Question

The initial conditions of a particle executing SHM are as shown in the figure. Find the phase constant $\varphi$ of the particle.
Given: Range of $\varphi$ is $[0,2\pi ]$

• A. $\frac{7\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{5\pi }{4}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is C $\frac{5\pi }{4}$

Equation of a particle executing SHM, is given by $x=A\sin (\omega t+\varphi )......\left(1\right)$
where $\varphi$ is in the range of $[0,2\pi ]$
From the given figure, we can deduce that, at $t=0\to x=\frac{-A}{\sqrt{2}}$
Now from equation $\left(1\right)$, we can write that
$\frac{-A}{\sqrt{2}}=A\sin \varphi \Rightarrow \sin \varphi =\frac{-1}{\sqrt{2}}\Rightarrow \varphi =\frac{5\pi }{4}\text{or}\frac{7\pi }{4}$
Differentiating $\left(1\right)$ with respect to time on both sides, we get
$v=A\omega \cos (\omega t+\varphi )$
Also, from the figure, we can deduce that the particle is moving towards negative extreme position. So, we can conclude that at $t=0$,
$v=A\omega \cos \varphi <0\Rightarrow \cos \varphi <0\Rightarrow \varphi =\frac{5\pi }{4}$
Thus, option (c) is the correct answer.