A sine wave propagates through a medium. What is the minimum...

Let the equation of the propagating sine wave be $y = A \sin(\omega t - kx)$.

The velocity of a particle at position $x$ and time $t$ is given by:

$v = \dfrac{\partial y}{\partial t} = A \omega \cos(\omega t - kx)$

The speed of the particle is the magnitude of its velocity:

$|v| = A \omega |\cos(\omega t - kx)|$

For two particles at positions $x_1$ and $x_2$ to always possess the same speed, their speeds must be equal for all values of time $t$:

$|\cos(\omega t - kx_1)| = |\cos(\omega t - kx_2)|$

This condition implies:

$\omega t - kx_1 = \pm (\omega t - kx_2) + n\pi$

Since this must hold true for all time $t$, the coefficient of $t$ on both sides must be the same. Thus, we must take the positive sign:

$\omega t - kx_1 = \omega t - kx_2 + n\pi$

$k(x_2 - x_1) = n\pi$

Substituting $k = \dfrac{2\pi}{\lambda}$ and $\Delta x = x_2 - x_1$:

$\dfrac{2\pi}{\lambda} \Delta x = n\pi$

$\Delta x = \dfrac{n\lambda}{2}$

To find the minimum distance between two distinct particles, we take $n = 1$:

$\Delta x = \dfrac{\lambda}{2}$