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 is first time passing through the equilibrium position is equal to

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

Answer 1

Answer: (B)

Phase of the motion is $(\omega t+\phi)$
Using $x=A \sin (\omega t+\phi)$
and $V=A \omega \cos (\omega t+\phi)$
for conditions at $t=0 \rightarrow x=A$ and $V=0$ then $\phi=\frac{\pi}{2}$
When it passes equilibrium position for the first time
$t=\frac{T}{4}$
Phase $\frac{2 \pi}{T} \cdot \frac{T}{4}+\frac{\pi}{2}=\pi$