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:
$E= \frac{1}{2}kA^2$
From which we find
$A= \sqrt{ \frac{2E}{k} }= \sqrt{ \frac{2 \cdot 5.5 J}{1200 N/m} }=0.096 m$