Question

A particle executing simple harmonic motion has a time period of $4s$. After how much interval of time from $t=0$ will its displacement be half of its amplitude ?

• A. $\frac{1}{3}s$
• B. $\frac{1}{2}s$
• C. $\frac{2}{3}s$
• D. $\frac{1}{6}s$

Answer 1

Answer: (A)

The displacement equation for SHM is
$y=a\sin \omega t^{\prime }$
where $\omega$ is angular velocity $\left(\omega =\frac{2\pi }{T}\right)$ and $a$ the amplitude.
Given,
$y=\frac{a}{2},t^{\prime }=\frac{t}{2}$
$\frac{a}{2}=a\sin \frac{2\pi t}{4}$
$\Rightarrow \frac{1}{2}=\sin \frac{\pi t}{2}$
$\therefore \sin \frac{\pi }{6}=\sin \frac{\pi t}{2}$
$\Rightarrow \frac{\pi }{6}=\frac{\pi t}{2}$
$\Rightarrow t=\frac{1}{3}s$