Question

A particle executes simple harmonic motion with amplitude of 3.00cm. At what position does its speed equal half of its maximum speed?

Answer 1

$x=Acos\omega t,$

$v=-A\omega sin\omega t,$

$v_{max}=A\omega,$

$v=-v_{max}sin \omega t,$

$\frac{v}{v_{max}}=-sin \omega t=\frac 12,$

$sin\omega t=-\frac 12,~~$ $\Rightarrow~~$ $cos\omega t=\frac{\sqrt 3}{2},$

$x_1=\frac{A\sqrt3}{2}\approx2.60~\text{cm}.$