Question

A mass is oscillating with amplitude A at the end of a spring.

How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

Answer 1

x=A•sinωt
v =dx/dt = A•ω•cosωt
PE = KE
kx^2/2 = mv^2/2
k = m•ω^2
m•ω^2• (A•sinωt)^2/2 = m•( A•ω•cosωt)^2/2
(sinωt)^2 = (cosωt)^2
(tan ωt)^2 = 1
ωt = π/4
x=A•sinωt = A•sin(π/4) =0.707•A

Answer 2

It is not clear