Energy density in 1D elastic wave

[Andy365]

Hello,
consider a 1D elastic wave which have the amplitude:
$$A=cos(x)$$

What is the energy density: $$\frac{dE}{dx}$$ of this wave?

I seem to recall that the energy of a wave is proportional to the square of the amplitude:
$$E \propto A^2$$

That seem to mean that $$\frac{dE}{dx} \propto cos(x)^2$$

However the energy density should be constant for all x in this case, since there is no loss!?

If I instead define $$A=e^{ix}$$, and use that $$E \propto |A|^2$$
things work better since $$|A|^2 = 1$$, which is independent of x.

So what is happening here?


Thanks in advance for any answers!

 

[Bill_K]

Andy, To write down the energy you need to be more explicit about what kind of wave it is. You say the wave is one-dimensional, but you didn't say what type of medium is involved, and whether the wave is longitudinal or transverse. Let's suppose it's transverse, and described by a transverse displacement a(x,t). The medium (string?) will be described by a mass per unit length ρ and a restoring force per unit length T (string tension?).

The wave equation is ρ ∂²a/∂t² - T ∂²a/∂x² = 0, with solution a(x,t) = A sin(kx - ωt) with phase velocity ω/k = √(T/ρ).

Then the energy per unit length is E = ½ρ(∂a/∂t)² + ½T(∂a/∂x)² = ρA²ω² cos²(kx - ωt)

(Note the last two terms are equal, i.e. the potential energy density equals the kinetic energy density.) This result confirms you're right, the energy density is proportional to A². And it is proportional to cos². It is *not* constant, it is maximum where the displacement is zero, because that's the point where the wave has the largest kinetic and potential energy.

 

[Andy365]

Andy, To write down the energy you need to be more explicit about what kind of wave it is. You say the wave is one-dimensional, but you didn't say what type of medium is involved, and whether the wave is longitudinal or transverse. Let's suppose it's transverse, and described by a transverse displacement a(x,t). The medium (string?) will be described by a mass per unit length ρ and a restoring force per unit length T (string tension?).

The wave equation is ρ ∂²a/∂t² - T ∂²a/∂x² = 0, with solution a(x,t) = A sin(kx - ωt) with phase velocity ω/k = √(T/ρ).

Then the energy per unit length is E = ½ρ(∂a/∂t)² + ½T(∂a/∂x)² = ρA²ω² cos²(kx - ωt)

(Note the last two terms are equal, i.e. the potential energy density equals the kinetic energy density.) This result confirms you're right, the energy density is proportional to A². And it is proportional to cos². It is *not* constant, it is maximum where the displacement is zero, because that's the point where the wave has the largest kinetic and potential energy.

Thank you Bill_K for your answer!
Yes I am talking about a transverse wave.

So the energy per unit length varies between 0 and ρA²ω² along the period of the wave.
(It seems we usually define the energy to be an average over a whole period?)
But the wave transports energy in the propagation direction. This "flow of energy" must be constant along x. How do we define this?

 

[olivermsun]

(Note the last two terms are equal, i.e. the potential energy density equals the kinetic energy density.) This result confirms you're right, the energy density is proportional to A². And it is proportional to cos². It is *not* constant, it is maximum where the displacement is zero, because that's the point where the wave has the largest kinetic and potential energy.

I must not be reading a part correctly -- are you saying that the potential energy be largest when the displacement is zero?

 

[Bill_K]

I must not be reading a part correctly -- are you saying that the potential energy be largest when the displacement is zero?

That is correct. Because in a vibrating string the potential energy stored at a particular spot is determined by how much the string deviates from the horizontal, namely ∂a/∂x. A region where the string is perfectly flat, a = const ≠ 0, for example, contains *no* potential energy. Do you agree? For a vibrating string it's only places where ∂a/∂x ≠ 0 that the potential energy is stored. This holds true for both transverse and longitudinal vibrations. Imagine a slinky being stretched longitudinally. Again, you'll store no energy where the displacement a = const, only where the spring is actually being stretched, i.e. where ∂a/∂x ≠ 0.

This property is different for different types of wave, which is why I asked Andy very first thing what kind of wave he meant. There are other types of waves that behave the way you're thinking. For example sound waves in a crystal lattice would, in which each atom has an "at rest" location, and is attached to it by a little spring. In that case, like you said, the potential energy depends on a rather than ∂a/∂x.

 

[olivermsun]

That is correct. Because in a vibrating string the potential energy stored at a particular spot is determined by how much the string deviates from the horizontal, namely ∂a/∂x. A region where the string is perfectly flat, a = const ≠ 0, for example, contains *no* potential energy. Do you agree?

I do agree. The spring here is the string itself, which is stretched when dy/dx ~= 0. I was confusing myself by thinking about a slightly different kind of wave. ;)