Question

A horizontal wire is stretched with a tension of 94.0 N, and the speed of transverse waves for the wire is 406 m/s. What must the amplitude of a traveling wave of frequency 69.0 Hz be for the average power carried by the wave to be 0.365 W?

Short answer

The amplitude of a traveling wave of frequency 69.0 Hz be for the average power carried by the wave to be 0.365 W should be 0.0041m.

Step-by-step solution

Determination of the formula of Mechanical Waves

The average power carried by the wave is:

$P=\frac{1}{2}\mu A^2{\omega }^{2}v$

Also,

localid="1664344211351" $$\begin{gathered} v=\sqrt{\frac{T}{\mu }} \\ \mu =\frac{T}{v^2} \\ w=2\pi f \end{gathered}$$

Substitute the values of$\mu$ and w in formula for power we get

$P=\frac{\left(T{A}^2\right)4{\pi }^2{f}^2}{2v}$

Application of the formula of Mechanical Waves

$$\begin{gathered} A=\sqrt{\frac{Pv}{2{\pi }^2{f}^{2}T}} \\ A=\sqrt{\frac{0.365\times 406}{19.72\times 4761\times 94}} \\ A=0.0401m \end{gathered}$$

Therefore, the amplitude of a traveling wave of frequency 69.0 Hz be for the average power carried by the wave to be 0.365 W should be 0.0041m.