Calculating the energy of a harmonic wave

Consider an ideal string stretched along the x direction and oscillating transversely in the y direction. The string has tension T and linear density ρ (mass per unit length, in kg/m). In this first section the string carrying the harmonic wave is so long that reflections at the ends can be neglected; the string of length L fixed at both ends is treated in the section on normal modes.

We want to calculate the total energy of the string. The total energy density is the sum of its kinetic energy and its potential energy

${\displaystyle d{E}_{\mathrm{t}\mathrm{o}\mathrm{t}}(x,t)=d{E}_{\mathrm{c}\mathrm{i}\mathrm{n}}(x,t)+d{E}_{\mathrm{p}\mathrm{o}\mathrm{t}}(x,t)}$.

From this, the total energy is obtained by summing over the whole string and averaging over one period of the oscillation

${\displaystyle E_{\mathrm{t}\mathrm{o}\mathrm{t}}=⟨E_{\mathrm{c}\mathrm{i}\mathrm{n}}(t)⟩+⟨E_{\mathrm{p}\mathrm{o}\mathrm{t}}(t)⟩}$.

The result must be that, without friction, the total energy is a constant of the motion, that is, it does not depend on time.

The calculation is particularly simple in an important case, namely when the wave propagating along the string is harmonic. In this case the motion is

${\displaystyle y(x,t)=A\sin (kx-\omega t)}$.

Since trigonometry tells us that

${\displaystyle {\cos }^2(x)=\frac{1+\cos (2x)}{2}}$,

all the averages are very simple to calculate. Over one period the cosine function has zero average, so the only term that contributes is the constant term (equal to 1/2). So, in the particular case of harmonic waves, not only the total energy is constant, but also the energy per unit length and the energy per unit time.

This result is also particularly important because, through Fourier's theorem, it allows us to break down the energy of a complex wave into the sum of the energies of its harmonic components.

Kinetic energy

A small piece of string of length dx has mass ρdx. This small piece can be treated in every respect as a point mass, so its kinetic energy is

${\displaystyle \mathrm{d}{E}_{\mathrm{c}\mathrm{i}\mathrm{n}}(x,t)=\frac{1}{2}m{v}_y^2=\frac{1}{2}\rho v_y^2(x,t)dx}$

From the solution ${\displaystyle y(x,t)=A\sin (kx-\omega t)}$ we can easily obtain the transverse velocity of the string

${\displaystyle v_y(x,t)=-\omega A\cos (kx-\omega t)}$.

Note: v_y is NOT the speed of sound in the string, which we denote instead by c.

The kinetic energy of a piece of string as long as the wavelength λ is therefore found by adding up the kinetic energies of the small elements of string of length dx along the whole string.

${\displaystyle E_{\mathrm{c}\mathrm{i}\mathrm{n}}(t)={\displaystyle \underset{0}{\overset{\lambda }{\int }}}\mathrm{d}{E}_{\mathrm{c}\mathrm{i}\mathrm{n}}(x,t)=\frac{1}{4}{A}^2\rho {\omega }^2\lambda }$

Now we take the average over one period

${\displaystyle E_{\mathrm{c}\mathrm{i}\mathrm{n}}=\frac{\omega }{2\pi }{\displaystyle \underset{0}{\overset{\frac{2\pi }{\omega }}{\int }}}{E}_{\mathrm{c}\mathrm{i}\mathrm{n}}(t)\mathrm{d}t=\frac{1}{4}{A}^2\rho {\omega }^2\lambda }$

Potential energy

The element of potential energy for the oscillating string is

${\displaystyle d{E}_{\mathrm{p}\mathrm{o}\mathrm{t}}(x,t)=\frac{1}{2}T{\left(\frac{\partial y(x,t)}{\partial x}\right)}^2\mathrm{d}x}$,

where T is the tension.

