Question

If a particle executes shm with frequency f then frequency of osscilation of kinetic energy

Answer 1

Question :

If a particle executes shm with frequency f then frequency of osscilation of kinetic energy

Answers :

Let , a particle in SHM which Follows the equation :

$\sf\:x=A\sin\omega\:t$

and frequency of particle ,$\sf\:f=\dfrac{\omega}{2\pi}...(1)$

We know that ,

Velocity in SHM: It is defined as the time rate of change of the displacement of the particle at the given instant .

$\sf\:v=\dfrac{dx}{dt}$

$\sf\implies\:v=A\omega\cos\omega\:t....(2)$

We have to find ,frequency of osscilation of kinetic energy:

$\sf\:K=\dfrac{1}{2}\times\:mv^2$

Put the value of v form equation (2) ,then

$\sf\:K=\dfrac{1}{2}mA^2\omega^2\cos^2\omega\:t$

We know that :cos2x = 2 cos²x-1

$\sf\implies\:K=\dfrac{1}{2}mA^2\omega^2(\dfrac{\cos2\omega\:t+1}{2})$

$\sf\implies\:K=\dfrac{1}{4}mA^2\omega^2+\dfrac{1}{4}A^2\omega^2\:m\cos2\omega\:t$

Here , $\sf\dfrac{1}{4}mA^2\omega^2$ is constant .

Then ,

Kinetic energy is depend upon:

$\sf\implies\:K=\dfrac{1}{4}A^2\omega^2\:m\cos2\omega\:t$

Thus , $\sf\omega_k=2\omega...(3)$

Frequency of osscilation of Kinteic energy:

$\sf\:f'=\dfrac{\omega_k}{2\pi}$

Form equation (3)

$\sf\implies\:f'=\dfrac{2\omega}{2\pi}$

$\sf\implies\:f'=2f$

Therefore, the frequency with which its kinetic energy oscillates is 2f.

$\rule{200}2$

More About the topic:

Simple harmonic oscillation (SHM)

It is the simplest form of vibratory motion.

Answer 2

The frequency with which its kinetic energy oscillates is. Solution : During one complete oscillation, the kinetic energy will become maximum twice. Therefore the frequency of kinetic energy will be (2 f)