Why is energy in a wave proportional to amplitude squared

I'm a mathematics student trying to grasp some basics about wave propagation. A sentence I find very often in introductive physics textbooks is the following:

In a wave, energy is proportional to amplitude squared.

This is something I would like to understand better in the case of mechanical (linear) waves.

The simplest model is the vibrating string of mass density $\mu$ and tension $T$: here an element of string of rest length $dx$ and vertical displacement $y(x, t)$ possesses a kinetic energy $\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2$ and a potential elastic energy $\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2$. So we have the total energy

$$dE=\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2+\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2,$$

which completely explains the sentence. Can we obtain a similar formula for higher-dimensional waves? How should I modify the above formula for a (simplified model of an) elastic membrane, for example? I would expect something like

$$dE=\text{some constant}\left(\frac{\partial z}{\partial t}\right)^2+\text{some other constant}\left\lvert\nabla z\right\rvert^2,$$

where $z=z(x, y, t)$ is the vertical displacement. Am I right?

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Excellent question :-) – David Z Nov 9 '11 at 19:35
$$dE=\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2+\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2,$$ Where in this equation is amplitude squared? – Muno Aug 27 '15 at 12:59
@Muno: Amplitude appears if you consider a monochromatic wave $y(t)=Ae^{i(\omega t - k x)}$. – Giuseppe Negro Aug 27 '15 at 20:45

You are correct, the equation is generalizable to higher dimensions. The equation you gave is simply the sum of the various forms of energy. In the equation

$$dE = \frac{\mu}{2} dx \left( \frac{\partial y}{\partial t}\right)^2 + \frac{T}{2} dx \left(\frac{\partial y}{\partial x} \right)^2$$

The first term on the right hand side is the kinetic energy and the second term is the elastic potential energy. As Mike said, the kinetic energy is just $p^2 / 2m$. The elastic potential energy at any location on the string is given by adding up all of the force it took to get that piece there (given by Hooke's law); that is

$$F_s (x) = -T y(x)$$ $$U = \int_0 ^y Ty(x) dy = \frac{T}{2} y(x)^2$$

The total energy is then just the sum of the kinetic and potential energies with appropriate modification using the mass density times a infinitesimal length as the mass.

When you move to higher dimensions, you have to account for that in the kinetic and potential energies. The kinetic energy of a portion of the surface is the square of the momentum of that portion ($mv$) divided by the mass of that portion. The potential energy is the spring constant of the membrane divided by 2, multiplied by the square of the displacement from zero.

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Thank you, I got the idea: in the end, energy formulas like those are consequences of Hooke's law. – Giuseppe Negro Nov 16 '11 at 18:45
I have come again to this very good answer after some time. I think that the formula you got here for the elastic potential energy is wrong, since it should depend on $\frac{\partial y}{\partial x}$ and not on $y$. I think that the error is in formula $F_s=-Ty(x)$; I would say that, by Hooke's law, one has $$F_s=-Tdy=-T\frac{\partial y}{\partial x}dx.$$ – Giuseppe Negro Apr 10 '14 at 20:16

It's not true in general that the energy of a wave is always proportional to the square of its amplitude, but there are good reasons to expect this to be true in most cases, in the limit of small amplitudes. This follows simply from expanding the energy in a Taylor series, $E=a_0+a_1 A+a_2 A^2+\ldots$ We can take the $a_0$ term to be zero, since it would just represent some potential energy already present in the medium when there was no wave excitation. The $a_1$ term has to vanish, because otherwise it would dominate the sum for sufficiently small values of $A$, and you could then have waves with negative energy for an appropriately chosen sign of $A$. That means that the first nonvanishing term should be $A^2$. Since we don't expect the energy of the wave to depend on phase, we expect that only the even terms should occur, $E=a_2A^2+a_4A^4+\ldots$ So it's only in the limit of small amplitudes that we expect $E\propto A^2$.

