Question

A simple pendulum of length $1\text{m}$ is oscillating with an angular frequency of $10\text{rad/s}.$ The support of the pendulum starts oscillating up and down with a small angular frequency of $1\text{rad/s}$ and an amplitude of ${10}^{-2}\text{m}$. The relative change in the angular frequency of the pendulum is best given by:

• A. ${10}^{-3}\text{rad/s}$
• B. $1\text{rad/s}$
• C. ${10}^{-1}\text{rad/s}$
• D. ${10}^{-5}\text{rad/s}$

Answer 1

The correct option is A ${10}^{-3}\text{rad/s}$

Angular frequency of the pendulum is given by,

$\omega =\sqrt{\frac{g}{l}}$

$\therefore$ relative change in its angular frequency is,

$\frac{d\omega }{\omega }=\frac{1}{2}\frac{dg}{g}\left[\text{as length remains constant}\right]$

For the vertical oscillations of the support, ${\omega }_s=\sqrt{\frac{g}{A}}\Rightarrow g=A{\omega }_s^2$

$\Rightarrow dg=2A{\omega }_s$

Here, $[{\omega }_s=\text{angular frequency of support and A = amplitude]}$

$\Rightarrow \frac{d\omega }{\omega }=\frac{1}{2}\times \frac{2A{\omega }_s}{g}$

$\therefore d\omega =\frac{1}{2}\times \frac{2\times {10}^{-2}\times 1}{10}={10}^{-3}\text{rad/s}$

Hence, $\left(A\right)$ is the correct answer.