91Ó°ÊÓ

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:

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

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;

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

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.

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:

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

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;

localid="1664342954484" $\begin{array}{l}{\mathrm{f}}_{engine}=\left(2064\mathrm{Hz}\right)×(\frac{(344\mathrm{m}/\mathrm{s})+(30\mathrm{m}/\mathrm{s})}{(344\mathrm{m}/\mathrm{s}+(20\mathrm{m}/\mathrm{s})}\\ {\mathrm{f}}_{engine}=2120.7\mathrm{Hz}\end{array}$

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

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:

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

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

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