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The total energy density in a harmonic wave on a stretched string is given by

$$\frac{1}{2}p A^2 \omega^2 sin^2(kx-\omega t).$$

We can see that this energy oscillates between a maximum and a minimum. So the energy is maximum at 0 displacement when the string is stretched and at its maximum speed (both KE and PE density are maximum at the same time) and minimum when the displacement is maximum as it is unstretched and doesnt have any velocity.

This makes sense but I am having trouble merging this with SHM oscillations. In SHM the KE and PE are not in phase. And if we consider each particle of the wave acting as a shm oscillator then would the PE not be maximum at the maximum displacement?

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  • $\begingroup$ That expression looks like a kinetic energy (if "p" is the mass density). $\endgroup$ – Pieter Jan 22 '18 at 23:21
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PE and KE that we are talking here are of a small part of string. PE and KE are maximum when the element passes through its mean position as velocity is maximum and string part is most stretched. At crest (or trough) velocity is zero and string part is not stretched so both PE and KE are zero of that string part.

Now to relate it with SHM, let us take SHM of spring-block system. In the string, the string part is like a block and rest of the string is like the spring. PE that we talk in SHM is of the spring and not stored in the block, whereas in the wave on string we are talking of PE stored in the string part i.e. block. That is why the two PE we are talking are different. The PE stored in the rest of the string is behaving as PE stored in the spring. Hope it helps!

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Since net energy (potential and kinetic) in a stretched string is a constant in space and time for a uniformly travelling wave, the total energy density must also be a constant.

However, the expression you have written is for kinetic energy density. Working from $y= A \cos(kx-\omega t)$ with $\mu$ as the mass density, we can write $$\mathrm{d}K = \frac{1}{2}\mu v^2 \mathrm{d}x \\\frac{\mathrm{d}K}{\mathrm{d}x} = \frac{1}{2}\mu \omega^2 A^2 \sin^2(kx-\omega t)$$

Similarly, potential energy density can be derived by applying our knowledge of spring and finding the effective spring constant for a stretched string. You will find $$\frac{\mathrm{d}U}{\mathrm{d}x} = \frac{1}{2}k^2FA^2\cos^2(kx-\omega t)$$

Upon calculations, you will find $k^2F = \mu \omega^2$

The total energy density is a constant since$$\frac{\mathrm{d}E}{\mathrm{d}x} = \frac{\mathrm{d}U}{\mathrm{d}x} + \frac{\mathrm{d}K}{\mathrm{d}x} = \frac{1}{2}\mu\omega^2A^2$$

It just switches back and forth between potential and kinetic energy twice every cycle. Since the average of either $\cos^2 \theta$ or $\sin^2 \theta$ is $1/2$, the energy density is on average shared equally between kinetic and potential energy.

You can see this is relatable to SHM. As you expect, the potential energy density is maximum for maximum displacement.

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In a travelling wave the total energy of a piece of string between $x$ and $x+dx$ is not constant. This is because each piece of of string is doing work on it's neighbour to the right at a rate $$ P= - T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}. $$
The local version of the energy conservation law is then $$ \frac{\partial}{\partial t}\left(\frac 12 \rho \left(\frac{\partial y}{\partial t}\right)^2+ \frac 12 T \left(\frac{\partial y}{\partial x}\right)^2\right)+ \frac{\partial }{\partial x}\left( - T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}\right)=0. $$ The expression in parenthesis in the first term is the total energy density (KE+PE). This equation says the local time-rate-of-change of the total energy between $x$ and $x+dx$ is equal to rate-work being done on the string in the interval at $x$ minus the rate at which energy is flowing out due to work being done at $x+dx$. Assuming I have made no typos, this energy equation can be verified by using the equation of motion $$ \rho\frac{\partial^2 y}{\partial t^2}- T\frac{\partial^2 y}{\partial t^2}=0. $$

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This answer is a complete rewrite of my previous answer, which was simply wrong. I fell into a trap – as others have done.

For small amplitude waves (such that $A \ll \lambda$), we can show the following:

• Changes in the tension, T, are negligible compared to T.

• Longitudinal displacements of the string are negligible.

It is with these assumptions that, using Newton's Second law, we can derive the wave equation, $$\frac{\partial^2 y}{\partial t^2}=\frac{T}{\mu}\frac{\partial^2 y}{\partial x^2}.$$ A solution is the simple harmonic wave, for which the displacement at time t and distance x along the string, is $$y=A \cos (\omega t-kx)$$ in which the wave speed is $$\frac {\omega}{k} =v=\sqrt \frac{T}{\mu}.$$ The KE per unit length is simply $$E_k=\tfrac12 \mu \left[\frac{\partial y}{\partial t}\right]^2=\tfrac12 \mu \left[\frac{\partial}{\partial t}A \cos (\omega t-kx)\right]^2=\tfrac12 \mu A^2 \omega^2 \sin^2 (\omega t-kx)$$

The PE per unit length needs to be derived from first principles. Consider a small portion of wire of length $dx$ when the string is straight but under tension T. If during the passage of the wave it acquires a slope $\frac{\partial y}{\partial x}$ (noting $\frac{\partial y}{\partial x}\ll 1$ since $A \ll \lambda$), then its new length is $$ds=\sqrt{1+\left[\frac{\partial y}{\partial x}\right]^2}dx \approx \left[1+\frac 12 \left(\frac{\partial y}{\partial x}\right)^2 \right]dx$$ So the extension due to the wave of $dx$ is $$ds-dx=\frac 12 \left(\frac{\partial y}{\partial x}\right)^2 dx.$$ Remembering that the tension T is effectively constant during the stretching of the element due to the wave, the wave-associated potential energy stored per unit length is $$E_p =\frac{T(ds -dx)}{dx} = \tfrac 12 T\left(\frac{\partial y}{\partial x}\right)^2=\tfrac 12 T\left[\frac{\partial}{\partial x} A \cos(\omega t-kx)\right]^2$$ So, using the wave speed equation, $\frac{\omega}{k}=\sqrt \frac{T}{\mu},$ in the final step, $$E_p =\tfrac 12 TA^2k^2\sin^2 (\omega t-kx)=\tfrac 12 \mu A^2\omega^2\sin^2 (\omega t-kx)$$ We have found identical equations for the PE and the KE, so these have the same peak value and vary in phase with each other. But for a single particle oscillating with a restoring force, such as a mass-spring system, the KE and the PE are in anti phase. If this discrepancy seems unacceptable...

  1. Try deriving the formula for a particle's PE using just the sinusoidal variation of displacement. You will find that you also need to use information about the system of which the particle forms a part. The phase relationship between PE and KE does not follow from the sinusoidal (simple harmonic) variation of displacement alone. There is therefore no logical objection to the phase relationships between PE and KE being different for different systems (e.g. string wave and mass-spring oscillator).

  2. Compare with a sinusoidal e-m wave in free space. B and E are in phase and so are the energies stored in these fields. [For oscillations in a simple LC circuit (analogous to a mass-spring system), Q and I are in quadrature and therefore so are E and B, so the field energies, proportional to $E^2$ and $B^2$, are in anti phase.]

  3. The last sentence of Umang's answer (not to mention the rest of the answer!) contains a most useful insight.

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  • $\begingroup$ @Jake Rose You probably sorted this out a long time ago, but just in case you haven't, this totally rewritten answer gives a detailed treatment. $\endgroup$ – Philip Wood May 3 at 15:53

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