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Train whistles; moving/stationary sound and velocity

  • Thread starter gmmstr827
  • Start date
  • #1
86
1

Homework Statement



Two trains emit 516-Hz whistles. One train is stationary. The conductor on the stationary train hears a 3.5-Hz beat frequency when the other train approaches. What is the speed of the moving train?

b = beat
f'_b = 3.5 Hz
f = 516 Hz

Homework Equations



f_b = |f_1 - f_2|
v_sound = 343 m/s (speed of sound in 20°C air)
Doppler equation for "source moving toward stationary observer":
f' = f/(1+(v_source/v_sound))

The Attempt at a Solution



First I'll tweak the beat frequency equation to solve for what the stationary train conductor hears as the frequency of the moving train's whistle.
f'_b = |f' - f| <<< f' > f since the train is moving TOWARD him.
3.5 Hz = |f' - 516 Hz|
f' = 519.5 Hz

Now I'll substitute all values into the Doppler equation to find the velocity of the moving train.
519.5 Hz = 516 Hz / (1 - (v_source / 343 m/s) )
v_source = 2.31 m/s

Though possible, this seems like a low speed for a moving train. Did I do everything correctly?
Thank you!
 

Answers and Replies

  • #2
1,384
0
Yes you did everything fine.
 
  • #3
I may be a little late, but I think I got the error.
The beat frequency is:
f_b = (f1 - f2) / 2
You forgot to divide by 2. Then you get a sommewhat higher velocity.
 
  • #4
86
1
Actually you don't divide by 2 in that formula. I got the answer right. But thanks for checking for me!

You may be thinking of the formula to check if a pipe is closed by looking at the frequency of sound traveling through it:
f_closed = (fn1-fn2)/2
n_closed = fn1/f1closed
 

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