Question

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

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

Answer 1

Answer: (D)

Angular frequency of simple pendulum,
$\omega =\sqrt{\frac{g}{l}}$
As the support oscillates up and down, effective $g$ changes
$\Delta g=2.{\omega }_0^{2}A=2\times 1^2\times 10^{-2}m{s}^{-2}$
$=0.02m{s}^{-2}$
Change in angular frequency $d\omega$ of the pendulum is given by
$\frac{d\omega }{\omega }=\frac{-1}{2}\cdot \frac{dg}{g}$
$\frac{|\Delta \omega |}{\omega }=\frac{1}{2}\frac{|\Delta g|}{g}$
$=\frac{1}{2}\times \frac{0.02}{10}$
$=0.001rad{s}^{-1}$