Chapter 5.5, Problem 71E

a.

To determine

Use a half-angle formula to rewrite the equation in terms of $\cos \theta$ .

Answer to Problem 71E

  $\boxed{\cos \theta =\frac{M^2-2}{M^2}}$

Explanation of Solution

Given information:

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 form a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\theta /2\right)=1/M$ .

Calculation:

The match number is related to the apex angle $\theta$ of the cone by

  $\sin \left(\frac{\theta }{2}\right)=\frac{1}{M}$

From the half angle formula

  $\sin \frac{u}{2}=\pm \sqrt{\frac{1-\cos u}{2}}\mathrm{......}\left(1\right)$

From the formula (1)

  $\begin{array}{l}\sin \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos }{2}}\\ \frac{1}{M}=\pm \sqrt{\frac{1-\cos \theta }{2}}\\ \frac{1}{M^2}=\pm \sqrt{\frac{1-\cos \theta }{2}}\\ \frac{2}{M^2}=-\frac{1-\cos \theta }{2}\\ \frac{2}{M^2}=1-\cos \theta \\ \cos \theta =1-\frac{1}{M^2}\\ \end{array}$

  $=\frac{M^2-2}{M^2}$

Hence, $\boxed{\cos \theta =\frac{M^2-2}{M^2}}$

b.

To determine

Find the angle $\theta$ that corresponds to a Mach number of $2$ .

Answer to Problem 71E

  $\boxed{\pi }$

Explanation of Solution

Given information:

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 form a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\theta /2\right)=1/M$ .

Calculation:

Let $M=1$ in the result of (a)

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

Hence, the angle $\theta$ that correspondence to the number $M=1$ is $\boxed{\pi }$

c.

To determine

Find the angle $\theta$ that corresponds to a Mach number of $4.5$ .

Answer to Problem 71E

  $\boxed{0.4482}$

Explanation of Solution

Given information:

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 form a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\theta /2\right)=1/M$ .

Calculation:

Let $M=4.5$ in the results of (a)

  $\begin{array}{l}\cos \theta =\frac{{4.5}^2-2}{{4.5}^2}\\ =\frac{18.25}{20.25}\\ \theta \approx 0.4482\end{array}$

Hence, the angle $\theta$ that correspondence to the number $M=4.5$ is $\boxed{0.4482}$

d.

To determine

Determine the speed of an object with the Mach numbers from parts (b) and (c).

Answer to Problem 71E

  $\boxed{760mi/h}$ and $\boxed{3,420mi/h}$

Explanation of Solution

Given information:

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 form a cone behind the airplane. The Mach number is related to the apex angle $\theta$ of the cone by $\sin \left(\theta /2\right)=1/M$ .

Calculation:

Let the match number is $1$

  $\begin{array}{l}\frac{Speedofobject}{Speedofsound}=M\\ \frac{Speedofobject}{760}=1\\ Speedofobject=760\end{array}$

Hence, if the match number is $1$ then speed is $\boxed{760mi/h}$

Let the match number is $4.5$

  $\begin{array}{l}\frac{Speedofobject}{Speedofsound}=M\\ \frac{Speedofobject}{760}=4.5\\ Speedofobject=760\times 4.5\end{array}$

  $=3,420$

Hence, if the match is $4.5$ then speed is $\boxed{3,420mi/h}$