Energy of the Frictionless Harmonic Oscillator
For the frictionless oscillator, mechanical energy is conserved. The total energy is the sum of kinetic energy and spring potential energy:
$$ E = T + U $$Using
$$ x(t) = A\cos(\omega t - \delta) $$ $$ \dot{x}(t) = -A\omega\sin(\omega t - \delta) $$the kinetic energy is
$$ T = \frac{1}{2}m\dot{x}^2 $$ $$ T = \frac{1}{2}mA^2\omega^2\sin^2(\omega t - \delta) $$Since \( \, \displaystyle \omega^2 = \frac{k}{m} \, \), this becomes
$$ T = \frac{1}{2}kA^2\sin^2(\omega t - \delta) $$The spring potential energy is
$$ U = \frac{1}{2}kx^2 $$ $$ U = \frac{1}{2}kA^2\cos^2(\omega t - \delta) $$Therefore the total energy is
$$ E = \frac{1}{2}kA^2\sin^2(\omega t - \delta) + \frac{1}{2}kA^2\cos^2(\omega t - \delta) $$ $$ E = \frac{1}{2}kA^2\left(\sin^2(\omega t - \delta) + \cos^2(\omega t - \delta)\right) $$ $$ E = \frac{1}{2}kA^2 $$Although \(T\) and \(U\) change with time, their sum remains constant. Energy flows back and forth between kinetic energy and potential energy: the oscillator has maximum potential energy at maximum displacement, and maximum kinetic energy as it passes through equilibrium.
Over one period,
$$ T = \frac{2\pi}{\omega} $$the time-averaged kinetic and potential energies are equal:
$$ \langle T \rangle = \frac{1}{T}\int_0^T \frac{1}{2}kA^2\sin^2(\omega t - \delta)\,dt $$ $$ \langle T \rangle = \frac{1}{4}kA^2 $$ $$ \langle U \rangle = \frac{1}{T}\int_0^T \frac{1}{2}kA^2\cos^2(\omega t - \delta)\,dt $$ $$ \langle U \rangle = \frac{1}{4}kA^2 $$Thus, averaged over a full cycle, half of the oscillator's energy is kinetic and half is stored in the spring.