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Q.
A body executing simple harmonic motion has a periodic time of $3 s$. After how much time from $t=0$, its displacement will be half of its amplitude?
Oscillations
Solution:
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,
$T=3 s$
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} $
$\frac{2 \pi t}{3} =\frac{\pi}{6}$
$t =\frac{1}{4} s$