Question

A particle executes SHM of frequency $f$. The frequency of its kinetic energy is:

• A. $f$
• B. $\frac{f}{2}$
• C. $2f$
• D. zero

Answer 1

The correct option is C $2f$

$KE=\frac{1}{2}m{\omega }^2(x_0^2-x^2)=\frac{1}{2}m{\omega }^2(x_0^2-x_0^2{\sin }^2\omega t)=\frac{1}{2}m{\omega }^2{x}_0^2(1-{\sin }^2\omega t)=\frac{1}{2}m{\omega }^2{x}_0^2{\cos }^2\omega t=\frac{1}{2}m{\omega }^2{x}_0^2\left(\frac{1+\cos 2\omega t}{2}\right)=\frac{1}{4}m{\omega }^2{x}_0^2+\frac{1}{4}m{\omega }^2{x}_0^2\cos 2\omega t$
frequency of above i.e. frequency of K.E. is $2\omega$ or $2f$
alternate solution: We know that in SHM particle has minimum(zero) velocity at extreme points and maximum velocity at mean point. With the help of diagram we can easily see that in one time period of pendulum (starting from left extreme point) K.E. reaches zero to maximum two times i.e. frequency of
K.E. will be twice of the SHM frequency. So the correct answer is 'D'