Wavelength and Frequency — Question 7

Question 7

The shortest common repeat can exceed every visible subspacing

A profile is $y(x,0)=A\cos(2\pi x/L)+B\cos(6\pi x/L)$ with $A,B\ne0$. In time, both terms translate unchanged at the same speed $v$.

1. Find the fundamental spatial repeat and the two component frequencies.

2. Why is $L/3$ not a wavelength of the complete profile even though one component repeats at that spacing?

Question 7 - Solution

The shortest common repeat can exceed every visible subspacing

Method and mathematical working

A repeat distance must return every nonzero harmonic to its original phase. Find a common period, then test whether a smaller proposed shift leaves the complete sum unchanged.$$\begin{aligned}k_1\Delta x&=2\pi m,\qquad 3k_1\Delta x=2\pi n.\\[4pt]\Delta x=L&:\quad(\Delta\Phi_1,\Delta\Phi_3)=(2\pi,6\pi),\\\Delta x=L/3&:\quad(\Delta\Phi_1,\Delta\Phi_3)=(2\pi/3,2\pi).\end{aligned}$$

(a) Calculation and conclusion

The component wavelengths are $L$ and $L/3$. Their shortest common repeat is $L$, because the first nonzero component requires a phase advance of $2\pi$. Frequencies are $v/L$ and $3v/L$.

(b) Interpretation and limits

After a shift $L/3$, the second term repeats but the first advances only $2\pi/3$. The complete profile therefore does not generally repeat. Internal ripples and component wavelengths must not be confused with the fundamental period of the whole pattern.

Physical check

The nonzero fundamental forces the shortest repeat to be its own wavelength. Removing that component would change the conclusion, which is why both nonzero amplitudes are part of the hypothesis.