Answer
$10\;W$
Work Step by Step
The average rate at which energy is transported by the wave to the opposite end of the cord is given by
$\Big(\frac{dK}{dt}\Big)_{avg}=\frac{1}{2}\mu v \omega^2 y_m^2$
or $\Big(\frac{dK}{dt}\Big)_{avg}=\frac{1}{2}\mu \sqrt {\frac{T}{\mu}} \omega^2 y_m^2$
or, $\Big(\frac{dK}{dt}\Big)_{avg}=\frac{1}{2}\sqrt {T\mu} \omega^2 y_m^2$
Substituting the given values
$\Big(\frac{dK}{dt}\Big)_{avg}=\frac{1}{2}\sqrt {(1200\times2\times10^{-3})} \times(1200)^2\times(3\times10^{-3})^2$
or, $\Big(\frac{dK}{dt}\Big)_{avg}\approx10\;W$
$\therefore$ The average rate at which energy is transported by the wave to the opposite end of the cord is $10\;W$.
