Chapter 5.5, Problem 126E

(a)

To determine

To calculate: The angle $\theta$ that corresponds to a given Mach number.

Answer to Problem 126E

The angle $\theta$ that corresponds to a given Mach number is $180°$ .

Explanation of Solution

Given information:

The following relation:

  $\sin \frac{\theta }{2}=\frac{1}{M}$

Where $M$ correspond to Mach number and $M=1$

Calculation:

The angle $\theta$ that corresponds to the given Mach number is as follows:

  $\begin{array}{l}\sin \frac{\theta }{2}=\frac{1}{M}\text{ [Given]}\\ \sin \frac{\theta }{2}=\frac{1}{1}\text{ [Substitute]}\\ \sin \frac{\theta }{2}=1\text{ [Simplify]}\\ \frac{\theta }{2}=\arcsin (1)\text{ [Inverse trigonometric function]}\\ \frac{\theta }{2}=90°\text{ [Value of inverse]}\\ \theta =180°\text{ [Multiply both sides by 2]}\end{array}$

(b)

To determine

To calculate: The angle $\theta$ that corresponds to a given Mach number.

Answer to Problem 126E

The angle $\theta$ that corresponds to a given Mach number is $25.68°$ .

Explanation of Solution

Given information:

The following relation:

  $\sin \frac{\theta }{2}=\frac{1}{M}$

Where $M$ correspond to Mach number and $M=4.5$

Calculation:

The angle $\theta$ that corresponds to the given Mach number is as follows:

  $\begin{array}{l}\sin \frac{\theta }{2}=\frac{1}{M}\text{ [Given]}\\ \sin \frac{\theta }{2}=\frac{1}{4.5}\text{ [Substitute]}\\ \frac{\theta }{2}=\arcsin \left(\frac{1}{4.5}\right)\text{ [Inverse trigonometric function]}\\ \frac{\theta }{2}=12.84°\text{ [Value of inverse]}\\ \theta =25.68°\text{ [Multiply both sides by 2]}\end{array}$

(c)

To determine

To calculate: The speed of an object having the Mach numbers in part(a) and (b).

Answer to Problem 126E

For $M=1$ , speed of object is $760\text{ miles per hour}$ .

For $M=4.5$ , speed of object is $\text{3240 miles per hour}$ .

Explanation of Solution

Given information:

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

Speed of sound in air = $760\text{ miles per hour}$ .

The following relation:

  $\sin \frac{\theta }{2}=\frac{1}{M}$

Where $M$ correspond to Mach number.

Calculation:

As given,

  $M=\frac{V_o}{V_s}$

Where $V_o$ is the speed of object and $V_s$ is the speed of sound.

For $M=1$ , speed of object is calculated as:

  $\begin{array}{l}M=\frac{V_0}{V_s}\text{ [Given]}\\ 1=\frac{V_0}{760}\text{ [Substitute]}\\ V_0=760\text{ miles per hour [Simplify] }\end{array}$

For $M=4.5$ , speed of object is calculated as:

  $\begin{array}{l}M=\frac{V_0}{V_s}\text{ [Given]}\\ 4.5=\frac{V_0}{760}\text{ [Substitute]}\\ V_0=3420\text{ miles per hour [Simplify] }\end{array}$

(d)

To determine

To write: The given equation as a trigonometric function of $\theta$ .

Answer to Problem 126E

The given equation as a trigonometric function of $\theta$ can be written as $\cos \theta =1-\frac{2}{M^2}$ .

Explanation of Solution

Given information:

The following relation:

  $\sin \frac{\theta }{2}=\frac{1}{M}$

Where $M$ correspond to Mach number.

Formula used:

The following half angle formula:

  $\sin \frac{u}{2}\text{=}\pm \sqrt{\frac{1-\cos u}{2}}$

Calculation:

The given trigonometric equation as a function of $\theta$ can be written as:

  $\begin{array}{l}\sin \frac{\theta }{2}\text{=}\frac{1}{M}\text{ [Given]}\\ \pm \sqrt{\frac{1-\cos \theta }{2}}\text{=}\frac{1}{M}\text{ [Half angle formula]}\\ \frac{1-\cos \theta }{2}=\frac{1}{M^2}\text{ [Square both sides]}\\ 1-\cos \theta =\frac{2}{M^2}\text{ [Multiply both sides by 2]}\\ \cos \theta =1-\frac{2}{M^2}\text{ [Simplify]}\end{array}$