Question

A particle performing SHM has a velocity $v_0$ at mean position. Find the velocity of the particle, when it reaches a point which is at a distance $\frac{A}{2}$ from the mean position.
[$A=$ amplitude of SHM]

• A. $\frac{v_0}{4}$
• B. $\frac{\sqrt{3}}{2}{v}_0$
• C. $\frac{3v_0}{2}$
• D. $\frac{v_0}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}{v}_0$

Velocity-displacement relationship for SHM is $v=\omega \sqrt{A^2-x^2}$
At mean point, $x=0\&v=\omega A=v_0$
When the particle is at $x=\frac{A}{2},$
$v=\omega \sqrt{A^2-{\left(\frac{A}{2}\right)}^2}=\omega \sqrt{\frac{3A^2}{4}}$
$\Rightarrow v=\frac{\sqrt{3}A\omega }{2}\Rightarrow v=\frac{\sqrt{3}{v}_0}{2}$