A simple pendulum of length $1$ m is oscillating with an angular frequency $10$ rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of $1$ rad/s and an amplitude of $10^{-2}$ m. The relative change in the angular frequency of the pendulum is best given by:
Answer: (A) $10^{-3}$ rad/s
In the frame of the support, the effective gravity is $g_{\text{eff}} = g + a$, where the support's acceleration has maximum value
$$a_{\max} = \omega_s^2A = 1^2\times10^{-2} = 10^{-2}\ \text{m/s}^2$$
Since $\omega \propto \sqrt{g_{\text{eff}}}$:
$$\frac{\Delta\omega}{\omega} = \frac12\frac{\Delta g}{g} = \frac12\times\frac{10^{-2}}{10} = 5\times10^{-4}$$
So the change is of the order of $10^{-3}$ (as a fraction; $\Delta\omega = 10\times5\times10^{-4} \approx 5\times10^{-3}$ rad/s), the best of the given choices.
Solution by Sreeraj P, M.Sc Physics