14 Jun 2025

Simple Harmonic Motion and Energy

The SHM equation and solution, phase, energy exchange, and time-averaged energies.

oscillations simple-harmonic-motion energy

Simple harmonic motion occurs when the restoring force is directly proportional to the displacement from stable equilibrium and points toward that equilibrium. For a displacement $x$,

\[F=-kx, \qquad k>0.\]

Newton’s second law gives

\[m\ddot x=-kx.\]

After division by $m$,

\[\boxed{\ddot x+\omega_0^2x=0}, \qquad \boxed{\omega_0=\sqrt{\frac{k}{m}}}.\]

Solution of the SHM equation

Try a solution $x=e^{\lambda t}$. Substitution gives

\[(\lambda^2+\omega_0^2)e^{\lambda t}=0.\]

Since $e^{\lambda t}\ne0$,

\[\lambda^2+\omega_0^2=0, \qquad \lambda=\pm i\omega_0.\]

The real general solution is therefore

\[x=C\cos\omega_0t+D\sin\omega_0t.\]

Writing $C=A\cos\phi$ and $D=-A\sin\phi$ gives

\[\boxed{x=A\cos(\omega_0t+\phi)}.\]

Here $A\ge0$ is the amplitude and $\phi$ is the initial phase. One complete cycle corresponds to an increase of $2\pi$ in phase, so

\[\omega_0T=2\pi,\]

and hence

\[\boxed{T=\frac{2\pi}{\omega_0}=2\pi\sqrt{\frac{m}{k}}}.\]

Differentiation gives

\[v=\dot x=-A\omega_0\sin(\omega_0t+\phi),\] \[a=\ddot x=-A\omega_0^2\cos(\omega_0t+\phi) =-\omega_0^2x.\]

Thus the speed is greatest at equilibrium and zero at either turning point, whereas the acceleration magnitude is greatest at a turning point and zero at equilibrium.

Kinetic, potential, and total energies

Since $F=-dU/dx$,

\[-\frac{dU}{dx}=-kx \quad\Longrightarrow\quad \frac{dU}{dx}=kx.\]

Choosing $U=0$ at $x=0$ and integrating,

\[U=\int_0^x kx'\,dx'=\frac12kx^2.\]

Substitution of the SHM displacement gives

\[\boxed{U(t)=\frac12kA^2\cos^2(\omega_0t+\phi)}.\]

The kinetic energy is

\[K=\frac12mv^2 =\frac12mA^2\omega_0^2\sin^2(\omega_0t+\phi).\]

Because $m\omega_0^2=k$,

\[\boxed{K(t)=\frac12kA^2\sin^2(\omega_0t+\phi)}.\]

Adding the two energies and using $\sin^2\alpha+\cos^2\alpha=1$,

\[E=K+U =\frac12kA^2\left(\sin^2(\omega_0t+\phi)+\cos^2(\omega_0t+\phi)\right),\]

so

\[\boxed{E=\frac12kA^2=\text{constant}}.\]

Time-average energies

For any starting time $t_0$, the average over one full period is

\[\left\langle\cos^2(\omega_0t+\phi)\right\rangle =\frac1T\int_{t_0}^{t_0+T}\cos^2(\omega_0t+\phi)\,dt.\]

Using $\cos^2\alpha=(1+\cos2\alpha)/2$,

\[\left\langle\cos^2(\omega_0t+\phi)\right\rangle =\frac12+\frac{1}{2T} \left[\frac{\sin(2\omega_0t+2\phi)}{2\omega_0}\right]_{t_0}^{t_0+T}.\]

Since $2\omega_0T=4\pi$, the sine has the same value at both limits. Therefore

\[\left\langle\cos^2(\omega_0t+\phi)\right\rangle=\frac12.\]

The same calculation gives $\langle\sin^2(\omega_0t+\phi)\rangle=1/2$. Hence

\[\boxed{\langle K\rangle=\frac14kA^2}, \qquad \boxed{\langle U\rangle=\frac14kA^2},\]

and, because the total energy is constant,

\[\boxed{\langle E\rangle=\frac12kA^2}.\]

The differential equation, energy conservation, and full-period averages are verified in the Maxima worksheet; every printed residual is zero.