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 = $\frac{A}{2}$ will be

[MP PMT 1987]

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Velocity of a particle executing S.H.M is given by

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