A transverse wave with an amplitude of 0.50 mm and a frequency of...

The average power transmitted by a transverse wave on a stretched string is given by:

$P_{avg} = \dfrac{1}{2} \mu \omega^2 A^2 v$

The wave speed is related to tension and linear mass density by $v = \sqrt{\dfrac{T}{\mu}}$, which gives $\mu = \dfrac{T}{v^2}$.

Substituting $\mu$ into the power equation:

$P_{avg} = \dfrac{1}{2} \left( \dfrac{T}{v^2} \right) \omega^2 A^2 v = \dfrac{T \omega^2 A^2}{2v}$

Given values are:
$T = 100 \text{ N}$
$v = 100 \text{ m s}^{-1}$
$f = 100 \text{ Hz} \Rightarrow \omega = 2 \pi f = 200 \pi \text{ rad s}^{-1}$
$A = 0.50 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$

Substituting these values:

$P_{avg} = \dfrac{100 \times (200 \pi)^2 \times (0.5 \times 10^{-3})^2}{2 \times 100}$

$P_{avg} = \dfrac{1}{2} \times 40000 \pi^2 \times 0.25 \times 10^{-6}$

$P_{avg} = 5000 \pi^2 \times 10^{-6} \text{ W}$

$P_{avg} = 5 \pi^2 \times 10^{-3} \text{ W}$

Using $\pi^2 \approx 9.87$:

$P_{avg} \approx 5 \times 9.87 \times 10^{-3} \text{ W} = 49.35 \text{ mW}$

The closest value is $49 \text{ mW}$.