Question

A particle executes simple harmonic motion with a frequency $v$. The frequency with which the kinetic energy oscillates is

• A. $v/2$
• B. $v$
• C. $2v$
• D. $Zero$

Answer 1

The correct option is C $2v$

It is given that, the a particle execute SHM with frequency v. The equation of particle executing SHM is

$x=A\sin \omega t$

Velocity of particle is

$v=\frac{dx}{dt}$

$v=\frac{d}{dt}\left(A\sin \omega t\right)$

$v=A\omega \cos \omega t$

So, the kinetic energy is:

$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(A\omega \cos \omega t\right)}^2=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$

$K=\frac{1}{2}m{A}^2{\omega }^2\left(\frac{1+\cos 2\omega t}{2}\right)$

$K=\frac{1}{4}m{\omega }^2{A}^2(1+\cos 2\omega t)$

So, from equation the frequency of oscillation of kinetic energy is

${\omega }_K=2\omega$

Frequency of oscillation of kinetic energy is double as that of oscillation of particle.