Question

A spring with a spring constant of 1200 N/m has a 55-g ball at its end. The energy of the system is 5.5J. What is the amplitude of vibration?

Answer 1

The energy of the system is E=5.5 J. This energy is the same at every moment of the oscillation. When the stretch x of the spring is maximum (so, when the stretch is equal to the amplitude: x=A) the velocity of the spring is zero, so all this energy is just elastic potential energy of the spring:

`E= (1)/(2)kA^2`
From which we find

`A= \sqrt{ (2E)/(k) }= \sqrt{ (2 \cdot 5.5 J)/(1200 N/m) }=0.096 m`