Simple Harmonic Motion 5 Question 3
3. A simple pendulum of length $1 m$ is oscillating with an angular frequency $10 rad / s$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 rad / s$ and an amplitude of $10^{-2} m$. The relative change in the angular frequency of the pendulum is best given by
(a) $1 rad / s$
(b) $10^{-5} rad / s$
(c) $10^{-3} rad / s$
(d) $10^{-1} rad / s$
(2019 Main, 11 Jan II)
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Answer:
Correct Answer: 3. (c)
Solution:
- We know that time period of a pendulum is given by
$$ T=2 \pi \sqrt{\frac{l}{g}} $$
So, angular frequency $\omega=\frac{2 \pi}{T}=\sqrt{\frac{g}{l}}$
Now, differentiate both side w.r.t $g$
$$ \begin{aligned} \therefore \quad \frac{d \boldsymbol{\omega}}{d g} & =\frac{1}{2 \sqrt{g} \sqrt{l}} \\ d \boldsymbol{\omega} & =\frac{d g}{2 \sqrt{g} \sqrt{l}} \end{aligned} $$
By dividing Eq. (ii) by Eq. (i), we get
$$ \frac{d \omega}{\omega}=\frac{d g}{2 g} $$
Or we can write
$$ \frac{\Delta \omega}{\omega}=\frac{\Delta g}{2 g} $$
As $\Delta g$ is due to oscillation of support.
$$ \therefore \quad \Delta g=2 \omega^{2} A $$
$\left(\omega _1 \rightarrow 1 rad / s\right.$, support)
Putting value of $\Delta g$ in Eq. (iii) we get
$$ \begin{aligned} & \frac{\Delta \omega}{\omega}=\frac{1}{2} \cdot \frac{2 \omega _1^{2} A}{g}=\frac{\omega _1^{2} A}{g} ;\left(A=10^{-2} m^{2}\right) \\ \Rightarrow \quad & \frac{\Delta \omega}{\omega}=\frac{1 \times 10^{-2}}{10}=10^{-3} rad / s \end{aligned} $$
