A simple pendulum of length 1m is oscillating with an angular frequency  $10{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}$ and an amplitude of $10{}^{\text{-2}}m$ . The relative change in the angular frequency of the pendulum is best given by(take acceleration due to gravity g=10m/$s^2$)

$\omega =10{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}\text{,}g=10{\text{\hspace{0.17em}m\hspace{0.17em}s}}^{\text{-2}}$

$$\begin{gathered} {\text{g}}_{\max }=\text{g\hspace{0.17em}+\hspace{0.17em}a} \\ {\text{g}}_{\min }=\text{g\hspace{0.17em}-\hspace{0.17em}a} \\ difference of above two equations is \Delta g, \\ \Delta g=2a=( 2)acceleration,and a={\omega }_{support}^{2}A, here A is amplitude, \\ \Delta g=2{\omega }_s^{2}A---\left(1\right) \\ {\omega }_{\text{support}}={\omega }_{\text{s}}=1{\text{\hspace{0.17em}rad.s}}^{\text{-1}} \\ \text{A}={10}^{\text{-2}}m \\ \omega =\sqrt{\frac{g_{eff}}{l}} \\ differentiate above equation \\ \Rightarrow \frac{\Delta \omega }{\omega }=\frac{1}{2}\frac{\Delta g}{g} \\ \Delta \omega =\frac{1}{2}\frac{\Delta g}{g}\times \omega \\ substitute equation \left(1\right), \\ \Delta \omega =\frac{1}{2}\times \frac{2{\omega }_s^{2}A}{g}\times \omega \\ \therefore \Delta \omega =\frac{1\times {10}^{-2}}{10}\times 10={10}^{-2}{\text{rad\hspace{0.17em}s}}^{\text{-1}} \\ \Delta \omega change in angular frequency of pendulum \end{gathered}$$