 Solution

For a right-moving sinusoidal string displacement $y=A\cos (kx-\omega t)$, the instantaneous energy flux of a transverse string wave is $P(x,t)=-T{y}_x{y}_t=T{A}^{2}k\omega {\sin }^2(kx-\omega t)$. Averaging over one period gives $⟨{\sin }^2⟩=1/2$. Since $c=\omega /k$ and $T=\rho c^2$,The displayed power is a cycle average, not the time-dependent instantaneous power. Here $\rho$ denotes linear mass density, rather than the volume density used in the surface-wave calculation.