Question

A particle executes linear $\text{SHM}$ of amplitude $A$ along $x-$ axis. At $t=0$ the position of the particle is $x=\frac{A}{2}$ and it moves along $+x$ direction. Find the phase constant $\left(\delta \right)$ if the equation is written as $x=\text{A sin}(\omega t+\delta )$

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${120}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

Given, at $t=0,;x=\frac{A}{2}$
from, $x=\text{A sin}(\omega t+\delta )$
$\frac{A}{2}=\text{A sin}\delta$
$\Rightarrow \text{sin}\delta =\frac{1}{2}\Rightarrow \delta ={30}^{\circ }$ or ${150}^{\circ }$
But for $\delta ={150}^{\circ }$, velocity $(v=A\omega \cos (\omega t+\delta \left)\right)$ will be negative.
Hence $\delta ={30}^{\circ }$