Q54E - University Physics with Modern Physics

Question

The siren of a fire engine that is driving northward at 30.0 m/s emits a sound of frequency 2000 Hz. A truck in front of this fire engine is moving northward at 20.0 m/s. (a) What is the frequency of the siren’s sound that the fire engine’s driver hears reflected from the back of the truck? (b) What wavelength would this driver measure for these reflected sound waves?

Step-by-Step Solution

Answer

a) Frequency of reflected siren’s sound is $f_{engine}=2120.7Hz$

b) Wavelength of reflected sound waves is ${\lambda }_{engine}=0.176m$

Step 1:

a) Determining the frequency detected by the truck and then considering the truck as a source emitting that frequency towards the fire engine

So, first, the fire engine is the source, the truck is the observer, and the frequency detected by the truck is described as:

$f_t=f_s\left(\frac{v\pm v_t}{v+v_s}\right)$

As given;

The speed of sound is $v=344m/s$

Speed of truck is $v_t=20m/s$

Speed of the fire engine $V_s=30m/s$

Frequency of sound emitted by fire engine $f_s=2000\mathrm{Hz}$

Frequency of sound detected by the truck $f_t$

On putting the values;

$\begin{array}{l}{\mathrm{f}}_{\mathrm{t}}=\left(2000\mathrm{Hz}\right)\times (\frac{(344\mathrm{m}/\mathrm{s})-(20\mathrm{m}/\mathrm{s})}{(344\mathrm{m}/\mathrm{s}-(30\mathrm{m}/\mathrm{s})}\\ =2064\mathrm{Hz}\end{array}$

Used the minus sign at the numerator as the truck is receding from the fire engine, and the minus sign at the denominator as the fire engine approaches the truck.

Step 2:

Now, the sound emitted by the fire engine will strike the truck and reflect back to the fire engine, so the truck will be considered a source emitting a frequency of 2064Hz, so the equation is:

$f_{engine}=f_s\left(\frac{v\pm v_{engine}}{v\pm v_{truck}}\right)$

Used the plus sign at the numerator as the fire engine approaches the truck as it is a source and plus sign at the denominator as the truck receding from the fire engine;

$\begin{array}{l}{\mathrm{f}}_{engine}=\left(2064\mathrm{Hz}\right)\times (\frac{(344\mathrm{m}/\mathrm{s})+(30\mathrm{m}/\mathrm{s})}{(344\mathrm{m}/\mathrm{s}+(20\mathrm{m}/\mathrm{s})}\\ {\mathrm{f}}_{engine}=2120.7\mathrm{Hz}\end{array}$

Hence, the frequency of the reflected siren’s sound is ${\mathrm{f}}_{engine}=2120.7\mathrm{Hz}$

Step 3:

b)  The truck is moving away from the fire engine, indicating that the fire engine is detecting the waves behind the truck as well as the wavelength of the waves behind the source. So, it is expressed as:

${\lambda }_{behind}=\frac{v+v_S}{f_S}$

On putting the values;

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{behind}}=\frac{\left(344\mathrm{m}/\mathrm{s}\right)+\left(20\mathrm{m}/\mathrm{s}\right)}{2064\mathrm{Hz}} \\ {\mathrm{\lambda }}_{\mathrm{behind}}=0.176\mathrm{m} \end{gathered}$$

Hence, the wavelength of reflected sound waves is ${\mathrm{\lambda }}_{\mathrm{behind}}=0.176\mathrm{m}$