Question

A particle starts from mean position ${}^{\prime }{O}^{\prime }$ where $x=0$ and oscillates to and fro about this point. When it is passing through ${}^{\prime }{O}^{\prime }$ while moving towards negative extreme position , find the phase constant $\varphi$ of the particle at this point.

• A. $\varphi =0^{\circ }$
• B. $\varphi =\frac{\pi }{2}$
• C. $\varphi =\pi$
• D. $\varphi =\frac{\pi }{4}$

Answer 1

The correct option is C $\varphi =\pi$


Since the particle starts at mean position,
we know that, the displacement can be written as
$x=A\sin (\omega t+\varphi )\dots \dots \left(1\right)$
From this by differentiating we get,
$V=A\omega \cos (\omega t+\varphi )\dots \dots \left(2\right)$
Since, particle was initially at $x=0$, from equation $1$ we get,
$0=A\sin \varphi \Rightarrow \sin \varphi =0^{\circ }\Rightarrow \varphi =n\pi$ where $n=0,1,2,\mathrm{3.....}$
$\varphi$ can be $0$ , $\pi$ , $2\pi .....$
Since the particle is moving towards negative extreme position, at $t=0$, Particle velocity is negative
$A\omega \cos \varphi <0$
$\Rightarrow \cos \varphi <0$
If, $\varphi =0^{\circ },2\pi ,4\pi ...$ then $\cos \varphi >0$
But, If $\varphi =\pi ,3\pi ,5\pi ....$ then $\cos \varphi <0$
So, we can conclude that $\varphi =\pi ,3\pi ,5\pi ....$.
Hence option (c) is the correct answer.