Question 222: Oscillations

Question

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

Accepted Answer

Step 1: Standard expression for velocity in SHM
For SHM:
$ x(t) = a \sin(\omega t) $
or equivalently,
$ v = \frac{dx}{dt} = \omega \sqrt{a^2 - x^2}, $
where $\omega = \frac{2\pi}{T}$.
Step 2: Put $x = 	frac{a}{2}$
$ v = \omega \sqrt{a^2 - \left(\frac{a}{2}ight)^2} = \omega \sqrt{a^2 - \frac{a^2}{4}} = \omega \sqrt{\frac{3a^2}{4}}. $
$ v = \omega \cdot \frac{\sqrt{3}}{2} a. $
Step 3: Express $\omega$ in terms of $T$
$ \omega = \frac{2\pi}{T}. $
So:
$ v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}\pi a}{T}. $
Step 4: Compare with options
This matches
Option D
:
$ \boxed{\frac{\sqrt{3}\pi a}{T}} $

Answer

$\frac{3 \pi^2 a}{T}$

Answer

$\frac{\sqrt{3} \pi a}{2 T}$

Answer

$\frac{\pi a}{T}$

Answer

$\frac{\sqrt{3} \pi \mathrm{a}}{\mathrm{T}}$

Oscillations

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

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