Chapter 7.2, Problem 58E

(a)

To determine

To Find: An ambulance siren has a frequency of $2000hz$ . Write two equations modeling the frequency heard when the ambulance is approaching and when the ambulance is moving away.

Answer to Problem 58E

  $f_s=\frac{1480000}{740-r}$ , $f_l=\frac{1480000}{740+r}$

Explanation of Solution

Given:

  

Given:

  $\begin{array}{l}{f}_s=200\\ f_l=\frac{740f_s}{740+r}\\ f_s=\frac{740fs}{740-r}\end{array}$

Then

When the ambulance is approaching

  $\begin{array}{l}{f}_l=\frac{740(2000)}{740+r}\\ f_l=\frac{1480000}{740+r}\end{array}$

When the ambulance is moving away

  $\begin{array}{l}{f}_s=\frac{740(2000)}{740-r}\\ f_s=\frac{1480000}{740-r}\end{array}$

Hence, $f_s=\frac{1480000}{740-r}$ , $f_l=\frac{1480000}{740+r}$

(b)

To determine

To Graph: The equation using the domain $0\le r\le 60$

Answer to Problem 58E

  

Explanation of Solution

Given:

  

Given:

  $f_s=\frac{1480000}{740-r}$ , $f_l=\frac{1480000}{740+r}$

The graph of both the functions in the domain $0\le r\le 60$

  

(c)

To determine

To Find: For any speed how does the frequency heard for an approaching sound source compare with the frequency heard when the source moves away.

Answer to Problem 58E

  $f_l<f_s$

Explanation of Solution

Given:

  

Given:

  $f_s=\frac{1480000}{740-r}$ , $f_l=\frac{1480000}{740+r}$

Since, $0\le r\le 60$

i.e.

For any value of $r$

We have

  $f_l<f_s$