Chapter 17, Problem 17.11OQ

To determine

The effect on observed frequency due to wind.

Answer to Problem 17.11OQ

The rank of the observed frequency from highest to lowest is $\left(b\right)>\left(d\right)>\left(a\right)>\left(c\right)>\left(e\right)$.

Explanation of Solution

Given info: The speeds of the source and observer are $25\text{m}/\text{s}$ respectively.

Formula to calculate the observed frequency by observer is,

$f\text{'}=\left(\frac{c+v_{\text{o}}}{c-v_{\text{s}}}\right)f$ (1)

Here,

$f\text{'}$ is the observed frequency.

$c$ is the speed of the sound in the vacuum.

$f$ is the emitted frequency.

$v_{\text{o}}$ is the speed of observer.

$v_{\text{s}}$ is the speed of the source.

For case $\left(\text{a}\right)$:

The source and observer are stationary. Hence, the speed of the source and observer are $0$ respectively.

Substitute $0$ for $v_{\text{o}}$ and $0$ for $v_{\text{s}}$ in equation (1) to find $f\text{'}$.

$\begin{array}{c}f\text{'}=\left(\frac{c+0}{c-0}\right)f\\ f\text{'}=f\end{array}$

Thus, the speed of the observed frequency is equal to the real frequency.

For case $\left(\text{b}\right)$:

The source is moving toward the stationary observer. Hence, the speed of the source and observe are $\text{25}\text{m}/\text{s}$ and $0$.

Substitute $0$ for $v_{\text{o}}$ and $\text{25}\text{m}/\text{s}$ for $v_{\text{s}}$ in equation (1) to find $f\text{'}$.

$\begin{array}{c}f\text{'}=\left(\frac{c+0}{c-\text{25}\text{m}/\text{s}}\right)f\\ f\text{'}=\left(\frac{c}{c-\text{25}\text{m}/\text{s}}\right)f\end{array}$ (2)

Assume the observed frequency is $f_1$ instead of $f\text{'}$.

Substitute $f\text{'}$ for $f_1$ in equation (2).

$f_1=\left(\frac{c}{c-\text{25}\text{m}/\text{s}}\right)f$ (3)

For case $\left(\text{c}\right)$:

The source is moving away from stationary observer. Hence, the speed of the source and observe are $-\text{25}\text{m}/\text{s}$ and $0$.

Substitute $0$ for $v_{\text{o}}$ and $-\text{25}\text{m}/\text{s}$ for $v_{\text{s}}$ in equation (1) to find $f\text{'}$.

$\begin{array}{c}f\text{'}=\left(\frac{c+0}{c-\left(-\text{25}\text{m}/\text{s}\right)}\right)f\\ f\text{'}=\left(\frac{c}{c+\text{25}\text{m}/\text{s}}\right)f\end{array}$ (4)

Assume the observed frequency is $f_2$ instead of $f\text{'}$.

Substitute $f\text{'}$ for $f_2$ in equation (4).

$f_2=\left(\frac{c}{c+\text{25}\text{m}/\text{s}}\right)f$ (5)

For case $\left(\text{d}\right)$:

The observer is moving toward the stationary source. Hence, the speed of the observer and source are $\text{25}\text{m}/\text{s}$ and $0$.

Substitute $\text{25}\text{m}/\text{s}$ for $v_{\text{o}}$ and $0$ for $v_{\text{s}}$ in equation (1) to find $f\text{'}$.

$\begin{array}{c}f\text{'}=\left(\frac{c+\text{25}\text{m}/\text{s}}{c-0}\right)f\\ f\text{'}=\left(\frac{c+\text{25}\text{m}/\text{s}}{c}\right)f\end{array}$ (6)

Assume the observed frequency is $f_3$ instead of $f\text{'}$.

Substitute $f\text{'}$ for $f_3$ in equation (6).

$f_3=\left(\frac{c+\text{25}\text{m}/\text{s}}{c}\right)f$ (7)

For case $\left(\text{e}\right)$:

The observer is moving away from stationary source. Hence, the speed of the observer and source are $-\text{25}\text{m}/\text{s}$ and $0$.

Substitute $-\text{25}\text{m}/\text{s}$ for $v_{\text{o}}$ and $0$ for $v_{\text{s}}$ in equation (1) to find $f\text{'}$.

$\begin{array}{c}f\text{'}=\left(\frac{c+\left(-\text{25}\text{m}/\text{s}\right)}{c-0}\right)f\\ f\text{'}=\left(\frac{c-\text{25}\text{m}/\text{s}}{c}\right)f\end{array}$ (8)

Assume the observed frequency is $f_4$ instead of $f\text{'}$.

Substitute $f\text{'}$ for $f_4$ in equation (8).

$f_4=\left(\frac{c-\text{25}\text{m}/\text{s}}{c}\right)f$ (9)

Divide the equation (3) by equation (7).

$\begin{array}{l}\frac{f_1}{f_3}=\frac{\left(\frac{c}{c-\text{25}\text{m}/\text{s}}\right)f}{\left(\frac{c+\text{25}\text{m}/\text{s}}{c}\right)f}\\ =\frac{v^2}{\left(v^2-625{\text{m}}^2/{\text{s}}^2\right)}\end{array}$

Thus, observed frequency for case $\left(\text{b}\right)$ is greater than case $\left(\text{d}\right)$.

$f_1>f_3$

Divide the equation (5) by equation (9).

$\begin{array}{c}\frac{f_2}{f_4}=\frac{\left(\frac{c}{c+\text{25}\text{m}/\text{s}}\right)f}{\left(\frac{c-\text{25}\text{m}/\text{s}}{c}\right)f}\\ =\frac{v^2}{v^2-625{\text{m}}^2/{\text{s}}^2}\end{array}$

Thus, observed frequency for case $\left(\text{c}\right)$  is greater than case $\left(\text{e}\right)$.

$f_2>f_4$

The rank of the observed frequency from highest to lowest is,

$\begin{array}{c}{f}_1>f_3>f\text{'}>f_2>f_4\\ \left(b\right)>\left(d\right)>\left(a\right)>\left(c\right)>\left(e\right)\end{array}$

Conclusion:

Therefore, the rank of the observed frequency from highest to lowest is $\left(b\right)>\left(d\right)>\left(a\right)>\left(c\right)>\left(e\right)$