Ride the Wave

Consider the following function of position and time:

$\large{f = sin(a \, x - t)}$

As it turns out, this is the equation for a forward-propagating traveling wave. You can think of this as though you were dragging a sinusoidal wire along the x-axis. If you follow the wave at its speed, the amplitude at your position will always look the same. In order to follow the wave, how fast must we move? To answer this, think of time advancing by $\Delta t$ . In order to ride the wave, the argument of the sine function must remain constant.

$\large{a \, x_0 - t_0 = a \, (x_0 + \Delta x) - (t_0 + \Delta t) \\ 0 = a \, \Delta x - \Delta t \\ \implies \frac{\Delta x}{\Delta t} = \frac{1}{a}}$

The traveling wave propagates with speed $\large{\frac{1}{a}}$ . In this problem, the quantity $\large{\frac{1}{a}}$ happens to be exactly the particle's speed. Therefore, the particle maintains the same position relative to the traveling wave, and the force on the particle is always zero (as it is at the start). The particle's speed remains constant throughout. I also confirmed this result with numerical simulations.

At $t=0, F_x$ is clearly zero, and it will remain zero for all values of $t$ . Just using an intuitive approach, consider the situation after an infinitesimally small amount of time $\Delta t$ . Effectively $\dot{x}(\Delta t) = 1$ as it would not have yet been affected by any potential $F_x$ , so $x(\Delta t) = \dot{x}(\Delta t) \cdot \Delta t = \Delta t$ , and $F_x = \sin \left( \dfrac{\Delta t}{1} - \Delta t \right) = 0$ , meaning $F_x$ will never grow beyond zero.

Not sure how to prove this with formal mathematics.

Disclaimer: I know to solve this problem, it must be shown that the force remains zero. I'm not sure if the following justifies that, but I think it does.

To do this we must show that $\ddot{x} = 0$ , and when $\ddot{x}=0$ , the force also equals zero, and we must also require that the change in the force remains zero while the force is zero. $\frac{dF_x}{dt} = \frac{x\ddot{x}\cos\big(\frac{x}{\dot{x}}-t\big)}{\dot{x}^2}$ $0 = \frac{x\ddot{x}\cos\big(\frac{x}{\dot{x}}-t\big)}{\dot{x}^2}$ $0 = \frac{x\ddot{x}}{\dot{x}^2} \Rightarrow \ddot{x} = 0, \, \, \, \, x(t)>0, t>0 .$ This information reduces the difficulty of the problem greatly, and we can now reduce in the following way: $0 = \sin\Big(\frac{x}{\dot{x}} - t\Big) \Rightarrow \frac{x}{\dot{x}} - t = 0$ which is a simple differential equation to solve. Rearranging and integrating gives $\frac{\dot{x}}{x} = \frac{1}{t} \Rightarrow \ln(x(t)) = \ln(t) + C$ $x(t) = kt, \, \, \, \, k = e^{C}$ where $x'(t) = k$ and since we know $x'(0) = 1$ it must be that $k=1$ . So putting this all together we have $x(t) = t \Rightarrow x(10) = 10.$