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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
$\frac{\pi a\sqrt{3}}{T}$
B
$\frac{\pi a\sqrt{3}}{2T}$
C
$\frac{\pi a}{T}$
D
$\frac{3\pi^2a}{T}$
Detailed Solution
Speed in SHM at displacement x: $v = \omega\sqrt{a^2 - x^2}$, with $\omega = \frac{2\pi}{T}$
At $x = \frac{a}{2}$: $v = \frac{2\pi}{T}\sqrt{a^2 - \frac{a^2}{4}}$
$v = \frac{2\pi}{T}\sqrt{\frac{3a^2}{4}} = \frac{2\pi}{T}\times\frac{\sqrt{3}a}{2}$
$v = \frac{\pi a\sqrt{3}}{T}$
At $x = \frac{a}{2}$: $v = \frac{2\pi}{T}\sqrt{a^2 - \frac{a^2}{4}}$
$v = \frac{2\pi}{T}\sqrt{\frac{3a^2}{4}} = \frac{2\pi}{T}\times\frac{\sqrt{3}a}{2}$
$v = \frac{\pi a\sqrt{3}}{T}$