Chapter 25, Problem 89P

To determine

The frequency of the broadcast.

Answer to Problem 89P

The frequency of the broadcast is $1530\text{kHz}$.

Explanation of Solution

Sketch the diagram showing the distances.

Find the distance $d$ from the figure using Pythagoras theorem.

    $d=\sqrt{{\left(102\text{km}\right)}^2+h^2}$                                                                             (I)

Here, $d$ is the distance between the lane and the coast guard and $h$ is the height of the plane.

Find the equation for wavelength using equation (I).

    $\begin{array}{c}m\lambda =\Delta l\\ =\sqrt{{\left(102\text{km}\right)}^2+h^2}+h-102\text{km}\\ \lambda \text{=}\frac{\sqrt{{\left(102\text{km}\right)}^2+h^2}+h-102\text{km}}{m}\end{array}$                                                        (II)

Here, $m$ is the integer, $\Delta l$ is the path difference and $\lambda$ is the wavelength.

Find the equation for frequency.

    $f=\frac{c}{\lambda }$                                                                                                  (III)

Here, $f$ is the frequency and $c$ is the speed of light.

Conclusion:

Substitute $780\text{m}$ for $h$ and $m_1$ for $m$ in equation (II) to find $\lambda$.

    $\begin{array}{c}\lambda \text{=}\frac{\sqrt{{\left(102\text{km}\right)}^2+{\left(780\text{m}\right)}^2}+780\text{m}-102\text{km}}{m_1}\\ =\frac{\sqrt{{\left(\left(102\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2+{\left(780\text{m}\right)}^2}+780\text{m}-\left(102\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{m_1}\\ =\frac{783\text{m}}{m_1}\end{array}$        (IV)

Substitute $975\text{m}$ for $h$ and $m_2$ for $m$ in equation (II) to find $\lambda$.

    $\begin{array}{c}\lambda \text{=}\frac{\sqrt{{\left(102\text{km}\right)}^2+{\left(975\text{m}\right)}^2}+975\text{m}-102\text{km}}{m_2}\\ =\frac{\sqrt{{\left(\left(102\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2+{\left(975\text{m}\right)}^2}+975\text{m}-\left(102\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{m_2}\\ =\frac{980\text{m}}{m_2}\end{array}$           (V)

Substitute $1170\text{m}$ for $h$ and $m_3$ for $m$ in equation (II) to find $\lambda$.

    $\begin{array}{c}\lambda \text{=}\frac{\sqrt{{\left(102\text{km}\right)}^2+{\left(1170\text{m}\right)}^2}+1170\text{m}-102\text{km}}{m_3}\\ =\frac{\sqrt{{\left(\left(102\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^2+{\left(1170\text{m}\right)}^2}+1170\text{m}-\left(102\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{m_3}\\ =\frac{1170\text{m}}{m_3}\end{array}$       (VI)

Find the ratio $\frac{m_2}{m_1}$ from equations (IV) and (V).

    $\begin{array}{c}\frac{m_2}{m_1}=\frac{980\text{m}}{783\text{m}}\\ =\frac{5}{4}\end{array}$                                                                                               (VII)

Find the ratio $\frac{m_3}{m_1}$ from equations (V) and (VI).

    $\begin{array}{c}\frac{m_3}{m_1}=\frac{1177\text{m}}{783\text{m}}\\ =\frac{6}{4}\end{array}$                                                                                              (VIII)

From equations (VII) and (VIII), the value of $m_1$ is 4.

Substitute $4$ for $m_1$ in equation (IV) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{783\text{m}}{4}\\ =196\text{m}\end{array}$

Substitute $196\text{m}$ for $\lambda$ and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (III) to find $f$.

    $\begin{array}{c}f=\frac{3.00\times {10}^8\text{m/s}}{196\text{m}}\\ =\left(1530000\text{Hz}\right)\left(\frac{{10}^{-3}\text{kHz}}{1\text{Hz}}\right)\\ =1530\text{kHz}\end{array}$

Thus, the frequency of the broadcast is $1530\text{kHz}$.