Question

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude (1) and time period (T). The speed of the pendulum at $x = \frac{A}{2}$ will be

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

Answer 1

Answer: (A)

$\frac{\pi A\sqrt{3}}{T}$

Answer 2

$v = \omega\sqrt{a^{2} - y^{2}}$⇒ $v = \frac{2\pi}{T}\sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{\pi A\sqrt{3}}{T}$

[As y = A/2]