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Question

Physics Question on Oscillations

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 20m/s220 \, m/s^{-2} at a distance of 5m5 \,m from the mean position. The time period of oscillation is

A

1 s

B

2 π\pi s

C

2s

D

πs\pi s

Answer

πs\pi s

Explanation

Solution

a=ω2y\left|a\right| = \omega^{2}y
20=ω2(5)\Rightarrow 20 = \omega^{2} \left(5\right)
ω=2rads\Rightarrow \omega = 2 \frac{rad}{s}
T=2πω=2π2=πsT = \frac{2 \pi}{\omega} = \frac{2 \pi}{2} = \pi s