Answer to Question #284566 in Mechanics | Relativity for JoShua

Question

A mass is oscillating with amplitude 𝐴 at the end of the spring. How far is this mass from the equilibrium if the elastic potential energy equals 𝛾 times the kinetic energy?

Answer 1

total mechanical energy:

$\frac{kA^2}{2}=\frac{kx^2}{2}+\frac{mv^2}{2}$

we have:

$\frac{kx^2}{2}=\gamma\frac{mv^2}{2}$

then:

$\frac{kA^2}{2}=\frac{kx^2}{2}+\gamma\frac{kx^2}{2}$

$A^2=(1+\gamma)x^2$

displacement from the equilibrium:

$x=\frac{A}{\sqrt{1+\gamma}}$