Chapter 23, Problem 101A

(a)

To determine

The wavelength of the sound waves if airplane were not moving.

Answer to Problem 101A

  $1\text{ m}$

Explanation of Solution

Given:

Speed of sound, $v=340\text{m/s}$

Space between wave frontsin front of the plane, $0.50\text{m}$

Space between the wave fronts behind the plane, $1.50\text{m}$

Formula Used:

Wavelength can be obtained by:

$\lambda =\frac{v}{f}$

Here, $v$ is the speed of waves and f is their frequency.

Calculations:

Let the speed of the plane be $v_a$ .

$\begin{array}{l}\frac{v}{f}=\lambda \\ \frac{v-v_a}{f}=0.5\mathrm{...}\left(1\right)\\ \frac{v+v_a}{f}=1.5\mathrm{...}\left(2\right)\end{array}$

Add the two equations.

$\begin{array}{l}2\frac{v}{f}=2\text{m}\\ \lambda =\frac{v}{f}=1\text{m}\end{array}$

Conclusion:

Thus, the wavelength of the sound waves is 1 m.

(b)

To determine

The frequency of the sound waves.

Answer to Problem 101A

  $340\text{ Hz}$

Explanation of Solution

Given:

Speed of sound, $v=340\text{m/s}$

Wavelength of the sound wave, $\lambda =1\text{m}$ (calculated in previous part)

Formula Used:

The frequency of the sound wave can be obtained by:

$f=\frac{v}{\lambda }$

Here, $v$ is the speed of waves and $\lambda$ is their wavelength.

Calculations:

Substitute the values and solve:

$\begin{array}{c}f=\frac{v}{\lambda }\\ =\frac{340\text{m/s}}{1\text{m}}\\ =340\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of the sound waves is 340 Hz.

(c)

To determine

The speed of the airplane.

Answer to Problem 101A

  $170\text{ m/s}$

Explanation of Solution

Given:

Speed of sound, $v=340\text{m/s}$

Wavelength of the sound wave, $\lambda =1\text{m}$ (calculated in previous part)

Space between wave fronts in front of the plane, $0.50\text{m}$

Space between the wave fronts behind the plane, $1.50\text{m}$

Formula Used:

Wavelength can be obtained by:

$\lambda =\frac{v}{f}$

Here, $v$ is the speed of waves and f is their frequency.

Calculations:

Let the speed of the plane be $v_a$ .

$\begin{array}{l}\frac{v}{f}=\lambda \\ \frac{v-v_a}{f}=0.5\mathrm{...}\left(1\right)\\ \frac{v+v_a}{f}=1.5\mathrm{...}\left(2\right)\end{array}$

Substitute the known values in any one equation and solve for $v_a$ :

$\begin{array}{l}v-v_a=0.5f\\ 340\text{m/s}-v_a=0.5\text{m}\left(340\text{Hz}\right)\\ v_a=170\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the airplane is 170 m/s.

(d)

To determine

The frequency of the sound wave detected by an observer located directly in front of the airplane.

Answer to Problem 101A

  $680\text{ Hz}$

Explanation of Solution

Given:

Speed of sound, $v=340\text{m/s}$

Wavelength of the sound wave, $\lambda =1\text{m}$ (calculated in previous part)

Frequency of the sound, $f=340\text{Hz}$

Speed of the airplane, $v_a=170\text{m/s}$

Formula Used:

From Doppler effect, frequency of the sound wave when source is moving towards the observer can be obtained by:

$f\text{'}=\left(\frac{v+v_o}{v-v_s}\right)f$

Here, v is the speed of the sound, $v_o$ is the speed of the observer, $v_s$ is the speed of the source and f is the original frequency of the sound.

Calculations:

$v_o=0$

$v_s=v_a$

$\begin{array}{c}f\text{'}=\left(\frac{340}{340-170}\right)\left(340\text{Hz}\right)\\ =680\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of the sound wave detected by an observer located directly in front of the airplane is 680 Hz.

(e)

To determine

The frequency of the sound wave detected by an observer located directly behind the airplane.

Answer to Problem 101A

  $226.67\text{ Hz}$

Explanation of Solution

Given:

Speed of sound, $v=340\text{m/s}$

Wavelength of the sound wave, $\lambda =1\text{m}$ (calculated in previous part)

Frequency of the sound, $f=340\text{Hz}$

Speed of the airplane, $v_a=170\text{m/s}$

Formula Used:

From Doppler effect, frequency of the sound wave when source is moving away from the observer can be obtained by:

$f\text{'}=\left(\frac{v-v_o}{v+v_s}\right)f$

Here, v is the speed of the sound, $v_o$ is the speed of the observer, $v_s$ is the speed of the source and f is the original frequency of the sound.

Calculations:

$v_o=0$

$v_s=v_a$

$\begin{array}{c}f\text{'}=\left(\frac{340}{340+170}\right)\left(340\text{Hz}\right)\\ =226.67\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of the sound wave detected by an observer located directly in front of the airplane is $226.67\text{Hz}$ .