Answer
$10.59\;W$
Work Step by Step
The rate at which the energy enter the loop from each side is given by
$P_{avg}=\frac{1}{2}\mu v \omega^2 y_m^2$
or $P_{avg}=\frac{1}{2}\mu \sqrt {\frac{T}{\mu}} (2\pi f)^2 y_m^2$
or, $P_{avg}=2\pi^2\sqrt {T\mu}f^2 y_m^2$
Substituting the given values, we obtain
$P_{avg}=2\times\pi^2\times\sqrt {40\times\frac{0.125}{2.25}}\times (120)^2 (5\times10^{-3})^2\;W$
or, $P_{avg}=10.59\;W$
