The frequency at which kinetic energy changes | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t)$.The kinetic energy is $K.E. = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) = \frac{

Vedclass · Accepted Answer

$2f$

Vedclass · Answer

(C) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t)$.The kinetic energy is $K.E. = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) = \frac{1}{4} m \omega^2 A^2 (1 + \cos(2\omega t))$.The potential energy is $P.E. = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t) = \frac{1}{4} k A^2 (1 - \cos(2\omega t))$.Both kinetic and potential energies oscillate with an angular frequency of $2\omega$.Since the frequency $f = \frac{\omega}{2\pi}$,the frequency of energy oscillation is $f' = \frac{2\omega}{2\pi} = 2f$.Therefore,the frequency at which kinetic energy changes into potential energy is $2f$.

The frequency at which kinetic energy changes into potential energy in a simple harmonic motion ($S$.$H$.$M$.) with frequency $f$ is:

Similar Questions

Draw a graph to show the variation of $PE$,$KE$ and total energy of a simple harmonic oscillator with displacement.

Starting from the mean position,a body oscillates simple harmonically with a period of $2\,s$. After what time will its kinetic energy be $75\%$ of the total energy?

$A$ body is executing $S$.$H$.$M$. Its potential energy is '$P_1$' and '$P_2$' at displacements '$x$' and '$y$' respectively. The potential energy at displacement $(x+y)$ is

$A$ body performs linear simple harmonic motion of amplitude $A$. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?

The variation of kinetic energy $(KE)$ of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position $(A)$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module