Question

A body executing simple harmonic motion has a periodic time of $3s$. After how much time from $t=0$, its displacement will be half of its amplitude?

• A. $(1/8)s$
• B. $(1/6)s$
• C. $(1/4)s$
• D. $(1/3)s$

Answer 1

Answer: (C)

We know that, $y=a\sin \omega t\left(\because \omega =\frac{2\pi }{T}\right)$ where, $a$ is the amplitude, $y=a\sin \frac{2\pi }{T}t$
Given,
$T=3s$
So, when the displacement will be half of its amplitude, i.e.
$y=\frac{a}{2}\Rightarrow \frac{a}{2}=a\sin \frac{2\pi t}{3}$
$\frac{1}{2}=\sin \frac{2\pi t}{3}$
$\sin \frac{\pi }{6}=\sin \frac{2\pi t}{3}$
$\frac{2\pi t}{3}=\frac{\pi }{6}$
$t=\frac{1}{4}s$