Energy and Waves

Energy density of a plane progressive wave

By the energy density of a plane progressive wave we mean that the total energy(kinetic+potential) per unit volume of the medium through which the wave is passing. It is usually denoted by letter E. As we know, the displacement of a particle of the medium at distance \(x\) from the origin at an instant \(t\) is given by \(y=a\sin 2\pi /\lambda (vt-x)\) where the symbols have their usual meanings. The velocity of the particle is then \(u=\frac{\mathrm{dy} }{\mathrm{d} t}=\frac{2\pi v}{\lambda}a \cos \frac{2\pi}{\lambda}(vt-x)\) and acceleration of the particle is \(\frac{\mathrm{du} }{\mathrm{d} t}=\frac{\mathrm{d^2y} }{\mathrm{d} t^2}=\frac{-4\pi^2 v^2}{\lambda^2}a \sin \frac{2\pi}{\lambda}(vt-x)=\frac{-4\pi^2v^2}{\lambda^2}y\). Now, if we consider unit volume of the medium in the form of an extremely thin eleent of the medium, parallel to the wave front, we have, the mass of the element is \(\rho\), its density. Since, the layer is very thin, the velocity of all the particles in it may be assumed to be the same. Therefore, the KE per unit volume of the medium is,\begin{align*} =1/2\times mass \times velocity^2 \end{align*}\begin{align*} =1/2 \rho u^2 \end{align*}\begin{align*} =1/2\rho\frac{4\pi^2 v^2}{\lambda^2}a^2 \cos^2 \frac{2\pi}{\lambda}(vt-x)\end{align*}\begin{align*} =\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \cos^2 \frac{2\pi}{\lambda}(vt-x)---1 \end{align*}\begin{align*} \therefore Force \space acting \space on \space unit \space volume=mass \times acceleration \end{align*}\begin{align*} =\rho\frac{\mathrm{d^2y} }{\mathrm{d} t^2}=\frac{4\pi^2v^2\rho}{\lambda^2}y \end{align*} Here, we have ignored the negative sign which simply indicates that aceeleration is directed opposite to displacement. Therefore, work done in a small displacement \(dy\) of the layer is force times displacement \(\frac{4\pi^2v^2\rho}{\lambda^2}ydy\). Hence, work done during the whole displacement from \(0\) to \(y\) is,\begin{align*} \int_{0}^{y}\frac{4\pi^2v^2\rho}{\lambda^2}ydy\end{align*}\begin{align*} =\frac{4\pi^2v^2\rho y^2}{\lambda^22} \end{align*}\begin{align*} =\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \sin^2 \frac{2\pi}{\lambda}(vt-x)\end{align*} This work done must obviously stored up in the medium in the form of potential energy. So that, PE per unit volume of the medium is \(\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \sin^2 \frac{2\pi}{\lambda}(vt-x)\) Therefore, the total energy per unit volume of the medium or energy density of the plane progressive wave is the sum of KE and PE given by,\begin{align*} E=\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \cos^2 \frac{2\pi}{\lambda}(vt-x)+\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \sin^2 \frac{2\pi}{\lambda}(vt-x) \end{align*}\begin{align*} =\frac{2\pi^2 v^2\rho}{\lambda^2}a^2=2\pi^2(\frac{v}{\lambda})^2\rho a^2---2 \end{align*}\begin{align*} Also, E=2\pi^2f^2a^2\rho---3 \end{align*} where, f is the frequency of the wave. In case of cross section of the beam is unity, equations 2 and 3 give the total energy of the wave per unit length. Although, both kinetic and potential energies of the wave depend upon the values of \(x\) and \(t\), its total energy or the energy density is quite independent of either.

Distribution of energy in a plane progressive wave

To see how the energy of a plane progressive wave is distributed over a complete wavelength, let us find the average KE and PE over a complete wavelength. Average KE over a complete wavelength\((\lambda)\) is,\begin{align*} \frac{1}{\lambda}\int_{0}^{\lambda}\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \cos^2 \frac{2\pi}{\lambda}(vt-x)dx\end{align*}\begin{align*} =\frac{2\pi^2 f^2\rho a^2}{\lambda^2}\int_{0}^{\lambda}\frac{1}{2}[1+\cos \frac{4\pi}{\lambda}(vt-x)]dx\end{align*}\begin{align*} here, \int_{0}^{\lambda}\cos \frac{4\pi}{\lambda}(vt-x)dx=\frac{-\lambda}{4\pi}[\sin(\frac{4\pi}{\lambda}(vt-4x))-sin\frac{4\pi}{\lambda}vt]=0 \end{align*} So, the average KE of the wave over a complete wavelength is,\begin{align*} \frac{2\pi^2 f^2\rho a^2 \lambda}{2\lambda}=\pi^2f^2a^2\rho=1/2\space total\space energy \end{align*} Similarly, average PE of the wave over a complete wavelengths is,\begin{align*} \frac{1}{\lambda}\int_{0}^{\lambda}\frac{2\pi^2 v^2\rho}{\lambda^2}a^2 \sin^2 \frac{2\pi}{\lambda}(vt-x)dx\end{align*}\begin{align*} =\pi^2f^2a^2\rho=1/2\space total\space energy\end{align*} Thus, at any given instant, the energy of a plane progressive wave is on an average, half potential and half kinetic in form.

Energy current, Intensity of wave and spherical waves

We know that if the cross section of the beam or wave be taken to be unity, we may regard the total energy per unit volume or the energy density \(E=2\pi^2a^2f^2\rho\) as the total energy per unit length of the wave. And, since in progressive wave train, a new length \(v\) of the medium is set into motion every second, the energy transferred per second must be the energy contained in a length \(v\). this rate of flow of energy per unit area of cross section of wave front along the direction of wave propagation is called the energy current \(C\) or the energy flux of the wave and is equal to \(E \times V\). Thus, energy current of a plane progressive wave is \(C=2\pi^2f^2a^2\rho v\). Now, the intensity of the wave \(I\) is defined as the quantity of incident energy per unit area of the wave front per unit time. It is thus the same thing as energy current of wave. So, \(I=2\pi^2f^2a^2\rho v\) indicating that in any case, \(I\propto a^2\) where, a is the amplitude of the wave. Since, in amedium with little or no frictional resistance, a plane or one dimensional wave travels with its amplitude undiminished, the intensity of the wave remains the same throughout. However, in case of three dimensional or a spherical wave, such as the one emantating from a point source, the wave fronts are spherical shells of successively increasing radii and the intensity obey the well known inverse square law and thus goes on falling as the square of the distance \(r\) from the source. And, since, \(I\propto a^2\), we have \(a\propto 1/r\) ie. the amplitude in the case of spherical wave goes on decreasing progressively with distance from the source. Thus, whereas for a plane progressive wave, we have, \(y=a\sin 2\pi/\lambda(vt-x)\), for a spherical wave, \begin{align*} y=\frac{A}{r}\sin 2\pi/\lambda(vt-x)\end{align*} where, A is constant.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.