Editing Transverse wave (section)

=== Power in a transverse wave in string ===
(Let the linear mass density of the string be μ.)

The kinetic energy of a mass element in a transverse wave is given by:
<math display="block"> dK = \frac 1 2 \ dm \ v_y^2 = \frac12 \ \mu dx \ A^2 \omega^2 \cos^2 \left(\frac{2 \pi x}{\lambda} - \omega t\right)</math>

In one wavelength, kinetic energy
<math display="block"> K = \frac 1 2 \mu A ^2 \omega^2 \int ^\lambda _0 \cos^2 \left(\frac{2 \pi x}{\lambda} - \omega t\right) dx = \frac14 \mu A^2 \omega^2 \lambda</math>

Using [[Hooke's law]] the potential energy in mass element
<math display="block"> dU = \frac 1 2 \ dm \omega ^ 2 \ y ^ 2 = \frac 1 2 \ \mu dx \omega ^ 2 \ A^2 \sin^2 \left(\frac{2 \pi x}{\lambda} - \omega t\right)</math>

And the potential energy for one wavelength
<math display="block"> U = \frac 1 2 \mu A ^2 \omega^2 \int ^\lambda _0 \sin^2 \left(\frac{2 \pi x}{\lambda} - \omega t\right) dx = \frac 1 4 \mu A^2 \omega^2 \lambda</math>

So, total energy in one wavelength <math display="inline"> K + U = \frac 1 2 \mu A^2 \omega^2 \lambda</math>

Therefore average power is <math display="inline"> \frac 1 2 \mu A^2 \omega^2 v_x</math><ref>{{Cite web|title=16.4 Energy and Power of a Wave - University Physics Volume 1 {{!}} OpenStax|url=https://openstax.org/books/university-physics-volume-1/pages/16-4-energy-and-power-of-a-wave|access-date=2022-01-28|website=openstax.org|date=19 September 2016 |language=en}}</ref>