Question

What is the magnitude of the velocity when the elastic potential energy is equal to the kinetic energy? (Assume that U=0 at equilibrium.)

Answer 1

Answer:

Explanation:

General Equation of SHM is given by

$x=A\cos \omega t$

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

where x=position of particle

A=maximum Amplitude

$\omega =$angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. $\frac{1}{2}kA^2$

where k=spring constant

Potential Energy is given by $U=\frac{1}{2}kx^2$

also it is given that Potential Energy(U) is equal to Kinetic Energy(K)

Total Energy$=K+U$

Total$=2U=2\times \frac{1}{2}kx^2$

$\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2$

$x=\pm \frac{A}{\sqrt{2}}$

at $x=\frac{A}{\sqrt{2}}$

velocity is $v=\frac{A\omega}{\sqrt{2}}$