6.3 Energy in Simple Harmonic Motion

As we have seen back in chapter3, Hooke’s law is a conservative force, so the total energy is conserved. Let’s show this explicitly for the mass on a spring. From before, we know the total energy is given by

	$\displaystyle E$	$\displaystyle=K+U_{s}$		(6.32)
		$\displaystyle=\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}.$		(6.33)

Substituting equations 6.6 and 6.7, we get

$\displaystyle E$	$\displaystyle=\frac{1}{2}mx_{m}^{2}\omega^{2}\sin^{2}(\omega t+\phi)+\frac{1}{% 2}kx_{m}^{2}\cos^{2}(\omega t+\phi)$	(6.34)
	$\displaystyle=\frac{1}{2}mx_{m}^{2}\left(\frac{k}{m}\right)\sin^{2}(\omega t+% \phi)+\frac{1}{2}kx_{m}^{2}\cos^{2}(\omega t+\phi)$	(6.35)
	$\displaystyle=\frac{1}{2}kx_{m}^{2}\left(\sin^{2}(\omega t+\phi)+\cos^{2}(% \omega t+\phi)\right)$	(6.36)
	$\displaystyle=\frac{1}{2}kx_{m}^{2}.$	(6.37)

Where in the last line we have used the identity $\sin^{2}A+\cos^{2}A=1$ . This is independent of time, so the total energy is conserved as we found before. By using trigonometric identities for $\sin^{2}$ and $\cos^{2}$ , we have

	$\displaystyle K$	$\displaystyle=\frac{1}{2}kx_{m}^{2}\left(\frac{1}{2}-\frac{1}{2}\cos(2(\omega t% +\phi))\right)$		(6.38)
	$\displaystyle U$	$\displaystyle=\frac{1}{2}kx_{m}^{2}\left(\frac{1}{2}+\frac{1}{2}\cos(2(\omega t% +\phi))\right).$		(6.39)

This shows that the energies oscillate sinusoidally as well, but with a frequency that is twice the resonant frequency.

6.3.1 General Formulae for SHM

In general, for an system exhibiting SHM in the quantity $A$ , the equation of motion will be

\frac{\operatorname{\mathrm{d}}\!^{2}A}{\operatorname{\mathrm{d}}\!t^{2}}=-% \omega^{2}A,

(6.40)

and the potential and kinetic energies will have the form

U=\frac{1}{2}\alpha A^{2},\quad K=\frac{1}{2}\beta\left(\frac{\operatorname{% \mathrm{d}}\!A}{\operatorname{\mathrm{d}}\!t}\right)^{2},

(6.41)

where

\omega=\sqrt{\frac{\alpha}{\beta}}.

(6.42)

Example 6.5.

Let’s solve the pendulum using the energy approach.

The rotational kinetic energy of a pendulum is given by

K=\frac{1}{2}I\omega^{2}=\frac{1}{2}mL^{2}\omega^{2},

(6.43)

and the gravitational potential energy is

U=mgL(1-\cos\theta)\approx mgL\frac{\theta^{2}}{2},

(6.44)

where we have used the small-angle approximation $1-\cos\theta\approx\frac{\theta^{2}}{2}$ . Then using the formulae above (6.41 and 6.42), we get

\omega=\sqrt{\frac{mgL}{mL^{2}}}=\sqrt{\frac{g}{L}},

(6.45)

which is what we found before.

The following example is more complicated than anything we have seen so far.

Example 6.6.

Consider a small marble of mass $m$ and radius $r$ rolling without slipping in a large spherical dish of radius $R$ . Does the marble move with simple harmonic motion? If so, find the resonant frequency.

We will analyse the motion of the marble using energies, which is much simpler than solving this problem using forces. The height that the centre of mass changes by when it moves around in the dish is $h=(R-r)(1-\cos\theta)$ , so the potential energy is

U=mgh=mg(R-r)(1-\cos\theta)\approx mg(R-r)\frac{\theta^{2}}{2}.

(6.46)

The fact that $R\gg r$ justifies the use of the small-angle approximation. The marble has two different kinetic energies, its translation (which is actually rotation of its centre of mass around the pivot) and rotation around its own centre of mass. The marble is a solid sphere, so its moment of inertia about its centre is $\frac{2}{5}mr^{2}$ and the latter kinetic energy is given by

K_{\text{rot}}=\frac{1}{2}I\dot{\phi}^{2}=\frac{1}{2}\left(\frac{2}{5}mr^{2}% \right)\dot{\phi}^{2},

(6.47)

where $\dot{\phi}$ is the angular velocity of the marble around its centre of mass. The former kinetic energy is given by

K_{\text{tran}}=\frac{1}{2}I_{\text{pivot}}\dot{\theta}^{2}=\frac{1}{2}\left(m% (R-r)^{2}+\frac{2}{5}mr^{2}\right)\dot{\theta}^{2},

(6.48)

where we have used the parallel axis theorem to get the moment of inertia about the pivot point. We would like to find a relation between $\dot{\phi}$ and $\dot{\theta}$ to simplify the total kinetic energy. Note that when the marble rolls a distance $s=R\theta$ in the dish, which is equal to $r(\phi+\theta)$ . Then we have

	$\displaystyle\phi$	$\displaystyle=\frac{R-r}{r}\theta$		(6.49)
	$\displaystyle\implies\dot{\phi}$	$\displaystyle=\frac{R-r}{r}\dot{\theta}.$		(6.50)

Thus the total kinetic energy is

	$\displaystyle K=K_{\text{tran}}+K_{\text{rot}}$	$\displaystyle=\frac{1}{2}\left[m(R-r)^{2}\dot{\theta}^{2}+\frac{2}{5}mr^{2}% \dot{\theta}^{2}+\frac{2}{5}mr^{2}\left(\frac{R-r}{r}\dot{\theta}\right)^{2}\right]$		(6.51)
		$\displaystyle=\frac{1}{2}\left[\frac{7}{5}m(R-r)^{2}+\frac{2}{5}mr^{2}\right]% \dot{\theta}^{2}.$		(6.52)

These energies satisfy the forms in equation 6.41, so the marble does exhibit simple harmonic motion. The resonant frequency is given by equation 6.42:

\omega=\sqrt{\frac{mg(R-r)}{\frac{7}{5}m(R-r)^{2}+\frac{2}{5}mr^{2}}}.

(6.53)