Q.A wave is represented by an equation $y=A\sin(Bx+Ct)$. Given that the constants A, B and C are positive, can you tell in which direction the wave is moving?

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 $y = A\sin(kx - \omega t)$: $k = 0.02\ \text{cm}^{-1}$, $\omega = 4\ \text{rad/s}$.

$v = \frac{\omega}{k} = \frac{4}{0.02} = 200\ \text{cm/s}$

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.

For the two particles to have the same speed $|v|$ at every instant $t$, we need $|\cos\theta| = |\cos(\theta+\phi)|$ for all $\theta$. This holds only when $\phi = 0$ (particles identical, $\Delta x = 0$ or a whole wavelength) or $\phi = \pi$ (particles exactly out of phase, $\Delta x = \lambda/2$) — in the latter case the velocities are equal in magnitude but opposite in direction at every instant, so the speed (magnitude only) is always equal.

The smallest non-zero separation satisfying this is therefore $\Delta x = \lambda/2$.