Chapter 5.5, Problem 73E

a.

To determine

To rewrite: The equation in terms of $\cos \theta$

Answer to Problem 73E

  $\cos \theta =1-\frac{2}{M^2}$

Explanation of Solution

Given:The Mach number M of a supersonic airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}$ .

The relation of apex angle and Mach number is $\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}$ .

Now, change the formula to half-angle to complete angle in form of $\cos \theta$

  $\begin{array}{l}\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}\\ \text{Taking square both sides }\\ {\sin }^2\left(\frac{\theta }{2}\right)=\frac{1}{M^2}\left[\because {\sin }^2\left(\frac{\theta }{2}\right)=\frac{1-\cos \theta }{2}\right]\\ \frac{1-\cos \theta }{2}=\frac{1}{M^2}\\ \cos \theta =1-\frac{2}{M^2}\end{array}$

b.

To determine

To find: The angle $\theta$ .

Answer to Problem 73E

  $\theta =\pi$

Explanation of Solution

Given: The Mach number M of a supersonic airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}$ .

If the Mach number of 1.

In part (a) got an equation, $\cos \theta =1-\frac{2}{M^2}$

Put $M=1$

  $\begin{array}{c}\cos \theta =1-\frac{2}{1}\\ \cos \theta =1-2\\ \cos \theta =-1\\ \theta =\pi \end{array}$

c.

To determine

To find: The angle $\theta$ .

Answer to Problem 73E

  $\theta \approx {26}^{\circ }$

Explanation of Solution

Given: The Mach number M of a supersonic airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}$ .

If the Mach number of 4.5.

In part (a) got an equation, $\cos \theta =1-\frac{2}{M^2}$

Put $M=4.5$

  $\begin{array}{c}\cos \theta =1-\frac{2}{{4.5}^2}\\ \cos \theta =1-0.0988\\ \cos \theta =0.9012\\ \theta \approx {26}^{\circ }\end{array}$

d.

To determine

To find: The speed of an object with Mach numbers 1 and 4.5.

Answer to Problem 73E

For $M=1,v=760\text{mph}$

For $M=4.5,v=3420\text{mph}$

Explanation of Solution

Given: The Mach number M of a supersonic airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}$ .

Let speed of airplane be $v$ and speed of sound be $s$

The Mach number M of a supersonic airplane is the ratio of its speed to the speed of sound.

  $\therefore M=\frac{v}{s}$

For $M=1,s=760\text{ mph}$

  $\begin{array}{l}v=s\\ v=760\text{ mph}\end{array}$

For $M=4.5,s=760\text{ mph}$

  $\begin{array}{l}v=4.5s\\ v=4.5\times 760\\ \text{v=3420 mph}\end{array}$