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A simple pendulum is vibrating with an angular amplitude of $90 °$ . For what values of $θ$ with vertical,the acceleration is directed
1) vertically upwards 2) horizontally

3) vertically downwards

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\cos^{- 1} \frac{1}{\sqrt{3}}, 90 °, 0 °

0 °, \cos^{- 1} \frac{1}{\sqrt{3}}, 90 °

90 °, \cos^{- 1} \frac{1}{\sqrt{3}}, 0 °

\cos^{- 1} \frac{1}{\sqrt{3}}, 0 °, 90 °

answer is A.

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Detailed Solution

$T \cos θ = mg$

$T = \frac{mg}{\cos θ}$

At point ‘A’ $\frac{1}{2} {mv}_{A}^{2} = mg l$

$\frac{1}{2} {mv}_{A}^{2} = \frac{1}{2} {mv}_{P}^{2} + mg l (1 - \cos θ)$

$mg l = \frac{1}{2} {mv}_{P}^{2} + mg l - mg l \cos θ$

$\frac{1}{2} {mv}_{P}^{2} = mg l \cos θ$

$v_{P}^{2} = 2g l \cos θ$

$T - mg \cos θ = \frac{{mv}_{P}^{2}}{L}$

$\frac{mg}{\cos θ} - mg \cos θ = \frac{2 mg \cos θ}{L}$

$\frac{1}{\cos θ} = 3 \cos θ$

$\Rightarrow \cos θ = \frac{1}{\sqrt{3}}$

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