# Doppler Effect (moving source, moving listener)

1. Dec 2, 2007

1. The problem statement, all variables and given/known data
A man is traveling on a bicycle at 14 m/s along a straight road that runs parallel to some railroad tracks. He hears the whistle of a train that is behind him. The frequency emitted by the train is 840 Hz, but the frequency the man hears is 778 Hz. Take the velocity of sound to be 340 m/s.

a) What frequency is heard by a stationary observer located between the train and the bicycle?
b) What is the speed of the train, and is the train traveling away from or toward the bicycle?

velocity of man=14 m/s, frequency heard by man=778 Hz, velocity of train=unknown, frequency emitted by train=840 Hz, speed of sound=340 m/s

2. Relevant equations
$$f_{L}=\frac{v+v_{L}}{v+v_{S}}f_{S}$$

$$f_{L}$$=frequency heard by listener (man), $$f_{S}$$=frequency emitted by source (train), $$v_{L}$$=velocity of listener (man), $$v_{S}$$=velocity of source (train), v=velocity of sound

3. The attempt at a solution

I did part b first. Using the equation, and using direction of listener toward source as positive, I solved for velocity of the source (train) like so:
$$778=\frac{340-14}{340+v_{S}}840$$
Since the man hears the train behind him, it means he must be traveling away from the train. For velocity of the train, I got 12.0 m/s away from the man on the bicycle (since the velocity is positive).

Then, for part a, I used the same equation; velocity of listener I set to 0 m/s, and velocity of train (source) as 12 m/s, which I got from the previous part.
$$f_{L}=\frac{340+0}{340+12}840$$

I got 811 Hz as the frequency heard by a listener between the train and the bike (since the listener would be stationary, but the train would be moving away at 12 m/s).

I'd appreciate it if somebody would let me know if I did this correctly or not.

2. Dec 4, 2007