Answer:
The amplitude is [tex]A = 90.2 \ m[/tex]
Explanation:
From the question we are told that
The frequency of when sound is approaching observer is [tex]f = 392 Hz[/tex]
The frequency as the move away from observer is [tex]f_ a = 330 \ Hz[/tex]
The time between the pitch are [tex]t = 10 \ s[/tex]
Here you are the observer and your friends are the source of the sound
The period is mathematically evaluated as
[tex]T = 2 t[/tex]
as it is the time to complete one oscillation which from on highest pitch to the next highest pitch
Now T can also be mathematically represented as
[tex]T = \frac{2 \pi}{w}[/tex]
Where [tex]w[/tex] is the angular velocity
=> [tex]\frac{2 \pi}{w} = 2 * 10[/tex]
=> [tex]w = 0.314 \ rad/sec[/tex]
Now using Doppler Effect,
The source of the sound is approaching the observer
The
[tex]f = f_o (\frac{v}{v- wA} )[/tex]
[tex]392 = f_o (\frac{v}{v- wA} )[/tex]
Where A is the amplitude
So when the source is moving away from the observer
[tex]f_a = f_o (\frac{v}{v+ wA} )[/tex]
[tex]330 = f_o (\frac{v}{v+ wA} )[/tex]
Here [tex]f_o[/tex] is the fundamental frequency
Dividing the both equation we have
[tex]\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}[/tex]
[tex]1.1878 = \frac{v+wA}{v-wA}[/tex]
[tex]1.1878 v - 1.1878 wA = v+wA[/tex]
[tex]1.1878 v = 2.1878 wA[/tex]
=> [tex]A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}[/tex]
[tex]A = 90.2 \ m[/tex]
