Answer
(a) $P_{av} = 0.223~W$
(b) $P_{av} = 0.0557~W$
The average power decreases by a factor of $\frac{1}{4}$ when the amplitude is halved.
Work Step by Step
(a) We can find the average power carried by the wave.
$P_{av} = \frac{1}{2}\sqrt{\mu~F_T}~\omega^2~A^2$
$P_{av} = \frac{1}{2}\sqrt{(\frac{0.00300~kg}{0.800~m})~(25.0~N)}~[(2\pi)(120.0~Hz)]^2~(0.0016~m)^2$
$P_{av} = 0.223~W$
(b) We can find the average power carried by the wave when the amplitude is halved.
$P_{av} = \frac{1}{2}\sqrt{\mu~F_T}~\omega^2~(\frac{A}{2})^2$
$P_{av} = \frac{1}{2}\sqrt{(\frac{0.00300~kg}{0.800~m})~(25.0~N)}~[(2\pi)(120.0~Hz)]^2~(0.0008~m)^2$
$P_{av} = 0.0557~W$
The average power decreases by a factor of $\frac{1}{4}$ when the amplitude is halved.
