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JayKo

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## Homework Statement

Consider a traveling wave y=f(x-vt) on a string. Show that the kinetic and potential energies are always equal to each other. Remember that the potential energy of a wave on a string (over one wavelength) is given by

## Homework Equations

E(potential)= 1/2 * F * [tex]\int[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]x)^2 dx

E(kinetic) = 1/2 * [tex]\mu[/tex] * [tex]\int[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]t)^2 dx

F=[tex]\mu[/tex]v^2

ps:the integrand is integrate from 0 to lamba.

**3. The Attempt at a Solution ([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]t[tex]^{2}[/tex])=-[tex]\omega[/tex][tex]^{2}A[/tex] sin([tex]\omega[/tex]t-kx)**

([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]x[tex]^{2}[/tex])=-k[tex]^{2}A[/tex] sin([tex]\omega[/tex]t-kx)

so my question is, is it i just need to integrate both the 2nd order partial derivative that i will be show that both potential and kinetic are the same? thanks.

([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]x[tex]^{2}[/tex])=-k[tex]^{2}A[/tex] sin([tex]\omega[/tex]t-kx)

so my question is, is it i just need to integrate both the 2nd order partial derivative that i will be show that both potential and kinetic are the same? thanks.

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