Two train whistles, A and B, each have a frequency of 392 Hz. A is stationary and B is moving toward the right?
Two train whistles, A and B, each have a frequency of 392 Hz. Whistle A is stationary, while whistle B is moving to the right (away from A) at a speed of 35.0 m/s. A listener is positioned between the two whistles and is moving to the right at a speed of 15.0 m/s. No wind is blowing.
a) What frequency does the listener hear from whistle A?
b) What frequency does the listener hear from whistle B?
c) What is the beat frequency detected by the listener?
(Note: There are no question marks in the phrase "away from A," as it is a continuation of the thought.)
3 Answers
this is doppler effect.
observed frequency=f=((v+vl)/(v+vs))*fo
v=velocity of wave in medium=343m/s
vl=velocity of listener(+ if listener moving towards the source and – if moving away from source)
vs=velocity of source(+if source moving away from observer and – if source moving towards observer)
a)v=343,
vl= -15
vs=0
f0=392
fa can be found.
b)v=343
vl=15
vs=35
f0=392
fb can be found.
c)beat frequency=fa-fb
Feb 06, 2025
with respect to A the listener is opening the distance at 15 m/s
F = Fo ( V / ( V + Vl )
where V is speed of sound and Vl is relative speed of listener
F = 392 ( 343 / (343+15) = 375.6
with respect to B the source is moving away from the listener at 20 m/s
F = Fo ( ( V – Vs ) / V
where Vs is the speed of the source
F = 392 ( (343-20) / 343 ) = 369.1
the difference gives 6.5 beats/sec
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