Frequency, speed of train?

Question

An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 446 Hz when the train is approaching him and 442 Hz when the train is receding from him. From this he can find the speed of the train. What value does he find?
_______m/s (Assume the speed of sound is 345 m/s.)

Answer 1

this is Doppler effect:
http://en.wikipedia.org/wiki/Doppler_effect

The Doppler effect, named after Christian Doppler, is the change in frequency and wavelength of a wave as perceived by an observer moving relative to the source of the waves.

The relationship between observed frequency f' and emitted frequency f is given by:

f' = (v / (v +/- vsource)) f

where
v is the speed of waves in the medium
vsource is the velocity of the source (the object emitting the sound)
Because the detected frequency increases for objects moving toward the observer, the source's velocity must be subtracted when motion is moving toward the observer. (This is because the source's velocity is in the denominator.) Conversely, detected frequency decreases when the source moves away, and so the source's velocity is added when the motion is away.

f1 = 446 Hz f1 = f [v/(v - vsource)]

f2 = 442 Hz f2 = f [v/(v + vsource)]

f is the true frequency of the train whistle

f1/ f2 =(v + vsource) / (v - vsource)

(v-vsource)* f1/f2=(v + vsource)

v * f1/f2-vsource* f1/f2=v + vsource

v *( f1/f2-1)=vsource *(f1/f2+1)

vsource = v *( f1/f2-1)/( f1/f2+1)

vsource = 1.5541 m/s

Answer 2

The Doppler Effect is described by the equation

f_observed = f_emitted ( v / ( v + vs ) )

where v is the speed of sound and vs the relative speed of the source.

446 = f_emitted ( 345 / ( 345 - vs ) )
442 = f_emitted ( 345 / ( 345 + vs ) )

Two equations, two unknowns; solve for f_emitted (if desired; it'll be about 444 Hz) and the velocity of the train vs.

Answer 3

Since 2 situations are given we can find out both velocity of the train and also the original frequency of the train i.e. as observed by an observer in the ref frame of the train

vel of train=1.55m/s
original freq=443.399 Hz


f(1)= [345/(345-Vs)]f(o)

f(2)=[345/(345+Vs)]f(o)

solve these 2 eqns

Answer 4

The substitute in frequency is 551-466=one hundred and five Hz. this suggests the actually frequency is 551-one hundred and five/2=498.5 Hz. using fact the frequency has larger 10.fifty 3%, The practice is moving at 10.fifty 3% of 340 m/sec=35.8 m/sec.

Answer 5

http://en.wikipedia.org/wiki/Doppler_effect

Use the very first equation with your data and solve for velocity of the source.