Answer:
[tex]v_s=87m/s[/tex]
Explanation:
Let´s use Doppler effect, in order to calculate the speed of the vehicule. The Doppler effect equation for a general case is given by:
[tex]f_o=f_s*(\frac{v\pm v_o}{v\pm v_s})[/tex]
Where:
[tex]f_o=Observed\hspace{3}frequency[/tex]
[tex]f_s=Actual\hspace{3}frequency[/tex]
[tex]v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves[/tex]
[tex]v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]
[tex]v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer[/tex]
Now let's consider the next cases:
[tex]+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source[/tex]
[tex]-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source[/tex]
[tex]-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer[/tex]
[tex]+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer[/tex]
So, in this case:
[tex]v_o=25m/s[/tex]
[tex]f_o=2300Hz[/tex]
[tex]f_s=1600Hz[/tex]
[tex]v=343m/s[/tex]
Besides, the source and the observer both are in motion towards each other, hence:
[tex]f_o=f_s*(\frac{v+ v_o}{v- v_s})[/tex]
Solving for [tex]v_s[/tex]
[tex]v_s=v-(v+v_o)*\frac{f_s}{f_o}[/tex]
Finally, replacing the data provided:
[tex]v_s=343-(343+25)*\frac{1600}{2300} =87m/s[/tex]
The speed of the emergency vehicle is about 37 m/s
[tex]\texttt{ }[/tex]
Let's recall the Doppler Effect formula as follows:
[tex]\large {\boxed {f' = \frac{v + v_o}{v - v_s} f}}[/tex]
f' = observed frequency
f = actual frequency
v = speed of sound waves
v_o = velocity of the observer
v_s = velocity of the source
Let's tackle the problem!
[tex]\texttt{ }[/tex]
Given:
initial observed frequency = f'_1 = 2300 Hz
final observed frequency = f'_2 = 1600 Hz
velocity of the observer = v_o = 25 m/s
speed of sound in air = v = 343 m/s
Asked:
speed of the source = v_s = ?
Solution:
We will use the formula of Doppler Effect.
When the vehicle is approaching you,
[tex]f'_1 = \frac{v + v_o}{v - v_s} \times f[/tex]
[tex]2300 = \frac{343 + 25}{343 - v_s} \times f[/tex]
[tex]2300 = \frac{368}{343 - v_s} \times f[/tex]
[tex]2300(343 - v_s) = 368f[/tex]
[tex](343 - v_s) = 368f \div 2300[/tex]
[tex](343 - v_s) = \frac{4}{25}f[/tex] → Equation 1
[tex]\texttt{ }[/tex]
When the vehicle is past you,
[tex]f'_2 = \frac{v + v_o}{v - v_s} \times f[/tex]
[tex]1600 = \frac{343 + (-25)}{343 - (-v_s)} \times f[/tex]
[tex]1600 = \frac{318}{343 + v_s} \times f[/tex]
[tex]1600(343 + v_s) = 318f[/tex]
[tex](343 + v_s) = 318f \div 1600[/tex]
[tex](343 + v_s) = \frac{159}{800}f[/tex] → Equation 2
[tex]\texttt{ }[/tex]
Next, we will solve the two equations above by adding them ( Equation 1 + Equation 2 ):
[tex](343 - v_s) + (343 + v_s) = \frac{4}{25}f + \frac{159}{800}f[/tex]
[tex]686 = \frac{287}{800}f[/tex]
[tex]f = 686 \div \frac{287}{800}[/tex]
[tex]f = 1912\frac{8}{41} \texttt{ Hz}[/tex]
[tex]\texttt{ }[/tex]
[tex](343 - v_s) = \frac{4}{25}f[/tex]
[tex](343 - v_s) = \frac{4}{25}( 1912\frac{8}{41} )[/tex]
[tex]343 - v_s = 305\frac{39}{41}[/tex]
[tex]v_s = 343 - 305\frac{39}{41}[/tex]
[tex]v_s = 37\frac{2}{41} \texttt{ m/s}[/tex]
[tex]v_s \approx 37 \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Sound Waves
[tex]\texttt{ }[/tex]
Keywords: Sound, Wave , Wavelength , Doppler , Effect , Policeman , Stationary , Frequency , Speed , Beats
