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

2 s

\pi \text{ s}

2\pi \text{ s}

1 s

Step-by-Step Solution

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

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