# Why isn't the power transmitted by a travelling wave constant?

Let there be a wave with equation $y(x,t)=D\sin(kx-\omega t)$ which is travelling with speed $v$ through a string of density $\mu$. Then the power transmitted is $v \mu \omega^2 D^2 \cos^2(kx-\omega t)$.

I was thinking of a way of deducing this expression and thought of the following: since every point in the string moves in an harmonic fashion there is no change in their energy. Thus, the only reason power is transmitted is that every instant a new point wich wasn't moving, because it hadn't so far been reached by the wave, starts moving.

Since this point, with position $x_0=vt$, goes from being still to starting moving with speed $y'(vt,t)=-D\omega$, the change in energy equals its change in kinetic energy, that is, $\frac{1}{2}D^2 \omega ^2 \mu dx$. Hence the power transmitted should be $\frac{1}{2}D^2 \omega ^2 \mu \frac{dx}{dt}=\frac{1}{2}v \mu \omega^2 D^2$ which is the average of the correct expression over a full period.

Why is my argument wrong? I guess that it is not right that there is no change in the energy of the points which were already oscillating. However, they move like harmonic oscillators and in class we learnt that we could regard a string as something made of lots of tiny oscillators.