Question: A particle executes SHM of period 12 s. After two seconds, it passes through the centre of oscillati...

Question

A particle executes SHM of period 12 s. After two seconds, it passes through the centre of oscillation, the velocity is found to be 3.142 cm s–1. The amplitude of oscillations is

Answer 1

Solution

Let $y = A\sin\omega t$

Then $v = \frac{dy}{dt} = \omega A\cos\omega t$

$= \frac{2\pi}{T}A\cos\frac{2\pi}{T}t$ $\left( \because\omega = \frac{2\pi}{T} \right)$

$\therefore 3.142 = \frac{2 \times 3.142}{12} \times A\cos\frac{2\pi}{12} \times 2$

$\therefore A = 12cm$