The Speed of a Traveling Wave — Question 2

Question 2

A localized profile translates through its whole argument

At $t=0$ a pulse is $y(x,0)=A\exp[-(x-x_0)^2/a^2]$. It moves right at constant speed $v$ without distortion.

1. Write $y(x,t)$ and locate its maximum.

2. Find the full spatial width at half maximum and the duration between half-height crossings at a fixed detector.

Question 2 - Solution

A localized profile translates through its whole argument

Method and mathematical working

For a rigid rightward translation, hold $x-vt$ fixed. Solve the half-height condition relative to the moving center, then convert the separation of the two crossings into a time interval.$$\begin{aligned}\frac yA=\frac12&\Rightarrow -\frac{(x-x_0-vt)^2}{a^2}=-\ln2,\\x_\pm&=x_0+vt\pm a\sqrt{\ln2}.\\[4pt]w_{1/2}&=x_+-x_-=2a\sqrt{\ln2},\qquad\Delta t_{1/2}=\frac{2a\sqrt{\ln2}}v.\end{aligned}$$

(a) Calculation and conclusion

Replace $x$ by $x-vt$: $y=A\exp[-(x-x_0-vt)^2/a^2]$. The peak is at $x=x_0+vt$. Translation shifts the center without changing the width parameter.

(b) Interpretation and limits

At half height, $|x-x_0-vt|=a\sqrt{\ln2}$. The full width is $2a\sqrt{\ln2}$, and the detector interval is $2a\sqrt{\ln2}/v$. A pulse width is not a wavelength; this result concerns a single passing profile.

Physical check

The width has units of length and is independent of time. Doubling the translation speed leaves this spatial width unchanged while halving the duration measured at a stationary detector.