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}. $