Question

A particle is executing SHM with a period $\pi \sec$. When it is passing through the center of its path, its velocity is $0.1\, m/s$. What is its velocity when it is at a distance of $0.03 \,m$ from the mean position?

• A. 0.08 m/s
• B. 0.05 m/s
• C. 0.04 m/s
• D. 0.01 m/s

Answer 1

Answer: (A)

When the point is at a distance from the mean position its, velocity is given by $v=\omega \sqrt{a^{2}-y^{2}}$
Given, $T=\frac{2 \pi}{\omega}=\pi$
$\therefore \omega=25^{-1}$ and $v=a \omega 0.1=a \times 2$
$a=\frac{0.1}{2}=0.05\, m\,\,y =0.05\, m$
$\therefore $ Velocity, $v=2 \sqrt{(0.05)^{2}-(0.03)^{2}}$
$=0.08\, m / s$