Energy and Power of a Wave Traveling Along a String — Question 2

Question 2

An average power measurement determines the displacement amplitude

A sinusoidal wave travels on a string with $\mu=.020\,\mathrm{kg/m}$, $v=20\,\mathrm{m/s}$ and angular frequency $\omega=50\,\mathrm{rad/s}$. Mean power is $.050\,\mathrm{W}$. Use $\overline P=\mu v\omega^2A^2/2$.

1. Infer amplitude and maximum transverse particle speed.

2. Find mean energy per unit length and check power as energy density times propagation speed.

Question 2 - Solution

An average power measurement determines the displacement amplitude

Method and mathematical working

Invert the average-power formula to obtain amplitude. Use the resulting amplitude separately to calculate material velocity and mean energy per length.$$\begin{aligned}A&=\sqrt{\frac{2\overline P}{\mu v\omega^2}},\\u_{\max}&=A\omega,\qquad \overline{\mathcal E}=\frac{\overline P}{v}.\\[4pt]A&=\sqrt{\frac{2(0.050)}{(0.020)(20)(50)^2}}=0.010\,\mathrm m,\\u_{\max}&=0.50\,\mathrm{m/s},\quad\overline{\mathcal E}=0.050/20=0.0025\,\mathrm{J/m}.\end{aligned}$$

(a) Calculation and conclusion

$A=\sqrt{2\overline P/(\mu v\omega^2)}=.010\,\mathrm{m}$. The maximum transverse speed is $\omega A=.50\,\mathrm{m/s}$, much smaller than the wave speed here.

(b) Interpretation and limits

$\overline u=\mu\omega^2A^2/2=.0025\,\mathrm{J/m}$. Then $v\overline u=20(.0025)=.050\,\mathrm{W}$, as required. The units distinguish energy distributed along the string from energy crossing a point per second.

Physical check

Power divided by speed has units $\mathrm{J/m}$, as required. The material speed $A\omega$ is distinct from the speed at which the energy pattern propagates.