Standing Waves — Question 7

Question 7

An imperfect standing pattern retains a net traveling flux

Two opposing waves of amplitudes $A\ge B>0$ have the same $k,\omega$. The maximum and minimum spatial oscillation amplitudes of their sum are $10$ and $2\,\mathrm{mm}$.

1. Infer $A,B$ and determine whether any permanent displacement node exists.

2. Express the net mean rightward power in terms of $C=\mu v\omega^2/2$, taking $A$ as the rightward amplitude.

Question 7 - Solution

An imperfect standing pattern retains a net traveling flux

Method and mathematical working

Recover the two traveling amplitudes from the extrema of the standing envelope. Net mean flux is the difference of their directional powers.$$\begin{aligned}A+B&=10\,\mathrm{mm},\qquad A-B=2\,\mathrm{mm},\\\overline P_{\mathrm{net}}&=C(A^2-B^2)=C(A+B)(A-B).\\[4pt]A&=(10+2)/2=6\,\mathrm{mm},\quad B=(10-2)/2=4\,\mathrm{mm},\\\overline P_{\mathrm{net}}&=C[(0.006)^2-(0.004)^2]=2.0\times10^{-5}C.\end{aligned}$$

(a) Calculation and conclusion

The amplitude extrema are $A+B=10$ and $A-B=2\,\mathrm{mm}$, giving $A=6,B=4\,\mathrm{mm}$. Since the minimum is nonzero, there are no perfect nodes.

(b) Interpretation and limits

The net flux is $C(A^2-B^2)=C[(.006)^2-(.004)^2]=2.0\times10^{-5}C$, in watts when SI units are used. Opposing powers subtract; the residual flux is rightward. A spatially modulated envelope does not by itself imply zero net transport.

Physical check

The nonzero minimum excludes permanent displacement nodes. Convert millimeters to meters before squaring: $(0.010)(0.002)=2.0\times10^{-5}\,\mathrm{m^2}$.