Question

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude 'a' and time period $T$. The speed of the pendulum at $x = a/2$ will be

• A. $ \frac{ \pi a}{ T} $
• B. $ \frac{ 3 \pi^2 \, a}{ T} $
• C. $ \frac{ \pi a \sqrt 3 }{ T} $
• D. $ \frac{ \pi a \sqrt 3 }{ 2 T} $

Answer 1

Answer: (C)

For simple harmonic motion,
$v=\omega \sqrt{a^{2}-x^{2}}$.
When $x=\frac{a}{2}$,
When $x=\frac{a}{2},$
$ v=\omega \sqrt{a^{2}-\frac{a^{2}}{4}}=\omega \sqrt{\frac{3}{4} a^{2}}$.
As $\omega=\frac{2 \pi}{T}, $
$\therefore v=\frac{2 \pi}{T} \cdot \frac{\sqrt{3}}{2} a $
$\Rightarrow v=\frac{\pi \sqrt{3} a}{T}$.