Physics: doppler effect (observer & source in motion)

AI Thread Summary
A train is moving at 25 m/s while a car, initially 42 m behind and traveling at 49 m/s with an acceleration of 4 m/s², hears the train's whistle at a frequency of 460 Hz. The frequency perceived by the car's driver is calculated using the Doppler effect formula, but initial attempts yield incorrect results. The discussion emphasizes the need to consider the time delay for sound to reach the car and the car's acceleration when calculating the frequency after it has passed the train. It suggests that the effective wavelength changes due to the car's acceleration, requiring a more nuanced approach to the calculations. Understanding the Doppler effect in this context involves careful application of the relevant equations and consideration of the car's motion.
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Homework Statement


A train is moving parallel and adjacent to a highway with a constant speed of 25 m/s. A car, at time t0, is 42 m behind the train, traveling in the same direction as the train at 49 m/s, and accelerating at 4 m/s^2. The train's whistle blows at a frequency of 460Hz. The speed of sound in air is 343 m/s. At the moment t0, the car's driver hears a whistle from the train.

What frequency does the car's driver hear?

Part B. @ a time t, after the car has passed the train and is a distance 32 m ahead of the train, the whistle blows again. What frequency does the driver hear?


Homework Equations



f' =f [ (1+ v(observer) / v) / (1- v(source) / v)] <---equation 1

x(final) = x(initial) + v t +.5 a t2 <---equation 2

f'= f [1- (v(observer)/v)] <---equation 3

The Attempt at a Solution



Part A:

460[(1+(49/343)) / (1-(25/343)) = 567.044 = WRONG ANSWER


Part B:


I treated the train as stationary to find the time,I determined the car to be traveling 24 m/s
So using equation 2
74 = 0 + 24 t + ( .5 * 4 ) t2
v(final)=49 + ( 2.54 * 4 ) = 59.16 m/s



then using equation 3 I did

f'=460(1- (59.16/343) ) = 380.66=WRONG
 
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f' =f (1+ v(observer) / v) / (1- v(source) / v)] <---equation 1
Is this what you get on simplifying the above equation?

f&#039;=f(\frac{v+v_{obs}}{v-v_{s}} )

Try not to apply the equation directly. Since, the frequency is received by the car at that moment t_{0} surely the sound wave of that frequency which we have to find out must have left some time t_{b} ago for it to reach at that instant. Apply this, it might help.

As for computing the frequency during acceleration, you cannot simply apply the standard equations. For every time period T that the wave moves forward a length \lambda the car moves to cover a length of \frac{1}{2}aT^2. The effective wavelength should now become: \lambda&#039; = \lambda - \frac{1}{2}aT^2.
 
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