Wavelength and Frequency — Question 9

Question 9

A snapshot fixes wavelength but not temporal frequency

At $t=0$, two possible waves have the identical profile $A\cos(10x)$, with $x$ in meters. One has temporal angular frequency $20\,\mathrm{rad/s}$ and the other $50\,\mathrm{rad/s}$. Both travel right.

1. Write the two wave functions and compare wavelengths and speeds.

2. What one additional type of observation distinguishes the models? Explain why a more detailed single snapshot does not.

Question 9 - Solution

A snapshot fixes wavelength but not temporal frequency

Method and mathematical working

A snapshot fixes the coefficient of $x$ and the phase at one instant. The coefficient of time requires a temporal measurement or an independently known propagation law.$$\begin{aligned}\lambda&=\frac{2\pi}{k},\qquad v_1=\frac{\omega_1}{k},\qquad v_2=\frac{\omega_2}{k}.\\[4pt]y_1&=A\cos(10x-20t),\quad y_2=A\cos(10x-50t),\\\lambda_1=\lambda_2&=\frac{2\pi}{10},\quad v_1=2,\quad v_2=5\quad(\mathrm{m/s}).\end{aligned}$$

(a) Calculation and conclusion

The functions are $A\cos(10x-20t)$ and $A\cos(10x-50t)$. Both have $\lambda=2\pi/10=.6283\,\mathrm{m}$, but their speeds are $2$ and $5\,\mathrm{m/s}$.

(b) Interpretation and limits

A time record at a fixed point, or a second snapshot at a known later time with phase ambiguity resolved, distinguishes them. Arbitrarily detailed information at $t=0$ fixes the shape but not its rate of evolution without a dynamical law or velocity data.

Physical check

At $t=0$ both temporal terms disappear identically. Therefore no improvement in the spatial resolution of that same snapshot can recover which temporal coefficient applies.