Wavelength and Frequency — Question 6

Question 6

A phase can remain unchanged while both position and time change

A wave has $\lambda=.80\,\mathrm{m}$ and $T=.20\,\mathrm{s}$. Its phase is $2\pi(x/\lambda-t/T)$. Compare events separated by $\Delta x=.60\,\mathrm{m}$ and $\Delta t=.15\,\mathrm{s}$.

1. Find the phase difference and interpret the result.

2. If only the time separation becomes $.05\,\mathrm{s}$, find the phase difference and the relation between displacements.

Question 6 - Solution

A phase can remain unchanged while both position and time change

Method and mathematical working

Compute the signed spatial and temporal phase changes separately. Their subtraction determines whether the events reinforce or reverse the sinusoidal displacement.$$\begin{aligned}\frac{\Delta\Phi}{2\pi}&=\frac{\Delta x}{\lambda}-\frac{\Delta t}{T}.\\[4pt]\Delta\Phi&=2\pi\left(\frac{0.60}{0.80}-\frac{0.15}{0.20}\right)=0,\\\Delta t=0.050\,\mathrm s&\Rightarrow\Delta\Phi=2\pi(0.75-0.25)=\pi.\end{aligned}$$

(a) Calculation and conclusion

$\Delta\Phi=2\pi(.60/.80-.15/.20)=0$. The two events follow the same phase feature moving at $\lambda/T=4\,\mathrm{m/s}$. Spatial and temporal phase changes cancel.

(b) Interpretation and limits

Now $\Delta\Phi=2\pi(.75-.25)=\pi$, so the sinusoidal displacements are negatives of one another. Equal displacement cannot generally be judged from either the spatial or temporal separation alone.

Physical check

An event moving at $\Delta x/\Delta t=\lambda/T$ has zero phase change. This reproduces the speed of a fixed phase and checks the sign between the two contributions.