# 8.3.1 Derivation of TE Formula

To produce an oscillation with amplitude x0, the system must first be displaced from its equilibrium position to $\displaystyle x={{x}_{0}}$ . This requires an external force to pull or push against the restoring force. The work done by this external force corresponds to the (total) energy of oscillation, TE.

Recall that $\displaystyle a=-{{\omega }^{2}}x$  and the restoring force is $\displaystyle \displaystyle {{F}_{{net}}}=-m{{\omega }^{2}}x$ . Since the external force must match the restoring force at every position, $\displaystyle \displaystyle {{F}_{{ext}}}=m{{\omega }^{2}}x$ at every position.

Since the area under the Fx graph is work done, we can derive the formula for TE to be $\displaystyle TE=\frac{1}{2}({{x}_{0}})(m{{\omega }^{2}}{{x}_{0}})=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}$

If the restoring force is the only force acting on the system (i.e. no other dissipative forces), then the system will oscillate forever (after it has been set in motion). During the oscillation, the restoring force causes energy to be converted between the kinetic energy KE and potential energy PE of the system. From the principle of conservation of energy, we know that, $\displaystyle TE=KE+PE$

Maximum PE occurs at the extreme positions, where KE is zero.

Conversely, maximum KE occurs at the equilibrium position, where PE is zero. So

So $\displaystyle \frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}$ corresponds to not just TE, but also PEmax and KEmax. $\displaystyle TE=KE+PE=K{{E}_{{\max }}}=P{{E}_{{\max }}}=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}$