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A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s$^2$ at a distance of 5 m from the mean position. The time period of oscillation is
Explanation
$\omega = \sqrt{a/y} = 2$ rad/s, so $T = \pi$ s.
Detailed Solution
$|a| = \omega^2 y$
$20 = \omega^2(5) \Rightarrow \omega = 2$ rad/s
$T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$ s