The Speed of a Traveling Wave — Question 4

Question 4

An observer moving with the pattern sees no oscillation

A wave $y=A\cos(kx-\omega t)$ travels right at $v=\omega/k=8\,\mathrm{m/s}$ and has wavelength $.40\,\mathrm{m}$. An observer moves at constant signed speed $u$ along the line.

1. Find the frequency encountered by the observer.

2. Evaluate it for $u=3$, $8$, and $11\,\mathrm{m/s}$, and interpret the last case.

Question 4 - Solution

An observer moving with the pattern sees no oscillation

Method and mathematical working

Substitute the observer trajectory into the phase before extracting a frequency. The counting frequency is the magnitude of the signed phase rate divided by $2\pi$.$$\begin{aligned}\Phi(t)&=kx_0+(ku-\omega)t,\\f_{\mathrm{obs}}&=\frac{|ku-\omega|}{2\pi}=\frac{|u-v|}{\lambda}.\\[4pt]f_{\mathrm{obs}}(3)&=5/0.40=12.5\,\mathrm{Hz},\\f_{\mathrm{obs}}(8)&=0,\qquad f_{\mathrm{obs}}(11)=3/0.40=7.5\,\mathrm{Hz}.\end{aligned}$$

(a) Calculation and conclusion

Along $x=x_0+ut$, phase changes at rate $ku-\omega$. The observed frequency is $|\omega-ku|/(2\pi)=|v-u|/\lambda$. This is a sampling result for the prescribed wave and moving observer.

(b) Interpretation and limits

The frequencies are $12.5$, $0$, and $7.5\,\mathrm{Hz}$. At $u=v$ the observer stays at fixed phase. At $u>v$ the observer overtakes crests and encounters them in the opposite phase order; a negative signed phase rate is not a negative counted frequency.

Physical check

At $u=0$ this reduces to the laboratory frequency $20\,\mathrm{Hz}$. The limit $u=v$ gives no phase change, while crossing this speed reverses the order of phases encountered.