Energy and Power Carried by Waves
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the kinematics of a travelling wave \(y(x,t)=A\sin(kx-\omega t)\), the relationship between wave speed, tension, and linear mass density, and the concept of time-averaging a periodic function.
Why this matters
Understanding how energy moves through a wave is essential for interpreting everything from musical instruments to optical fibres.
In the previous lecture, we decomposed complex motions into simple sinusoidal modes. However, a mode is not just a shape; it is a carrier of energy. When you pluck a guitar string, you do not merely create a pattern of displacement; you inject mechanical energy into the system. This lecture quantifies that energy, showing how it is stored locally as kinetic and potential forms and how it propagates along the string. This concept of energy transport is the bridge between the static description of a wave and its dynamic role in physical systems, laying the groundwork for the impedance and reflection phenomena we will explore next.
The Physical Picture
Imagine a stretched string. When a wave travels along it, every small segment of the string is moving up and down. Because it is moving, it possesses kinetic energy. Simultaneously, the string is being stretched slightly as it bends into the wave shape. This stretching stores elastic potential energy, much like a compressed spring. A key insight, which we will formalise shortly, is that these two forms of energy are not independent. At any given point on the string, the kinetic and potential energies oscillate in perfect synchrony. At a crest or trough the segment is momentarily stationary and the string is locally flat, so both its kinetic and its stretching energy are zero. As the segment passes through the equilibrium position it moves fastest and the slope is steepest, so both energies are at their maximum. (This in-phase behaviour is special to travelling waves; in a standing wave the two are out of phase.) The total energy at any specific point oscillates, but the energy itself does not stay there; it flows along the string in the direction of wave propagation.
Local Energy Density
To quantify this, we consider a small element of the string of length \(\mathrm{d}x\) and mass \(\mathrm{d}m = \mu \mathrm{d}x\). The kinetic energy of this element depends on its transverse velocity \(\partial y / \partial t\). The potential energy depends on the local strain, which is related to the slope \(\partial y / \partial x\). For a travelling wave of the form \(y(x,t)=A\sin(kx-\omega t)\), the derivation (cited below) shows that the sum of these two contributions yields a local energy density \(u(x,t)\). This density is not constant in space or time; it pulses as the wave passes.
Notice that the energy density is proportional to the square of the amplitude \(A^2\) and the square of the angular frequency \(\omega^2\). This means that a high-frequency wave carries significantly more energy per unit length than a low-frequency wave of the same amplitude. The term \(\cos^{2}(kx-\omega t)\) indicates that the energy density oscillates between zero and a maximum value \(\mu\omega^2 A^2\) as the wave propagates.
Power Transport
Energy density tells us how much energy is present in a slice of the string, but we are often more interested in how fast that energy is being delivered to a distant point. This rate of energy flow is the power \(P\). For a wave on a string, the power is transmitted by the tension force \(T\) doing work on adjacent segments. The instantaneous power flowing past a point \(x\) is given by the product of the tension, the slope, and the transverse velocity.
The negative sign arises from the convention of work done by the left segment on the right segment. Substituting the derivatives of the sinusoidal wave into this expression reveals that the instantaneous power also oscillates in time and space. However, for most practical applications, such as the loudness of a sound or the intensity of a light beam, we care about the average power over one full cycle. The time-averaged power \(\langle P \rangle\) is the quantity that determines the steady rate of energy transfer.
Averaging and the Intensity Relation
When we average the oscillating energy density and power over one period, the trigonometric terms simplify. The average value of \(\cos^2\) or \(\sin^2\) over a full cycle is \(1/2\). This leads to the standard results for the mean energy density and mean power. The mean energy density \(\langle u \rangle\) is half the maximum value, and the mean power \(\langle P \rangle\) is directly proportional to the wave speed \(v\).
The final equality, \(\langle P \rangle = v \langle u \rangle\), is physically profound. It states that the power transported is simply the energy density multiplied by the speed at which that energy is moving. This is analogous to the relationship between current and charge density in electricity, or mass flow rate and mass density in fluid dynamics. It confirms that the wave acts as a conveyor belt for energy, moving at speed \(v\).
