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Energy Density

Consider again our little element of string at position x. We have shown that (for fixed x) the mass element will execute SHM as a function of time t. Therefore there is an effective LINEAR RESTORING FORCE in the y direction acting on the mass element $dm = \mu dx$:  $dF = F \, d\theta = F \left( \partial^2 y / \partial x^2 \right) dx$. But for a simple traveling wave we have^14.8   $y(x,t) = y_0 \cos(kx - \omega t)$  so   $\left( \partial^2 y / \partial x^2 \right) = - k^2 y$,  giving   $dF = - \left[ k^2 \, F \, dx \right] \, y$. In other words, the effective spring constant for an element of string  dx  long is   $\kappa_{\rm eff} = k^2 \, F \, dx$  where I have used the unconventional notation $\kappa$ for the effective spring constant to avoid confusing it with the wavenumber k, which is something completely different. Applying our knowledge of the potential energy stored in a stretched spring,   $dU = {1\over2} \kappa_{\rm eff} \, y^2$,  we have the elastic potential energy stored in the string per unit length,   $dU/dx = {1\over2} k^2 \, F \, y^2$  or, plugging in  y(x,t),

$${dU \over dx} \; = \; {1\over2} \, k^2 \, F \, y_0^2 \, \cos^2(kx - \omega t)$$ (14.22)

-- that is, the potential energy density is proportional to the amplitude squared.

What about kinetic energy? From SHM we expect the energy to be shared between potential and kinetic energy as each mass element oscillates through its period. Well, the kinetic energy  dK  of our little element of string is just   $dK = {1\over2} dm \, v_y^2$. Again   $dm = \mu \, dx$  and now we must evaluate  v_y. Working from   $y(x,t) = y_0 \, \cos(kx - \omega t)$  we have   $v_y = - \omega \, y_0 \, \sin(kx - \omega t)$,  from which we can write

$${dK \over dx} \; = \; {1\over2} \, \mu \, \omega^2 \, y_0^2 \, \sin^2(kx - \omega t) .$$ (14.23)

The total energy density is of course the sum of these two:

$${dE \over dx} \; = \; {dU \over dx} \; + \; {dK \over dx} \qquad \mbox{\rm or}$$

$${dE \over dx} \; = \; {1\over2} \, y_0^2 \, \left[ k^2 \, F . . . . . . s^2 \theta \; + \; \mu \; \omega^2 \, \sin^2 \theta \right]$$

where   $\theta \equiv kx - \omega t$. Using   $c = \omega / k = \sqrt{F/\mu}$  we can write this as

$${dE \over dx} \; = \; {1\over2} \, y_0^2 \, \left[ \mu \, \ . . . . . . \mu \, \omega^2 \, \sin^2 \theta \right] \quad \mbox{\rm or}$$

$${dE \over dx} \; = \; {1\over2} \, \mu \, \omega^2 \, y_0^2 .$$ (14.24)

You can use  $F \, k^2$  in place of   $\mu \, \omega^2$  if you like, since they are equal. [Exercise for the student.]

Note that the net energy density (potential plus kinetic) is constant in time and space for such a uniform traveling wave. 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.

If we want to know the energy per unit time (power P) transported past a certain point x by the wave, we just multiply dE/dx by c = dx/dt to get

$$P \; \equiv \; {dE \over dt} \; = \; {1\over2} \, \mu \, \omega^2 \, c \, y_0^2 .$$ (14.25)

Again, you can play around with the constants; instead of   $\mu \, \omega^2 \, c$  you can use   $\omega^2 \, \sqrt{F \mu}$  and so on.

Note that while the wave does not transport any mass down the string (all physical motion is transverse) it does transport energy. This is an ubiquitous property of waves, lucky for us!