Chapter 10, Problem 10.1P

To determine

The wave propagates at the same speed in water, mercury, gasoline, and glycerine.

Answer to Problem 10.1P

Below is the explanation on whether the wave propagates at the same speed in water, mercury, gasoline, and glycerine.

Explanation of Solution

Given Information:

$c^2=gy\left(1+\frac{\delta y}{y}\right)\left(1+\frac{\frac{1}{2}\delta y}{y}\right)$

$c_o{}^2=gy$

Shallow water wave propagation speed is given by:

$c_o=\sqrt{gy}$ ........ (Relation 1)

$c_o=$ Speed of wave

g = Acceleration due to gravity

y = depth of water

Relation 1 depends on viscosity and surface tension but does not depend on liquid physical properties.

A small wave propagation speed may be affected by the surface tension.

Reynolds number is given by:

$\mathrm{Re}=\frac{\rho vd}{\mu }$

Reynolds number depends on viscosity. The number is low for high viscous fluids. This in turn affects the wave propagation speed.

Hence, for water, mercury and gasoline the formulas are applicable, whereas for glycerine the formulas are not applicable as glycerine is highly viscous fluid

Conclusion:

Above is the explanation on whether the wave propagates at the same speed in water, mercury, gasoline, and glycerine.