Question

A particle is executing SHM .with amplitude A, along X-axis. Initially the particle is at $x=\frac{A}{2}$ , while moves along + X-direction. If the equation for the oscillation is given by $x=A\sin$ $(\omega t+\delta ),$ then $\delta$ is

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

Answer 1

Answer: (C)

Given, $x=A\sin \omega t+\delta$ At t = 0 (initially), $x=\frac{A}{2}$ $\Rightarrow$ $\frac{A}{2}=A\sin \delta \Rightarrow \sin \delta =\frac{1}{2}$ $\Rightarrow$ $\delta =\frac{\pi }{6}or\frac{5\pi }{6}$ ..(i) The velocity, $\frac{dx}{dt}=\frac{d(A\sin \omega t)}{dt}$ $\Rightarrow$ $v=A\omega \cos (\omega t+\delta )$ At $t=0,v=A\omega \cos \delta$ Now, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ and $\cos \frac{5\pi }{6}=\frac{\sqrt{3}}{2}$ At t = 0, v is positive, so $\delta$ must be equal to $\frac{\pi }{6}.$