Question

$y=A\sin (\omega t+{\varphi }_o)$ is the time – displacement equation of an SHM. At $t=0$, the displacement of the particle is $Y=\frac{A}{2}$ and it is moving along negative $x-$direction. Then, the initial phase angle ${\varphi }_o$ will be.

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{2\pi }{3}$
• D. $\frac{5\pi }{6}$

Answer 1

The correct option is D $\frac{5\pi }{6}$

$y=A\sin (\omega t+{\varphi }_o)$
At $t=0,x=\frac{A}{2}$
$\frac{1}{2}=\sin {\varphi }_o$
${\varphi }_o=\frac{\pi }{6},\frac{5\pi }{6}$

$v=\frac{dy}{dt}=A\omega \cos (\omega t+{\varphi }_o)$

At $t=0,v=A\omega \cos \left({\varphi }_o\right)$

For, ${\varphi }_o=\frac{\pi }{6},v\to +ve$, and,
For, ${\varphi }_o=\frac{5\pi }{6},v\to -ve$
$\therefore {\varphi }_o=\frac{5\pi }{6}$