2.2 Energy transmission by waves

Wave represents distributed energy. Except for standing waves, the wave transports energy along with it from one point to another. In the context of our course, we shall focus our attention to the transverse harmonic wave along a string and investigate energy being transported by it. But the discussion and results would be applicable to other transverse waves as well through a medium. For a base case, we shall consider an ideal case in which there is no loss of energy.

We supply energy by continuously vibrating the free end of taut string. This energy is transmitted by the small vibrating string element to the neighboring element following it. We can see that vibrating small elements (also referred as particle) possess energy as it oscillates (simple harmonic motion for our consideration) in the transverse direction. It can be easily visualized as in the case of SHM that the energy has two components i.e. kinetic energy (arising from motion) and potential energy (arising from position). The potential energy is elastic potential energy like that of spring. The small string element is subjected to tension and thereby periodic elongation and contraction during the cycle of oscillatory motion.

There are few important highlights of energy transport. The most important ones are :

• The energy at the extreme positions along the string corresponds to zero energy.
• The kinetic and elastic potential energies at the mean (equilibrium) position are maximum.

Kinetic energy

The velocity of string element in transverse direction is greatest at mean position and zero at the extreme positions of waveform. We can find expression of transverse velocity by differentiating displacement with respect to time. Now, the y-displacement is given by :

$$y=A\sin \left(kx-\omega t\right)$$

Differentiating partially with respect to time, the expression of particle velocity is :

$$v_p=\frac{\partial y}{\partial t}=-\omega A\cos \left(kx-\omega t\right)$$

In order to calculate kinetic energy, we consider a small string element of length “dx” having mass per unit length “μ”. The kinetic energy of the element is given by :

$$dK=\frac{1}{2}dm{v}_p^2=\frac{1}{2}\mu dx{\omega }^2{A}^2{\cos }^2\left(kx-\omega t\right)$$

This is the kinetic energy associated with the element in motion. Since it involves squared cosine function, its value is greatest for a phase of zero (mean position) and zero for a phase of π/2 (maximum displacement). Now, we get kinetic energy per unit length, “ $K_L$ ”, by dividing this expression with the length of small string considered :

$$K_L=\frac{dK}{dx}=\frac{1}{2}\mu {\omega }^2{A}^2{\cos }^2\left(kx-\omega t\right)$$

Rate of transmission of kinetic energy

The rate, at which kinetic energy is transmitted, is obtained by dividing expression of kinetic energy by small time element, “dt” :

$$\frac{dK}{dt}=\frac{1}{2}\mu \frac{dx}{dt}{\omega }^2{A}^2\cos {}^2\left(kx-\omega t\right)$$

But, wave or phase speed,v, is time rate of position i.e. $\frac{dx}{dt}$ . Hence,

$$\frac{dK}{dt}=\frac{1}{2}\mu v{\omega }^2{A}^2{\cos }^2\left(kx-\omega t\right)$$

Here kinetic energy is a periodic function. We can obtain average rate of transmission of kinetic energy by integrating the expression for integral wavelengths. Since only ${\cos }^2\left(kx-\omega t\right)$ is the varying entity, we need to find average of this quantity only. Its integration over integral wavelengths give a value of “1/2”. Hence, average rate of transmission of kinetic energy is :