Question

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

Answer 1

The moment in which both energies will have the same value, will be the moment in which in the path of the amplitude of the spring mass system, it is in the middle, that is to say if the two forces are equivalent

$KE = U$

they are in balance when,

$K = \frac{E}{2} \text{ and }U = \frac{E}{2}$

Therefore replacing the definition of Elastic potential energy for a Spring-Mass system, we have that

$U = \frac{1}{2} kx^2 = \frac{1}{4} kA^2$

Here,

k = Spring constant

x = Displacement

A = Amplitude

Rearranging to find the displacement:

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