Respuesta :
Answer:
39.45 m/s
Explanation:
This phenomenon is known as Doppler Effect, and it occurs when there is a source of a wave in motion with respect to an observer. In such situation, the frequency of the wave as perceived by the observer is shifted with respect to the real frequency.
In particular:
- The observed frequency is higher if the source is moving towards the observer
- The observed frequency is lower if the source is moving away from the observer
The formula to calculate the apparent frequency in the Doppler Effect is:
[tex]f'=\frac{v \pm v_o}{v\pm v_s}f[/tex]
where
f is the real frequency
f' is the observed frequency
v is the speed of the wave
[tex]v_s[/tex] is the velocity of the source, which is positive if the source is moving away, negative if it is moving towards the observer
[tex]v_o[/tex] is the velocity of the observer, which is negative if the observer is moving away, positive if it is moving towards the source
In this problem:
v = 340 m/s is the speed of sound
- When the ambulance is approaching you from behind,
f' = 320 Hz
[tex]v_o=-30 m/s[/tex] (car is moving away from ambulance)
So
[tex]320 = \frac{340-30}{340 -v_s}f[/tex] (1)
- When the ambulance overcomes you and moves away from you,
f' = 300 Hz
[tex]v_o=+30 m/s[/tex] (car is moving towards ambulance)
So
[tex]300 = \frac{340+30}{340 +v_s}f[/tex] (2)
Re-arranging both equations (1) and (2),
1)
[tex]320 = \frac{310}{340 -v_s}f\\320(340-v_s)=310f\\108,800-320v_s = 310 f[/tex]
2)
[tex]300 = \frac{370}{340 +v_s}f\\300(340+v_s)=370f\\102,000+300v_s = 370f[/tex]
Both equations can be rewritten as
[tex]f=\frac{108,000-320v_s}{310}\\f=\frac{102,000+300v_s}{370}[/tex]
And by equating them we find:
[tex]\frac{108,000-320v_s}{310}=\frac{102,000+300v_s}{370}\\37(10,800-32v_s)=31(10,200+30v_s)\\399,600-1184v_s=316,200+930v_s\\83,400=2114v_s\\v_s=\frac{83400}{2114}=39.45 m/s[/tex]