1. Wave Speed on a Tense Rope.
Find the speed of a mechanical wave in a rope of tension \(100\,\text{N}\) and mass density \(20\, \text{g/cm}\text{.}\)
Section 14.6 Wave Speed and Medium
The speed of a mechanical wave in a medium depends on the elastic and inertial properties of the medium. We saw that the transverse wave on a string with tension \(T\) and mass per unit length \(\mu\) has the speed
\begin{equation}
v = \sqrt{\dfrac{T}{\mu}}.\tag{14.43}
\end{equation}
The wave on a more tense string will move faster on the string - this is due to better coupling in the motion of adjacent particles. On the other hand, if mass density of the string is high, the motion of the wave will be impeded by the inertia of the string since progression of wave requires chaning momentum of subsequenct parts of the string.
The same competition between elastic properties and inertia would occur in every medium through which a mechanical wave passes. For instance, when sound wave moves through air, which is a longitudinal wave, the volume of air would be compressed and rarified. The elasticity associated with change in volume is characterized by the bulk modulus of air. The inertia of the air particles will be represented by the density of air. Representing bulk modulus by \(B\) and density by \(\rho\text{,}\) we will expect the speed of sound \(v_s\) in air to be
\begin{equation}
v_s = \sqrt{\dfrac{B}{\rho}}.\tag{14.44}
\end{equation}
Numerically speaking, the bulk modulus of air has approximate value of \(1.42\times 10^{5}\text{ N/m}^2\) and the density is \(1.2\text{ kg/m}^3\text{.}\) Therefore, speed of sound in air will be
\begin{equation*}
v_s = \sqrt{\dfrac{1.42\times 10^{5}}{1.2}} = 344\text{ m/s}.
\end{equation*}
This value will depend on temperature and pressure since both the bulk modulus and the densty will change. But, even so, the value of \(344\text{ m/s}\) is quite representative of the speed of sound in air.
The speed of sound in water is different than that in air. With density of water \(1000\text{ kg/m}^3\) and bulk modulus \(2.1\times 10^{9}\text{ N/m}^2\text{,}\) we get
\begin{equation*}
v_s(\text{in water}) = \sqrt{\dfrac{2.1\times 10^{9}}{1000}} = 1449\text{ m/s}.
\end{equation*}
You can see that sound travels about four times faster in water than it does in air.
Exercises Exercises
2. Speed of Mechanical Wave in Air.
Find the speed of a mechanical wave in air of density \(1.2\, \text{kg/m}^3\) and bulk modulus \(10^5\,\text{N/m}^2\text{.}\)
3. Speed of Sound Wave in Water.
Find the speed of a mechanical wave in water of density 1 g/cc and bulk modulus \(2\times 10^9\, \text{N/m}^2\text{.}\)
4. Using Sonar to Map Ocean’s Surface.
A sonar is an ultrasound device that is used to map the surface of the ocean. At a particular place the echo is heard 2 seconds after the sonar sends an ultrasound. How deep is the ocean there? Use 1500 m/s for the speed of sound in salt water.
5. Speed and Wavelength of Ultrasound Emitted by a Bat.
A bat emits an ultrasound of frequency \(50,000\, \text{Hz}\) which bounces off a wall and returns to the bat in \(1\, \text{ms}\text{.}\) (a) Ignoring the speed of the bat, how far away is the bat from the wall? (b) What is the wavelength of the ultrasound wave. Use density of air \(1.3\, \text{kg/m}^3\) and bulk modulus \(1 \times 10^5\, \text{Pa}\text{.}\)
Solution. a
From the bulk modulus \(B\) and the density \(\rho\) the speed \(v\) of the speed,
\begin{equation*}
v = \sqrt{\frac{B}{\rho}} = \sqrt{\frac{1 \times 10^5\ \text{Pa} }{1.3 \text{kg/m}^3}} = 277\ \text{m/s}.
\end{equation*}
Using this speed we can estimate the distance \(D\) of the bat from the wall
\begin{equation*}
\frac{2D}{t} = v\ \ \Longrightarrow\ \ D = \frac{vt}{2} = \frac{277\ \text{m/s}\times0.001\ \text{s} }{2} = 0.139\ \text{m}.
\end{equation*}
\noindent (b) The wavelength will be
\begin{equation*}
\lambda = \frac{v}{f} = \frac{277\ \text{m/s}}{50,000\ \text{Hz}} = 5.54\ \text{mm}.
\end{equation*}
