Chapter 12.7, Problem 10P

To determine

To calculate: The distance between the observer and the source of the signal, if the strength of a radio signal is inversely propositional to the square of the distance from the source of the signal, also given when the observer which is 200 m away from a source moves closer to the source then the signal at the second position is 16 times as staring as the first position.

Answer to Problem 10P

The distance between observer and source of the signal in second position is $d_2^{}=50$ .

Explanation of Solution

Given information:

The strength of a radio signal is inversely propositional to the square of the distance from the source of the signal

Let, Strength of the signal is $s_1^{}$ and after moving closer is $s_2^{}$ .

Distance between observer and source of the signal is $d_1^{}$ and after moving closer is $d_2^{}$ .

So $s\prec \frac{1}{d^2}$

Let k be a constant

Before moving

  $s_1^{}=\frac{k}{d_1^2}$...... (1)

After moving closer

  $s_2^{}=\frac{k}{d_2^2}$...... (2)

Also given $s_2^{}=16\cdot s_1^{}$ .

Before moving

  $s_1^{}=\frac{k}{d_1^2}$

  $s_1^{}=\frac{k}{{\left(200\right)}^2}$...... (3)

After moving closer

  $\begin{array}{l}{s}_2^{}=\frac{k}{d_2^2}\\ 16\cdot s_1^{}=\frac{k}{d_2^2}\end{array}$...... (4)

Divide equation 3 and 4

  $\frac{s_1^{}}{16\cdot s_1^{}}=\frac{d_2^2}{{\left(200\right)}^2}$

  ${\left(\frac{d_2^{}}{200}\right)}^2=\frac{1}{16}$

  $\frac{d_2^{}}{200}=\sqrt{\frac{1}{16}}$

Solve further,

  $\frac{d_2^{}}{200}=\frac{1}{4}$

  $d_2^{}=50$