Question

A monochromatic cosine wave with frequency \(0.24\ \text{Hz}\) and wavelength \(16\ \text{km}\) interferes with another monochromatic cosine wave with frequency \(0.3\ \text{Hz}\) and wavelength \(10\ \text{km}\). The group velocity of the resulting wave in km/s is _________ (rounded off to one decimal place).

Hint:

For two close frequencies/wavelengths, \(v_g=\dfrac{\Delta f}{\Delta(1/\lambda)}\). Using Hz and km directly yields \(v_g\) in km/s (no \(2\pi\) factors remain).

Answer 1

Step 1: Use the group-velocity definition from a two-wave packet.
For two nearby waves, the group velocity is \[ v_g=\frac{\Delta \omega}{\Delta k} = \frac{2\pi\,\Delta f}{2\pi\,\Delta(1/\lambda)} = \frac{\Delta f}{\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right)}. \] Step 2: Substitute the numbers.
\[ \Delta f = f_2-f_1 = 0.30-0.24 = 0.06\ \text{Hz}, \] \[ \Delta\!\left(\frac{1}{\lambda}\right) = \frac{1}{10}-\frac{1}{16}=0.1000-0.0625=0.0375\ \text{km}^{-1}. \] Step 3: Compute \(v_g\).
\[ v_g=\frac{0.06}{0.0375}=1.6\ \text{km/s}. \] \[ \boxed{1.6\ \text{km/s}} \]