= Solution
For a right-moving sinusoidal string displacement $y=A\cos(kx-\omega t)$, the instantaneous <energy flux of a transverse string wave> is $P(x,t)=-T y_x y_t=T A^2k\omega\sin^2(kx-\omega t)$. Averaging over one <period> gives $\langle\sin^2\rangle=1/2$. Since $c=\omega/k$ and $T=\rho c^2$,
$$
\boxed{\overline P=\frac12T A^2k\omega
=\frac12c\rho A^2\omega^2.}
$$
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.
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