Question 112 of 194

A particle executes S.H.M. of amplitude $A$ along $x$-axis. At $t=0$, the position of the particle is $x=\dfrac{A}{2}$ and it moves along positive $x$-axis. The displacement of particle in time $t$ is $x=A\sin \left(\omega t+\delta \right)$, then the value $\delta$ will be

A$\dfrac{\pi }{2}$B$\dfrac{\pi }{6}$C$\dfrac{\pi }{3}$D$\dfrac{\pi }{4}$

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Answer: (B) $\dfrac{\pi }{6}$

Given, the instantaneous equation of motion of the particle is
$x=A\sin \left(\omega t+\delta \right)...\left(1\right)$
As the particle is moving along positive $x$-axis, $v=\dfrac{dx}{dt}\gt 0$.
Substitute $0$ for $t$ and $\dfrac{A}{2}$ for $x$ into equation (1) and solve to calculate the required phase difference.
$\begin{matrix}\dfrac{A}{2} & = & A\sin \left(\omega \times 0+\delta \right) \\ & \Rightarrow & \sin \delta =\dfrac{1}{2} \\ & = & \sin \dfrac{\pi }{6} \\ & \Rightarrow & \delta =\dfrac{\pi }{6}\end{matrix}$