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} $
$\Rightarrow t=\frac{1}{3} s$