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A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle $θ$ with the vertical

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$\tan^{- 1} \frac{a}{g}$ in the forward direction

$\tan^{- 1} \frac{a}{g}$ in the backward direction

$\tan^{- 1} \frac{g}{a}$ in the backward direction

$\tan^{- 1} \frac{g}{a}$ in the forward direction

answer is B.

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Detailed Solution

In accelerated frame of reference, a fictitious force (pseudo force) ma acts on the bob of pendulum as shown in figure.
Hence, $tanθ = \frac{ma}{mg} = \frac{a}{g}$

$\Rightarrow θ = \tan^{- 1} (\frac{a}{g})$ in the backward direction.

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