Q.A progressive wave travels on a stretched string. A particle on this string takes 4.0 ms to move from its mean position to one of its extreme positions. The distance between two consecutive points on the string which are at their mean positions (at a certain time instant) is 2.0 cm. Find the frequency, wavelength and speed of the wave.

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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