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:

  1. (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 $



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