Q23E - University Physics with Modern Physics

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?

Step-by-Step Solution

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 1: 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,
 

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

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

 

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

Step 2: 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.