## Essential University Physics: Volume 1 (3rd Edition)

$T = 2\pi\sqrt{\frac{m}{k}}$
We find: $\frac{dx}{dt} = A\omega cos \omega t$ Thus, it follows: $\frac{d^2x}{dt^2} = -A\omega^2 sin \omega t$ Using equation 13.3 and the fact that $x=Asin\omega t$, we find: $m(-A\omega^2 sin \omega t)=-k Asin\omega t$ Since $T = \frac{2\pi}{\omega}$, we find: $T = 2\pi\sqrt{\frac{m}{k}}$