Oscillations - Result Question 35
39. 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
(a) $2 \pi s$
(b) $\pi s$
(c) $1 s$
(d) $2 s$
[2018]
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Answer:
Correct Answer: 39. (b)
Solution:
- (b) From question, acceleration, $a=20 m / s^{2}$, and displacement, $y=5 m$
$ |a|=\omega^{2} y $
$\Rightarrow 20=\omega^{2}(5)$
$\Rightarrow \omega=2 rad / s$
Time period of pendulum,
$ T=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi s $
