Chapter 4.3, Problem 52AYU

(a)

To determine

To write: a function $f^{\prime }\left(v_s\right)$

Answer to Problem 52AYU

The required function is $f^{\prime }\left(v_s\right)=\frac{461040}{772.4-v_s}$

Explanation of Solution

Given:

Traveling down the road at $45mph$

The actual pitch of the siren is $600hertz\left(Hz\right)$

Calculation:

Consider the function for the Doppler Effect given by the following relation:

  $f^{\prime }=f_a\left(\frac{\upsilon -{\upsilon }_o}{\upsilon -{\upsilon }_s}\right)$

Where

  

Since we are travelling at $45mph$ on the road, we will consider $v_o=45$ .

Also, the actual pitch of the siren is $600Hz$ , hence we will consider $f_a=600$ .

The function hence formed from the given information becomes:

  $\begin{array}{l}{f}^{\prime }\left(v_s\right)=600\left(\frac{772.4-45}{772.4-v_s}\right)\\ f^{\prime }\left(v_s\right)=\frac{461040}{772.4-v_s}\end{array}$

Conclusion:

Therefore, the required function is $f^{\prime }\left(v_s\right)=\frac{461040}{772.4-v_s}$

(b)

To determine

To find: the speed of the ambulance.

Answer to Problem 52AYU

The speed of the ambulance is $v_s=68.464\frac{m}{s}$

Explanation of Solution

Given:

  $f’=620Hz$

Calculation:

Put the value of $f^{\prime }=620$ then

  $620=600\left(\frac{772.4-45}{772.4-v_s}\right)$

By solving this equation we have the value

  $v_s=68.464\frac{m}{s}$

Conclusion:

The speed of the ambulance is $v_s=68.464\frac{m}{s}$

(c)

To determine

To graph: the functionby using a graphing utility

Answer to Problem 52AYU

  

Explanation of Solution

Calculation:

  

Conclusion:

Thus, the required graph is drawn.

(d)

To determine

To verify:the answer from part (b).

Answer to Problem 52AYU

The graph that the graphical value as well as the theoretical value of this is same at the point where $v=620Hz$

Explanation of Solution

Calculation:

It can be clearly seen from the graph that the graphical value as well as the theoretical value of this is same at the point where $v=620Hz$ .