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Energy

Clearly a wave can carry energy from one place to another. A cable we are coiling is hung up on a piece of wood. We flip a pulse onto the wire, it runs down to the piece of wood and knocks the wire free. Our lungs and larnyx create sound waves, and those waves trigger neurons in ears far away. The sun releases nuclear energy, and a few minutes later that energy, propagated to earth as a light wave, creates sugar energy stores inside a plant that are still later released while we play basketball. Since moving energy around seems to be important, perhaps we should figure out how a wave manages it.

Let us restrict our attention to a harmonic wave of known angular frequency $\omega$. Our results will still be quite general, because arbitrary wave pulses can be fourier decomposed as noted above. Consider a small piece of the string of length $dx$ and mass $dm = \mu
dx$. This piece of string, displaced to its position $y(x,t)$, will have potential energy:

$\displaystyle dU$ $\textstyle =$ $\displaystyle \frac{1}{2} dm \omega^2 y^2(x,t)$ (141)
  $\textstyle =$ $\displaystyle \frac{1}{2} \mu dx \omega^2 y^2(x,t)$ (142)
  $\textstyle =$ $\displaystyle \frac{1}{2} A^2 \mu \omega^2 \sin^2(kx - \omega t) dx$ (143)

We can easily integrate this over any specific interval. Let us pick a particular time $t = 0$ and integrate it over a single wavelength:
$\displaystyle U$ $\textstyle =$ $\displaystyle \int_0^\lambda \frac{1}{2} A^2 \mu \omega^2 \sin^2(kx) dx$ (144)
  $\textstyle =$ $\displaystyle \frac{1}{2k} A^2 \mu \omega^2 \int_0^\lambda \sin^2(kx) kdx$ (145)
  $\textstyle =$ $\displaystyle \frac{1}{2k} A^2 \mu \omega^2 \int_0^2\pi \sin^2(\theta) d\theta$ (146)
  $\textstyle =$ $\displaystyle \frac{1}{4} A^2 \mu \omega^2 \lambda$ (147)

Now we need to compute the kinetic energy in a wavelength at the same instant.

$\displaystyle dK$ $\textstyle =$ $\displaystyle \frac{1}{2} dm \left(\frac{dy}{dt}\right)^2$ (148)
  $\textstyle =$ $\displaystyle \frac{1}{2} A^2 \mu \omega^2 \cos^2(kx - \omega t) dx$ (149)

which has exactly the same integral:
\begin{displaymath}
K = \frac{1}{4} A^2 \mu \omega^2 \lambda
\end{displaymath} (150)

so that the total energy in a wavelength of the wave is:
\begin{displaymath}
E_{\rm tot} = \frac{1}{2} \mu \omega^2 A^2 \lambda
\end{displaymath} (151)

Study the dependences in this relation. Energy depends on the amplitude squared! (Emphasis to convince you to remember this! It is important!) It depends on the mass per unit length times the length (the mass of the segment). It depends on the frequency squared.

This energy moves as the wave propagates down the string. If you are sitting at some point on the string, all the energy in one wavelength passes you in one period of oscillation. This lets us compute the power carried by the string - the energy per unit time that passes us going from left to right:

\begin{displaymath}
P = \frac{E}{T} = \frac{1}{2} \mu \omega^2 A^2 \lambda f = \frac{1}{2}
\mu \omega^2 A^2 v
\end{displaymath} (152)

We can think of this as being the energy per unit length (the total energy per wavelength divided by the wavelength) times the velocity of the wave. This is a very good way to think of it as we prepare to study light waves, where a very similar relation will hold.


next up previous contents
Next: Sound Up: Waves Previous: Reflection of Waves   Contents
Robert G. Brown 2004-04-12