Question

Find out the phase constant $\varphi$ [in $\text{rad}$] of a particle performing SHM, if the initial conditions are as shown in the figure.

• A. $\frac{\pi }{3}$
• B. $\frac{4\pi }{3}$
• C. $\frac{5\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{5\pi }{3}$

Let's assume the equation of the particle executing SHM is
$x=A\sin (\omega t+\varphi )......\left(1\right)$
From the data given in the diagram, we can say that at $t=0,x=\frac{-\sqrt{3}A}{2}$
$\frac{-\sqrt{3}A}{2}=A\sin \varphi$
$\Rightarrow \varphi =\frac{4\pi }{3}\text{or}\frac{5\pi }{3}$
Differentiating $\left(1\right)$ with respect to time, we get
$v=A\omega \cos (\omega t+\varphi )$
Also from the figure, we can see that the particle is moving towards positive extreme position.
$\therefore$ At $t=0,v>0$
$\Rightarrow v=A\omega \cos \varphi >0\Rightarrow \cos \varphi >0\Rightarrow \varphi =\frac{5\pi }{3}\text{rad}$
Thus, option (c) is the correct answer.