 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q17.4-30PE

Expert-verified Found in: Page 629 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # (a) What frequency is received by a person watching an oncoming ambulance moving at $$110\;{\rm{km/h}}$$and emitting a steady $$800\;{\rm{Hz}}$$ sound from its siren? The speed of sound on this day is $$345\;{\rm{m/s}}$$. (b) What frequency does she receive after the ambulance has passed?

(a) The apparent frequency is $$877.58\;{\rm{Hz}}$$.

(b) The apparent frequency is $$735.01\;{\rm{Hz}}$$.

See the step by step solution

## The Concept of Apparent Frequency

The apparent frequency to a listener changes depending the sound source is moving away or toward the listener.

## Given Data

The frequency is $$f = 800\;{\rm{Hz}}$$.

The speed of the ambulance is $${v_s} = 100\;{\rm{km/h}} = 30.5\;{\rm{m/s}}$$.

The speed of sound is $$v = 345\;{\rm{m/s}}$$.

The speed of the listener is zero.

## Calculation of the frequency of the oncoming ambulance

(a)

The Doppler Effect tells the apparent frequency is,

$$f' = f\left( {\frac{{v - {v_o}}}{{v - {v_s}}}} \right)$$

Now, the apparent frequency is,

$$\begin{array}{c}f' = 800\left( {\frac{{345 - 0}}{{345 - 30.5}}} \right)\\ = 800 \times \frac{{690}}{{629}}\\ = 877.58\;{\rm{Hz}}\end{array}$$

Hence, the apparent frequency is $$877.58\;{\rm{Hz}}$$ .

## Calculation of the frequency of the passing away ambulance

(b)

Now, the apparent frequency of the passing away ambulance is,

$$\begin{array}{c}f'' = 800\left( {\frac{{345 - 0}}{{345 - \left( { - 30.5} \right)}}} \right)\\ = 800 \times \frac{{690}}{{751}}\\ = 735.01\;{\rm{Hz}}\end{array}$$

Hence, the apparent frequency is $$735.01\;{\rm{Hz}}$$ . ### Want to see more solutions like these? 