Question

A simple harmonic oscillator starts from extreme position and covers a displacement half of its amplitude in a time '$t$', the further time taken by it to reach mean position is

• A. $2t$
• B. $t$
• C. $t/\sqrt{2}$
• D. $t/2$

Answer 1

The correct option is D $t/2$

$x=Acos\left(\omega t\right)$
Taking $t=0$ at extreme position.
$x=\frac{A}{2}$ is reached in time t
$\omega t=\frac{\pi }{3};\omega =\frac{\pi }{3t}$
$x=0$ is reached at $\omega t_1=\frac{\pi }{2}$
$t_1=\frac{\pi }{2\omega }=\frac{\pi \times 3t}{2\times \pi }=\frac{3t}{2}$
$t_1-t=\frac{t}{2}$