Question

A particle is executing SHM. Find the phase constants of the particle, when it starts from the points $\text{A},\text{B}$ and $\text{C}$ respectively as shown below.
(Phase constant $\varphi$ is the range of $[0,2\pi ]$)

• A. ${\varphi }_A=\frac{3\pi }{2},{\varphi }_B=0,{\varphi }_C=\frac{\pi }{2}$
• B. ${\varphi }_A=\frac{\pi }{2},{\varphi }_B=\pi ,{\varphi }_C=\frac{\pi }{2}$
• C. ${\varphi }_A=\frac{3\pi }{2},{\varphi }_B=\pi ,{\varphi }_C=\frac{\pi }{2}$
• D. ${\varphi }_A=\frac{\pi }{2},{\varphi }_B=\pi ,{\varphi }_C=\frac{3\pi }{2}$

Answer 1

The correct option is C ${\varphi }_A=\frac{3\pi }{2},{\varphi }_B=\pi ,{\varphi }_C=\frac{\pi }{2}$

Let us assume the equation of the particle executing SHM is
$x=A\sin (\omega t+\varphi ).....\left(1\right)$
At $t=0$ and $x=0$, we get
$\sin \varphi =0\Rightarrow \varphi =n\pi$
where $n=0,1,2,3,\mathrm{4.....}$
Since $\varphi$ is in the range of $[0,2\pi ]$ [given], we can say that $\varphi =0,\pi \text{or}2\pi$
Differentiating $\left(1\right)$ with respect to time, we get
$v=A\omega \cos (\omega t+\varphi ).......\left(2\right)$

From the data given in the figure, for particle starting from point $\text{B}$, we can deduce that the particle is moving towards the negative extreme position.
$\therefore$ At $t=0$, velocity is negative.
$\Rightarrow v=A\omega \cos \varphi <0\Rightarrow \cos \varphi <0\Rightarrow {\varphi }_B=\pi$

For particle starting from point $\text{A}$, we can deduce that the particle is at its negative extreme position.
From $\left(1\right)$, we can write that
$-A=A\sin \left(\varphi \right)\Rightarrow \sin \varphi =-1\Rightarrow {\varphi }_A=\frac{3\pi }{2}$
For particle starting from point $\text{C}$, we can deduce that the particle is at its positive extreme position.
$\therefore$ From $\left(1\right)$, we can say that
$A=A\sin \varphi \Rightarrow \sin \varphi =1\Rightarrow {\varphi }_C=\frac{\pi }{2}$


Thus, option (c) is the correct answer.