A simple pendulum starts oscillating simple harmonically from its mean position ( $\mathrm{x}=0$ ) with amplitude ' a ' and periodic time ' T '. The magnitude of velocity of pendulum at $X=\frac{a}{2}$ is

Question

Aicharya · Accepted Answer

The velocity in simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$, where $\omega = \frac{2\pi}{T}$ is the angular frequency.

At displacement $x = \frac{a}{2}$, the velocity becomes:

$v = \frac{2\pi}{T} \sqrt{a^2 - \left(\frac{a}{2}\right)^2} = \frac{2\pi}{T} \sqrt{a^2 - \frac{a^2}{4}} = \frac{2\pi}{T} \sqrt{\frac{3a^2}{4}} = \frac{2\pi}{T} \cdot \frac{\sqrt{3}a}{2} = \frac{\sqrt{3}\pi a}{T}$

This corresponds to option D.

A simple pendulum starts oscillating simple harmonically from its mean position ( $\mathrm{x}=0$ ) with…

Solution