The Potential Energy of a Spring with a Constant Of k

Short Answer

The potential energy of a spring with a constant of $k$ , a maximum amplitude of $A$ , and an angular velocity of $\omega$ is given by $P=\frac{1}{2} k A^{2} \cos^{2} \omega t$
(joules) where $t$ is in seconds. Find the rate of change in the potential energy of a mass on a spring with a constant of $1.2 \mathrm{~N} / \mathrm{m}$ , a maximum amplitude of $0.53 \mathrm{~m}$ , and an angular velocity of $0.031 \mathrm{rad} / \mathrm{s}$ , when $t=5.8 \mathrm{~s}$ .