Answer:
A) 594 Hz
B) 512 Hz
Explanation:
Howdy!
This is a typical Doppler effect problem, whose formula is:
[tex]f = \frac{v+v_{r} }{v+v_{s}} f_{0}[/tex]
Where :
[tex]v_{r}[/tex]
is the velocity of the receiver (in our case is equal to zero)
[tex]v_{s}[/tex]
is the velocity of the source (in our case is 92 km/h) which is positive when the source is moving away (second case) and negative otherwise.
[tex]v_{s}[/tex]
Is the velocity of the wave in the medium.
Before we start calculating we need to have the velocity of the source in the same units as the velocity of the waves:
92km/h = 92*(1000)/3600 m/s = 25.5 m/s
A)
When the source is moving towards the receiver the sign of the velocity is negative, so:
[tex]f = \frac{343}{343-25.5} 550 Hz[/tex]
f = 594.1 Hz
B) Now the velocity of the source must change sign:
[tex]f = \frac{343}{343+25.5} 550 Hz[/tex]
f = 511.9 Hz
As a sanity check we know that when the source is moving towards the source the frequency is higher and when the source is moving away from the source the frequency is lower.
