Q.A sine wave of wavelength '$\lambda$' is travelling in a medium. What is the minimum distance between two particles of the medium which always have the same speed?

Compare y = 8sin(0.02x − 4t) with y = Asin(kx − ωt): speed = ω/k.

Comparing $y = 8\sin(0.02x - 4t)$ with the standard form $y = A\sin(kx - \omega t)$: $k = 0.02\ \text{cm}^{-1}$, $\omega = 4\ \text{rad/s}$ …

For a progressive wave $y = A\sin(kx-\omega t)$, the particle speed is $|v| = A\omega|\cos(kx-\omega t)|$; two particles have equal speed at every instant only if their phase difference is $0$ or $\pi$.

Let the wave be $y(x,t) = A\sin(kx - \omega t)$, so the transverse particle velocity is

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

For two particles at positions $x_1$ and $x_2 = x_1 + \Delta x$, their velocities are $-A\omega\cos\theta$ and $-A\omega\cos(\theta+\phi)$, where $\theta = kx_1-\omega t$ (which sweeps through all values as $t$ varies) and $\phi = k\Delta x$ is the constant phase difference between the particles.

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