Question

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

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

Answer 1

Answer: (D)

Angular frequency of simple pendulum,
$\omega =\sqrt{\frac{g}{l}}$
As the support oscillates up and down, effective $g$ changes
$\Delta g=2.\omega _{0}^{2}A=2\times 1^{2}\times 10^{- 2} \, ms^{- 2}$
$=0.02 \, ms^{- 2}$
Change in angular frequency $d\omega $ of the pendulum is given by
$\frac{d \omega }{\omega }=\frac{- 1}{2}\cdot \frac{d g}{g}$
$\frac{\left|\Delta \omega \right|}{\omega }=\frac{1}{2}\frac{\left|\Delta g\right|}{g}$
$=\frac{1}{2}\times \frac{0.02}{10}$
$=0.001 \, rads^{- 1}$