Answer to Question #177027 in Mechanics | Relativity for Lesego Malatsi

Question

We have a horizontal spring with a mass attached to oscillating with a maximum displacement of A on either side. At which intermediate displacement the kinetic energy would equal to the potential energy

Answer 1

Full energy is

$E=E_k+E_p=\frac{mv^2}{2}+\frac{kx^2}{2}=\frac{kA^2}{2},$

$E_k=E_p,\implies$

$2E_p=E,\implies$

$kx^2=\frac{kA^2}{2},\implies$

$x=±\frac{A\sqrt 2}{2}.$