The Speed of a Traveling Wave — Question 9

Question 9

A moving phase need not translate at a uniform speed

For $x>0$, consider $y=A\cos(ax^2-\omega t)$, with $a,\omega>0$. This is a prescribed field, not assumed to solve a uniform-string wave equation.

1. Find the local velocity of a fixed-phase feature.

2. Find the feature position as a function of time from $x(0)=x_0$. Explain why writing one global wavelength is inappropriate.

Question 9 - Solution

A moving phase need not translate at a uniform speed

Method and mathematical working

Differentiate the full position-dependent phase along a trajectory. Integrate the resulting first-order equation using the given initial position.$$\begin{aligned}2ax\dot x&=\omega,\\\frac{d(x^2)}{dt}&=\frac{\omega}{a},\qquad x^2=x_0^2+\frac{\omega t}{a}.\\[4pt]x(t)&=\sqrt{x_0^2+(\omega/a)t},\qquad \ddot x=-\frac{\omega^2}{4a^2x^3}<0.\end{aligned}$$

(a) Calculation and conclusion

Constant phase gives $2ax\,dx/dt-\omega=0$, so $c_{\mathrm{phase}}(x)=\omega/(2ax)$. The local spatial phase gradient is $2ax$, not a constant wave number.

(b) Interpretation and limits

$x(t)=\sqrt{x_0^2+(\omega/a)t}$. The feature slows as $x$ grows. Exact adjacent-phase locations satisfy $a(x_2^2-x_1^2)=2\pi$, so their spacing depends on position; $2\pi/(2ax)$ is a local phase-length scale, not a global repeat distance.

Physical check

Differentiating the resulting square-root trajectory returns $\omega/(2ax)$. The derivation is restricted to $x>0$; it does not assign a finite uniform speed at the origin.