Question 44 of 194

$Y=A\sin \left(\omega t+{\phi }_{0}\right)$ is the time-displacement equation of a $SHM.$ At $t=0$ the displacement of the particle is $Y=\dfrac{A}{2}$ and it is moving along negative $x$-direction. Then the initial phase angle ${\phi }_{0}$ will be:

A$\dfrac{2\pi }{3}$B$\dfrac{\pi }{6}$C$\dfrac{5\pi }{6}$D$\dfrac{\pi }{3}$

Show answer and solution▾

Answer: (C) $\dfrac{5\pi }{6}$

Initial phase $\dfrac{\pi }{2}+\dfrac{\pi }{3}=\dfrac{5\pi }{6}$.
Also it can be seen as below

The initial phase angle ${\phi }_{0}=\pi -\dfrac{\pi }{6}$
$=\dfrac{5\pi }{6}$