The derivation
Energy Density and Power Transport
For a transverse travelling wave \(y(x,t)=A\sin(kx-\omega t)\) on a stretched string of linear mass density \(\mu\) under tension \(T\) (wave speed \(v=\sqrt{T/\mu}\)), the local energy density is \(u=\mu\,\omega^{2}A^{2}\cos^{2}(kx-\omega t)\), with kinetic and potential parts equal at every point and instant; the instantaneous power transported past a point is \(P=-T\,\dfrac{\partial y}{\partial x}\dfrac{\partial y}{\partial t}\); and the time averages are \(\langle u\rangle=\tfrac12\mu\,\omega^{2}A^{2}\) and \(\langle P\rangle=\tfrac12\mu v\,\omega^{2}A^{2}=\tfrac12\sqrt{\mu T}\,\omega^{2}A^{2}=v\langle u\rangle\), so the transported power scales as amplitude squared times frequency squared.
A common misconception is that the energy density is uniform along the string. It is not; it varies as \(\cos^2(kx-\omega t)\). However, the average energy density \(\langle u \rangle\) is uniform in space for a pure travelling wave. If you were to measure the energy in a fixed box of length \(L\) over a long time, you would find the average energy is constant, even though the instantaneous energy sloshes in and out of the box. This distinction between instantaneous and time-averaged quantities is crucial when dealing with standing waves, where the energy does not propagate but instead oscillates between kinetic and potential forms at fixed locations. C
Worked example
Problem. A string has a linear mass density \(\mu = 0.050\ \mathrm{kg/m}\) and is under a tension \(T = 80\ \mathrm{N}\). A travelling wave on this string has an amplitude \(A = 2.0\ \mathrm{cm}\) and a frequency \(f = 100\ \mathrm{Hz}\). Calculate the average power transported by the wave.
- Determine the wave speed \(v\) using the relation \(v = \sqrt{T/\mu}\). \[ v = \sqrt{\frac{80\ \mathrm{N}}{0.050\ \mathrm{kg/m}}} = \sqrt{1600\ \mathrm{m^2/s^2}} = 40\ \mathrm{m/s} \]
- Determine the angular frequency \(\omega\) using \(\omega = 2\pi f\). \[ \omega = 2\pi (100\ \mathrm{Hz}) = 200\pi\ \mathrm{rad/s} \approx 628.3\ \mathrm{rad/s} \]
- Convert the amplitude to SI units: \(A = 2.0\ \mathrm{cm} = 0.020\ \mathrm{m}\).
- Calculate the average power using \(\langle P \rangle = \frac12 \mu v \omega^2 A^2\). \[ \langle P \rangle = \frac12 (0.050\ \mathrm{kg/m}) (40\ \mathrm{m/s}) (200\pi\ \mathrm{rad/s})^2 (0.020\ \mathrm{m})^2 \] \[ \langle P \rangle = 1.0 \times (40000\pi^2) \times 0.0004 \] \[ \langle P \rangle = 16\pi^2\ \mathrm{W} \approx 157.9\ \mathrm{W} \]
Units check. \(\mu\) is \(\mathrm{kg/m}\), \(v\) is \(\mathrm{m/s}\), \(\omega^2\) is \(\mathrm{rad^2/s^2}\) (dimensionless rad), and \(A^2\) is \(\mathrm{m^2}\). The product is \(\mathrm{kg/m} \cdot \mathrm{m/s} \cdot \mathrm{1/s^2} \cdot \mathrm{m^2} = \mathrm{kg \cdot m^2 / s^3} = \mathrm{J/s} = \mathrm{W}\). The units are consistent.
Three quick tests
- If the amplitude of a wave is doubled, by what factor does the average power increase?
Answer
The power is proportional to \(A^2\). Doubling \(A\) increases the power by a factor of \(2^2 = 4\). - Does the instantaneous energy density of a travelling wave depend on position?
Answer
Yes. For \(y = A\sin(kx-\omega t)\) both the kinetic and the potential (stretching) energy densities vary as \(\cos^2(kx-\omega t)\): the energy density is greatest where the string passes through equilibrium (largest speed and slope) and zero at the crests and troughs. - How does the average power depend on the frequency of the wave?
Answer
The average power is proportional to \(\omega^2\), which is proportional to \(f^2\). Therefore, doubling the frequency increases the power by a factor of 4.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.