The other issue to consider is that we had to assume that $E$ was a sufficiently smooth function of $A$ to allow it to be calculated using a Taylor series. This doesn't have to be true in general. As an easy example involving an oscillating particle, rather than a wave, consider a pointlike particle in a gravitational field, bouncing up and down elastically on an inflexible floor. If we define the amplitude as the height of the bounce, then we have $E \propto |A|$. But a realistic ball deforms, so the small-amplitude limit consists of the ball vibrating while remaining in contact with the floor, and we regain $E\propto A^2$.

You could also make up examples where $a_2$ vanishes and the first nonvanishing coefficient is $a_4$.

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It's just a sine wave. If frequency is constant, then velocity at the zero crossing is proportional to amplitude, and energy is proportional to velocity squared.

You need a simpler model, like a 1-dimension mass on a spring (or a small-angle pendulum). Its position x is a sine wave of a certain frequency (you can do the math to get the frequency). Its maximum x in one direction is its amplitude a. Its velocity v is dx/dt, which is 90 degrees out of phase with x. At the center of its swing, x = 0, and v = max. Then clearly if you double a, it will have twice as far to swing in the same time, so v will be doubled. I'm sure you got that.

Now your question is, why is energy $E$ equal $mv^2/2$, i.e. proportional to $v^2$? Well, that's a basic equation, but let me see if I can answer it anyway.

If you drop a weight w from a height h, it has initial potential energy wh, which is transformed into kinetic energy as it reaches the floor at velocity v. If it falls under the constant force of gravity, the distance it falls in a given time t is $gt^2/2$ (time-integral of velocity), and the velocity after that time is $v = gt$. So, if you want to double the velocity it has at the floor, you have to double $t$, right? And if you double $t$, you're going to quadruple the height. That quadruples the energy. I hope that answers the question.

Just thought of another explanation. If you have a spring whose force $f$ is $kx$ where $x$ is the displacement of the end of the spring, and $k$ is its stiffness. Since energy (work) is the integral of $fdx$, the energy $E$ stored in the spring, as a function of $x$ is $kx^2/2$. So there's your energy-amplitude relation.

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Unfortunately I don't find this answer very illuminating. First, I have some trouble understanding why you can neglect elastic energy like this. If I'm not mistaken, at the zero crossing elastic energy is not zero: on the contrary, it is maximum, because the string or membrane is maximally stretched. Second, you say: "energy is proportional to velocity squared", but this is what I'm trying to understand... Can you elaborate a bit on those points, please? Thank you. – Giuseppe Negro Nov 9 '11 at 19:00
@GiuseppeNegro: At a given point on the vibrating violin string, it can be modeled as a simple lateral spring. The tension in the string is orthogonal to that motion. It is exactly analogous to a small-angle pendulum, which acts as a simple spring. The fact that the "stiffness" is mg/r (r being the length of the pendulum wire) is only the mechanism by which the central force is created. – Mike Dunlavey Nov 9 '11 at 20:33
Thank you for this answer! I'm going to think about it a bit. – Giuseppe Negro Nov 9 '11 at 20:36
At the zero crossing the spring is minimally stretched. Although, at the extreemes where the spring is maximally stretched, there is an illuminating situation as well: The energy stored in a spring is $\frac{1}{2} k x^2$ where $k$ is the spring constant and $x$ is the distance from equilibrium. So again, energy is proportional to amplitude squared. – Colin K Nov 9 '11 at 20:37

Giuseppe Negro asks for understanding of a standard expression for kinetic energy, 1/2 x mass x speed squared. Please forgive the depth of my ignorance of how to display realistic algebra.

Consider a mass m moving with speed v. A force F acts to slow the mass to zero speed in distance s. The work done by F is Fs. After Newton, this is mas with a the acceleration. Notice a is negative - the mass is decelerated by F.

Applying an equation of (uniform) motion v squared - u squared = 2as with final speed zero and initial speed v gives v squared = 2as with the negatives disappearing, which rearranges to a = v squared/2s.

Recall that the work done by F is mas and put v squared/2s for a. s cancels, leaving 1/2 x m x v squared. This expression for the work done is the kinetic energy.

Purists will wince, but I hope this helps. My first (perhaps last?) visit here.

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