Chapter 1, Problem 18P

(a)

To determine

The show that the reflected wave has frequency of $f=f_{\text{source}}\frac{c+v}{c-v}$

Answer to Problem 18P

The reflected wave has frequency of $f=f_{\text{source}}\frac{c+v}{c-v}$ is shown.

Explanation of Solution

Write the expression for the frequency as seen by the car.

    $f_c=f_{\text{source}}\sqrt{\frac{c+v}{c-v}}$

Here, $f_c$ is the frequency as seen by the car, $f_{\text{source}}$ is the frequency of the source, $c$  is the speed of light and $v$ is the speed of the car.

Write the expression for the frequency of the reflected wave.

    $f=f_c\sqrt{\frac{c+v}{c-v}}$

Here, $f$ is the frequency of the reflected wave.

Conclusion:

Use $f_{\text{source}}\sqrt{\frac{c+v}{c-v}}$ for $f_c$ in above equation and rewrite.

    $\begin{array}{c}f=\left(f_{\text{source}}\sqrt{\frac{c+v}{c-v}}\right)\left(\sqrt{\frac{c+v}{c-v}}\right)\\ =f_{\text{source}}\left(\frac{c+v}{c-v}\right)\end{array}$

Therefore, the reflected wave has frequency of $f=f_{\text{source}}\frac{c+v}{c-v}$ is shown.

(b)

To determine

The show that the beat frequency can be written as $f_{\text{beat}}=2v/\lambda$

Answer to Problem 18P

The beat frequency can be written as $f_{\text{beat}}=2v/\lambda$ is shown.

Explanation of Solution

Use the expression for the frequency obtained in (a) and simplify.

    $\begin{array}{c}f\left(c-v\right)=f_{\text{source}}\left(c+v\right)\\ \left(f-f_{\text{source}}\right)c=\left(f+f_{\text{source}}\right)v\\ \approx 2f_{\text{source}}v\end{array}$

The expression for the beat frequency is,

    $f_{\text{beat}}=f-f_{\text{source}}$

Use $2f_{\text{source}}v$ for $f-f_{\text{source}}$ in the above expression.

    $\begin{array}{c}{f}_{\text{beat}}=\frac{2f_{\text{source}}v}{c}\\ =\frac{2v}{\lambda }\end{array}$

Conclusion:

Therefore, the beat frequency can be written as $f_{\text{beat}}=2v/\lambda$ is shown.

(c)

To determine

The beat frequency measured for a car speed of 30.0 m/s if the microwaves have frequency 10.0 GHz.

Answer to Problem 18P

The beat frequency measured for a car speed of 30.0 m/s if the microwaves have frequency 10.0 GHz is $2.00\text{ kHz}$

Explanation of Solution

Rewrite the expression for beat frequency

    $f_{\text{beat}}=\frac{2v}{\lambda }$

Use $\frac{c}{f_{\text{source}}}$ for $\lambda$ in above equation and rewrite.

    $f_{\text{beat}}=\frac{2v{f}_{\text{source}}}{c}$

Conclusion:

Substitute $30.0\text{m/s}$ for v , $10.0\text{GHz}$ for $f_{\text{source}}$ , and $3.00\times {10}^8\text{m}/\text{s}$ for c.

    $\begin{array}{l}{f}_{\text{beat}}=\frac{2\left(30.0\text{m/s}\right)\left(10.0\text{GHz}\right)}{\left(3.00\times {10}^8\text{m}/\text{s}\right)}\\ =\frac{2\left(30.0\text{m/s}\right)\left(10.0\text{GHz}\right)\left(\frac{{10}^9\text{ Hz}}{1\text{GHz}}\right)}{\left(3.00\times {10}^8\text{m}/\text{s}\right)}\\ =2.00\text{ kHz}\end{array}$

Therefore, the beat frequency measured for a car speed of 30.0 m/s if the microwaves have frequency 10.0 GHz is $2.00\text{ kHz}$

(d)

To determine

If the beat frequency measured is accurate to $\pm 5\text{Hz}$, how accurate is the velocity measured.

Answer to Problem 18P

The velocity measured is accurate to $0.0750\text{m}/\text{s}$

Explanation of Solution

The expression for change beat frequency is,

    $\Delta f_{\text{beat}}=\frac{2\Delta v{f}_{\text{source}}}{c}$

Rearrange in terms of v.

    $\Delta v=\frac{\Delta f_{\text{beat}}c}{2f_{\text{source}}}$

Conclusion:

Substitute $5\text{Hz}$ for $\Delta f_{\text{beat}}$, $10.0\text{GHz}$ for $f_{\text{source}}$, and $3.00\times {10}^8\text{m}/\text{s}$ for c.

    $\begin{array}{c}\Delta v=\frac{5\text{Hz}\left(3.00\times {10}^8\text{m}/\text{s}\right)}{2\left(10.0\text{GHz}\right)\left(\frac{{10}^9\text{ Hz}}{1\text{GHz}}\right)}\\ =0.0750\text{m}/\text{s}\end{array}$

Therefore, the velocity measured is accurate to $0.0750\text{m}/\text{s}$