Answer
$26.18\;W$
Work Step by Step
If two identical waves are sent along the same cord simultaneously in same direction, they interfere to produce a resultant sinusoidal wave traveling in that direction.
The mathematical form of the resultant sinusoidal wave is given by
$y(x, t) =[2y_m\cos\frac{\phi}{2}]\sin(kx-\omega t+\frac{\phi}{2}) $
The total average rate at which the resultant sinusoidal wave transport energy is given by
$P_{avg}=\frac{1}{2}\mu v \omega^2 (2y_m\cos\frac{\phi}{2})^2$
or $P_{avg}=\frac{1}{2}\mu \sqrt {\frac{T}{\mu}} \omega^2 (2y_m\cos\frac{\phi}{2})2$
or, $P_{avg}=2\sqrt {T\mu} \omega^2 y_m^2\cos^2\frac{\phi}{2}$
Substituting given values
$P_{avg}=2\sqrt {(1200\times2\times2^{-3})} \times (1200)^2 \times (3\times10^{-3})^2\cos^2(\frac{0.4\pi}{2})\;W$
$P_{avg}=26.18\;W$
$\therefore$ When the phase difference between two sinusoidal waves is $0.4\pi$, the total average rate at which they transport energy is $26.18\;W$
