60⁰

90⁰

$\cos^{- 1} (\frac{2}{3})$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{{mu}^{2}}{r} + mg = 2 mg \\ u = \sqrt{rg} \end{array}$
From COME
$\begin{array}{l} v^{2} = u^{2} - 2 gr (1 - \cos θ) \\ 0 = rg - 2 gr (1 - \cos θ) \\ \cos θ = \frac{1}{2}; θ = 60^{\circ} \end{array}$

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