A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple

Question

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/s² at a distance of 5 m from the mean position. The time period of oscillation is

Accepted Answer

For SHM, acceleration $|a| = \omega^2 y$. Given $|a| = 20 	ext{ m/s}^2$ and $y = 5 	ext{ m}$. So $20 = \omega^2(5) \implies \omega^2 = 4 \implies \omega = 2 	ext{ rad/s}$. The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi 	ext{ s}$.

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