Question 1.10.13: The angular frequency of the surface waves in a liquid is gi...

The angular frequency of the surface waves in a liquid is given in terms of the wave number k by \omega =\sqrt{gk+Tk^3/\rho },  where g is the acceleration due to gravity, \rho  is the density of the liquid, and T is the surface tension (which gives an upward force on an element of the surface liquid).
Find the phase and group velocities for the limiting cases when the surface waves have: (a) very large wavelengths and (b) very small wavelengths.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

The phase velocity can be found at once from (1.119) \upsilon _g=\frac{d\omega (k)}{dk}, \upsilon _{ph}= \frac{\omega (k)}{k};
:

\upsilon _{ph}=\frac{\omega}{k}=\sqrt{\frac{g}{k}+\frac{T}{\rho }k }=\sqrt{\frac{g\lambda }{2\pi }+\frac{2\pi T}{\rho \lambda } },                (1.210)

where we have used the fact that k=2\pi /\lambda ,\lambda   being the wavelength of the surface waves.
(a) If \lambda is very large, we can neglect the second term in (1.210); hence

\upsilon _{ph}=\sqrt{\frac{g\lambda }{2\pi } }=\sqrt{\frac{g}{k} }.              (1.211)

In this approximation the phase velocity does not depend on the nature of the liquid, since it depends on no parameter pertaining to the liquid such as its density or surface tension. This case corresponds, for instance, to deepwater waves, called gravity waves.

To obtain the group velocity, let us differentiate (1.211) with respect to k:d\upsilon _{ph}/dk=-(1/2k)\sqrt{g/k}=-\upsilon _{ph}/2k.   A substitution of this relation into (1.120) \upsilon _g=\frac{d\omega }{dk}=\upsilon _{ph}+k\frac{d\upsilon _{ph}}{dk}=\upsilon _{ph}-\lambda \frac{d\upsilon _{ph}}{d\lambda },   shows that the group velocity is half the phase velocity:

\upsilon _g=\frac{d\omega }{dk}=\upsilon _{ph}+k\frac{d\upsilon _{ph}}{dk}=\upsilon _{ph}- \frac{1}{2 }\upsilon _{ph}=\frac{1}{2}\upsilon _{ph}=\frac{1}{2}\sqrt{\frac{g\lambda }{2\pi } }.                (1.212)

The longer the wavelength, the faster the group velocity. This explains why a strong, steady wind will produce waves of longer wavelength than those produced by a swift wind.
(b) If \lambda  is very small, the second term in (1.210) becomes the dominant one. So, retaining only the second term, we have

\upsilon _{ph}=\sqrt{\frac{2\pi T }{\rho \lambda } }=\sqrt{\frac{T}{\rho }k },                (1.213)

which leads to d\upsilon _{ph}/dk =\sqrt{Tk/\rho }/2k=\upsilon _{ph}/2k.   Inserting this expression into (1.120), we obtain the group velocity

\upsilon _g =\upsilon _{ph}+k\frac{d\upsilon _{ph}}{dk}=\upsilon _{ph}+\frac{1}{2}\upsilon _{ph}=\frac{3}{2}\upsilon _{ph};              (1.214)

hence the smaller the wavelength, the faster the group velocity. These are called ripple waves; they occur, for instance, when a container is subject to vibrations of high frequency and small amplitude or when a gentle wind blows on the surface of a fluid.

Related Answered Questions