A simple pendulum of length 1, m is oscillati | vedclass

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Question

(A) The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g_{eff}}{\ell}}$.Taking the logarithmic derivative,we get $\frac{\Del

Vedclass · Accepted Answer

$10^{-3}$

Vedclass · Answer

(A) The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g_{eff}}{\ell}}$.Taking the logarithmic derivative,we get $\frac{\Delta \omega}{\omega} = \frac{1}{2} \frac{\Delta g_{eff}}{g_{eff}}$.When the support oscillates vertically with amplitude $A$ and frequency $\omega_s$,the effective acceleration due to gravity changes by $\Delta g = A \omega_s^2$.Here,$A = 10^{-2}\, m$ and $\omega_s = 1\, rad/s$.Thus,$\Delta g = 10^{-2} 	imes (1)^2 = 10^{-2}\, m/s^2$.Given $g \approx 10\, m/s^2$,the relative change in effective gravity is $\frac{\Delta g}{g} = \frac{10^{-2}}{10} = 10^{-3}$.Therefore,the relative change in angular frequency is $\frac{\Delta \omega}{\omega} = \frac{1}{2} 	imes \frac{\Delta g}{g} = \frac{1}{2} 	imes 10^{-3} = 0.5 	imes 10^{-3} = 5 	imes 10^{-4}$.Given the options provided,the closest order of magnitude is $10^{-3}$.

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