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:

We have a simple pendulum of length $$l = 1\ \text{m}$$. For small oscillations its natural angular frequency is given by the elementary formula for a simple pendulum,

$$\omega_0 \;=\; \sqrt{\dfrac{g}{l}}.$$

In the statement of the problem the natural or unperturbed frequency is already supplied as $$\omega_0 = 10\ \text{rad s}^{-1}.$$ Substituting this value and the given length $$l = 1\ \text{m}$$ into the formula, we can back-calculate the effective gravitational acceleration that is implied by the data:

$$\omega_0^2 \;=\; \dfrac{g}{l} \;\;\Longrightarrow\;\; g = \omega_0^2\,l = (10)^2\,(1) = 100\ \text{m s}^{-2}.$$

Next, the point of suspension itself oscillates up and down. Let its vertical displacement be written as

$$y(t)=a\cos(\omega_s t),$$

where the amplitude is $$a = 10^{-2}\ \text{m}$$ and the support’s angular frequency is $$\omega_s = 1\ \text{rad s}^{-1}.$$

The vertical acceleration of the support is therefore the second derivative of this displacement:

$$\ddot y(t) = -\,a\,\omega_s^2\cos(\omega_s t).$$

In the non-inertial frame attached to the support, this extra acceleration simply adds algebraically to the usual gravitational acceleration. Hence, at any instant the effective downward acceleration becomes

$$g_{\text{eff}}(t)=g+\ddot y(t) = g - a\,\omega_s^2\cos(\omega_s t).$$

A small-angle pendulum obeys the differential equation

$$\theta''(t) + \dfrac{g_{\text{eff}}(t)}{l}\;\theta(t)=0,$$

so the instantaneous (time-dependent) square of its angular frequency is

$$\omega^2(t)=\dfrac{g_{\text{eff}}(t)}{l} =\dfrac{g}{l}\Bigl(1-\dfrac{a\,\omega_s^2}{g}\cos(\omega_s t)\Bigr).$$

Let us denote the small fractional variation inside the bracket by

$$\varepsilon(t)=-\,\dfrac{a\,\omega_s^2}{g}\cos(\omega_s t),$$

so that $$\omega^2(t)=\omega_0^2\,[1+\varepsilon(t)].$$ Because $$|\varepsilon(t)|\ll 1,$$ we may expand the square root:

$$\omega(t) =\omega_0\,\sqrt{1+\varepsilon(t)} \approx\omega_0\Bigl(1+\dfrac{\varepsilon(t)}{2}\Bigr).$$

(Here we have used the first-order binomial expansion $$\sqrt{1+ \varepsilon}\approx 1+\varepsilon/2$$ for $$|\varepsilon|\ll 1.$$) Therefore the instantaneous change in angular frequency relative to the unperturbed value is

$$\Delta\omega(t)=\omega(t)-\omega_0 \approx\omega_0\,\dfrac{\varepsilon(t)}{2} = -\,\omega_0\,\dfrac{a\,\omega_s^2}{2g}\,\cos(\omega_s t).$$

The peak (maximum magnitude) of this change is obtained by setting $$|\cos(\omega_s t)|=1:$$

$$|\Delta\omega|_{\text{max}} = \omega_0\,\dfrac{a\,\omega_s^2}{2g}.$$

Substituting the numerical values,

$$|\Delta\omega|_{\text{max}} = (10\ \text{rad s}^{-1})\; \dfrac{(10^{-2}\ \text{m})\,(1\ \text{rad s}^{-1})^{2}} {2\,(100\ \text{m s}^{-2})}.$$

We simplify step by step:

$$a\,\omega_s^2 = (10^{-2})\times(1)^2 = 10^{-2}\ \text{m s}^{-2},$$

$$2g = 2\times 100 = 200\ \text{m s}^{-2},$$

$$\dfrac{a\,\omega_s^2}{2g} = \dfrac{10^{-2}}{200} = 5\times10^{-5},$$

$$|\Delta\omega|_{\text{max}} = 10 \times 5\times10^{-5} = 5\times10^{-4}\ \text{rad s}^{-1}.$$

This numerical value is $$0.0005\ \text{rad s}^{-1},$$ which is of the order of $$10^{-3}\ \text{rad s}^{-1}.$$ Among the alternatives supplied, the option that best matches this magnitude is $$10^{-3}\ \text{rad s}^{-1}.$$\

Hence, the correct answer is Option A.