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In my textbook, it says that the energy (I) of a wave, determined by the power of a wave (P) divided by area (A), is determined by the following formula:

$$I = \frac{1}{2}\rho v\omega^2A^2 = 2\pi^2 \rho vf^2A^2$$ where $\rho$ is density, $v$ is propagation speed, $\omega$ is angular speed ($\omega = 2\pi f$ and $A$ amplitude.

Can anyone show how it is like this?

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Minor notational comment to the question(v1): Usually one uses the Greek letter omega to denote angular speed, not double-u. –  Qmechanic Oct 21 '12 at 18:31
    
RRRR, did you try anything to prove this yourself? While this is not a bad question as is, it could be a lot better if you made an attempt and explained where you got stuck. –  David Z Oct 22 '12 at 8:17
    
And what kind of wave is this formula for? Wave of a string, surface water waves, etc.? –  Killercam Oct 22 '12 at 16:47

1 Answer 1

It looks like the expression for the energy of a surface water wave. Here's how it is derived - if this is not the case in question, the following should enable you to derive the relevent expression you require

A progressive wave of the surface of water can be expressed in terms of the elevation of the free surface $\eta$

$$\eta = a\sin (mx - nt),$$

where $a$ is the amplitude and the velocity is $v = n / m$. The velocity potentail for such a probagation can be written as

$$\phi = \frac{g a}{n}\frac{\cosh m(y + h)}{\cosh mh}\cos(mx - nt).$$

$y$ is the coordinate transverse to the direction of propagation. If we calculate the energy of the water between two virtical planes parralell to the direction of propogation a unit distance apart, we have, for a sinlge wave length the following expression for potential energy.

$$V = \frac{1}{2}g\rho \int^{\lambda}_{0}\eta^{2}\mathrm{d}x = \frac{1}{4}g\rho a^{2} \lambda,$$

where $\lambda = 2 \pi / m$.

The kenetic energy is given by

$$T = \frac{1}{2} \rho \int^{\lambda}_{0}\int^{0}_{-h} ((\frac{\partial \phi}{\partial x})^{2} + (\frac{\partial \phi}{\partial y})^{2})\mathrm{d}x\mathrm{d}y = -\frac{1}{2}\rho \int \phi \frac{\partial \phi}{\partial n}\mathrm{d}s,$$

where the RHS of this expression is the integral along the profile of a wavelength (this transformation should be easy to find on wikipedia), where $\partial n$ is measured along the normal to the water. This assumes we are dealing with small amplitudes we find

$$T = \frac{1}{2}\rho \int^{\lambda}_{0} (\phi \frac{\partial \phi}{\partial y})_{y = 0}\mathrm{d}x = \frac{1}{2}g \rho a^{2}\int^{\lambda}_{0} \cos^{2}(mx - nt)\mathrm{d}x,$$ $$T = \frac{1}{4}g \rho a^{2}\lambda.$$

So the total energy is $I = T + V$, so we get

$$I = T + V = \frac{1}{2}g \rho a^{2} \lambda.$$

Depending on the type of waves you want, this maybe able to be expressed in the form you have above. Alternatively, this should enable you to perform a simalar derivation for phenominon like waves on a string etc.

I hope this helps.

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