Substituting again ${\displaystyle A\sin (kx-\omega t)}$ for ${\displaystyle y(x,t)}$, we obtain

${\displaystyle \frac{\partial y(x,t)}{\partial x}=kA\cos (kx-\omega t)}$,

and therefore, summing over the whole string,

${\displaystyle E_{\mathrm{p}\mathrm{o}\mathrm{t}}(t)={\displaystyle \underset{0}{\overset{\lambda }{\int }}}\mathrm{d}{E}_{\mathrm{p}\mathrm{o}\mathrm{t}}(x,t)=\frac{1}{4}{A}^{2}T{k}^2\lambda }$.

Recalling that ${\displaystyle T=\rho c^2}$ and ${\displaystyle k=\frac{2\pi }{\lambda }}$, we see that the previous expression reduces to

${\displaystyle E_{\mathrm{p}\mathrm{o}\mathrm{t}}(t)=\frac{1}{4}{A}^2\rho {\omega }^2\lambda =E_{\mathrm{c}\mathrm{i}\mathrm{n}}(t)}$,

and consequently

${\displaystyle E_{\mathrm{p}\mathrm{o}\mathrm{t}}=E_{\mathrm{c}\mathrm{i}\mathrm{n}}}$.

Final result

In summary:

• for a harmonic wave, the total energy of the vibrating string is always divided equally between kinetic and potential energy
• the total energy of a piece of string of length λ is proportional to the square of the amplitude of the oscillation:

${\displaystyle E_{\mathrm{t}\mathrm{o}\mathrm{t}}=\frac{1}{2}{A}^2\rho {\omega }^2\lambda }$

Calculating the energy of a normal mode

The energy of a string of length L oscillating in its n-th normal mode can be calculated exactly as for the harmonic wave, of which the normal mode is just a special case.

The n-th normal mode of the string is characterised by a law of motion of the type

${\displaystyle y(x,t)=A_n\sin ({\omega }_{n}t)\sin (k_{n}x)}$.

The only difference is that now the sum along the string gives a time-dependent result, not a constant. However, the sum of the kinetic and potential terms must always be a constant without dissipative forces (friction).

Kinetic energy

${\displaystyle E_{\mathrm{c}\mathrm{i}\mathrm{n}}(t)={\displaystyle \underset{0}{\overset{L}{\int }}}\mathrm{d}{E}_{\mathrm{c}\mathrm{i}\mathrm{n}}(x,t)=\frac{1}{2}{A}_n^2\rho {\omega }_n^2{\cos }^2({\omega }_{n}t){\displaystyle \underset{0}{\overset{L}{\int }}}{\sin }^2(k_{n}x)\mathrm{d}x}$

Potential energy

${\displaystyle E_{\mathrm{p}\mathrm{o}\mathrm{t}}(t)={\displaystyle \underset{0}{\overset{L}{\int }}}\mathrm{d}{E}_{\mathrm{p}\mathrm{o}\mathrm{t}}(x,t)=\frac{1}{2}{A}_n^{2}T{k}_n^2{\sin }^2({\omega }_{n}t){\displaystyle \underset{0}{\overset{L}{\int }}}{\cos }^2(k_{n}x)\mathrm{d}x}$

Final result

Recalling that ${\displaystyle {\sin }^2(x)+{\cos }^2(x)=1}$ and ${\displaystyle T=\rho c^2}$, we see that the sum of the kinetic and potential energies is always equal to

${\displaystyle E_{\mathrm{t}\mathrm{o}\mathrm{t}}=\frac{1}{4}{A}_n^2\rho {\omega }_n^{2}L=\frac{1}{4}m{A}_n^2{\omega }_n^2}$,

where m is the mass of the string.

Power needed to generate a harmonic wave

The power is the energy that must be supplied per unit time to generate the wave. It can easily be obtained from the result for the harmonic wave, remembering that one period is needed to produce an oscillation of wavelength λ. So it is enough to divide the total energy of a piece of length λ by the period, that is, to multiply by the frequency

${\displaystyle f=\frac{c}{\lambda }}$,

obtaining

${\displaystyle W=\frac{1}{2}{A}^2\rho {\omega }^{2}c}$

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