Question

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is

• A. 1 m$s^{-1}$
• B. 1.414m$s^{-1}$
• C. 0.5m$s^{-1}$
• D. 0.75m$s^{-1}$

Answer 1

The correct option is D 0.75m$s^{-1}$

$v_{max}=A\omega =0.866\frac{m}{s}$
At $x=\frac{A}{2};v=\omega \sqrt{A^2-x^2}$
$v=\frac{\sqrt{3}}{2}\omega A=\frac{\sqrt{3}}{2}{v}_{max}$
$v=\frac{\sqrt{3}}{2}\times 0.866=0.75\frac{m}{s}$