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Q. 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

NEETNEET 2018Oscillations

Solution:

$\left|a\right| = \omega^{2}y$
$ \Rightarrow 20 = \omega^{2} \left(5\right)$
$\Rightarrow \omega = 2 \frac{rad}{s}$
$ T = \frac{2 \pi}{\omega} = \frac{2 \pi}{2} = \pi s $