Power of a Wave

Waves are oscillatory disturbances in physical quantities, like light waves, sound waves, or transverse oscillations of a string. These disturbances take energy to create and propagate, in order to move the constituent particles or change the electric/magnetic fields. The power of a wave is therefore energy transported per unit time by the oscillations of a particular wave. The derivation of a formula for the power depends on the medium -- for light waves, the power is measured by the poynting vector, whereas for oscillations on a string, the power can be computed directly by balancing forces using Newton's second law. However, for all types of waves, the formula and physical meaning of the power takes similar forms, typically depending on the square amplitude of the waves among other factors.

To derive the power carried by the oscillations of a string, first note that a general solution to the wave equation for the amplitude of transverse oscillations of the string is:

$y = A\sin { (\omega t-kx+\phi ) }$

where the constants above are amplitude $A$, angular frequency $\omega$, wavenumber $k$, and phase shift $\phi$.

From considering the forces acting on a small element of the string, one can find the power carried by that element and from there the power carried by the oscillations throughout the string. See below diagram:

A small piece of an oscillating string is accelerated by tension forces due to the displacements of nearby parts of the string.

The power carried by a particle simply obeys the formula $P = \vec{F} \cdot \vec{v}$. Since the force acting on a small part of the string is just the tension $T$ due to a nearby piece of the string, this power is:

$P = Tv \cos (90 - \theta) = Tv \sin \theta,$

where $\theta$ is the angle between the tension force and the horizontal as in the above diagram. For small oscillations that satisfy the wave equation, $\sin \theta \approx \theta \approx \tan \theta$ by the small-angle approximation. So the power can be rewritten in terms of the displacement $y$ of the small piece of string as:

$P = -T\frac { \partial y }{ \partial t } \frac { \partial y }{ \partial x }.$

Using the general solution to the wave equation written above, this is:

$P = -T\left( \omega A\cos { (\omega t-kx+\phi ) } \right) \left( -kA\cos { (\omega t-kx+\phi ) } \right) = Tk{ A }^{ 2 }\omega \cos ^{ 2 }{ (\omega t-kx+\phi ) } .$

Time-averaging over a period of oscillation, one finds the average power transmitted to be:

$\langle P \rangle = Tk{ A }^{ 2 }\omega \langle \cos ^{ 2 }{ (\omega t-kx+\phi ) } \rangle = \frac { Tk{ A }^{ 2 }\omega }{ 2 },$

since the average of cosine squared over an oscillation is $\frac12$.

This formula can also be written with different variables, using the fact that wave velocity is given in terms of mass density $\mu$ and tension $T$ by $v = \sqrt{\frac{T}{\mu}}$ and that by definition $k = \frac{\omega}{v}$. In this case the power is given by:

$\langle P \rangle = \frac12 \mu v \omega^2 A^2.$

When a wave travels along a string, energy is transmitted along the direction of propagation of the wave, in the form of potential energy and kinetic energy of the string oscillation. The total energy transferred in a given time is given in terms of the average power as an integral:

$E(t) = \int_0^t \langle P \rangle dt.$

Show that the energy transferred in one period of an oscillating wave on a string is equal to the energy associated with one wavelength of oscillation of a string wave:

$E_{\lambda} = \frac12m\omega^2 A^2 = \frac { \mu \lambda { \omega }^{ 2 }{ A }^{ 2 } }{ 2 },$

where $\mu$ is the mass per unit length. Note that the wave velocity is given in terms of $\mu$ and $T$ by $v = \sqrt{\frac{T}{\mu}}$ and that by definition $k = \frac{\omega}{v}$.

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Solution:

Computing the integral above with $T_0$ the period of oscillation, $T_0 = \frac{2\pi}{\omega}$:

$E(T_0) = \int_0^{T_0} \langle P \rangle dt = \int_0^{T_0} \frac12 T k A^2 \omega dt = \frac12 T k A^2 \omega T_0.$

Substituting in for $T$, $T_0$ and $k$, one finds

$E(T_0) = \frac12 \mu v^2 \frac{\omega}{v} A^2 \omega \frac{2\pi}{\omega} = \frac12 \mu v \omega A^2 (2\pi) = \frac12 \mu \left(\frac{\lambda}{2\pi}\right) \omega^2 A^2 (2\pi) = \frac12 \mu \lambda \omega^2 A^2.$

Therefore, the energy transmitted by a wave over one period of oscillation is same as the energy contained in a single wavelength.

This formula can also be proved using dimensional analysis. Note that the energy transmitted by the wave depends on the source of the waves. The source determines the amplitude of the wave as the source oscillates. The frequency, i.e., how rapid the oscillations are transmitted, also depends on the source. The mass density of the string determines the amount of inertia of the string; a heavier string oscillating at a particular frequency carries more energy because it takes more energy to move something with more inertia. Properties of the string likes its tension just characterize how massive the string is; the role of the tension is that it is the force internally transmitting the wave, so a greater tension corresponds to greater mass at a fixed velocity, consistent with the above formulae.

From the analysis, one can conclude that there are three fundamental quantities that fix the energy carried by a wave per unit wavelength. They are the:

1. Mass per unit length of the string
2. Amplitude of oscillation
3. Frequency of oscillation

Thus, the energy per unit wavelength $u$ is proportional to:

$u \propto \mu^{a} A^b \omega^c,$

for some exponents $a$, $b$, and $c$. By requiring that the right-hand side have units of energy density, these exponents can be fixed to be $a=1$, $b=2$, and $c=2$, implying $u \propto \mu \omega^2 A^2$, which is the correct energy per unit wavelength up to the prefactor of one-half.