Question

In a horizontal spring mass system, mass $m$ is released after being displaced towards right by some distance at $t=0$ on a frictionless surface. The phase angle of the motion in radian when it passes through the equilibrium position for the first time is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{3\pi }{2}$
• D. $0$

Answer 1

The correct option is B $\pi$

We know that equation of motion is given by, $x=A\cos \left(\omega t\right)$, when particle is at extreme position.
When it passes through mean position for the first time, the mass has covered $1/4^{\text{th}}$ of the complete oscillation.
Therefore, $t=T/4$.
Thus, phase difference$=\omega t=\frac{2\pi }{T}\times \frac{T}{4}=\frac{\pi }{2}$
Thus phase angle = phase difference + initial phase
$\therefore \frac{\pi }{2}+\frac{\pi }{2}=\pi$