Train whistles; moving/stationary sound and velocity

[gmmstr827]

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!

 

[zorro]

Yes you did everything fine.

 

[AndyUrquijo]

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.

 

[gmmstr827]

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

 

[rwisz]

Your calculations and approach seem correct. The low speed of 2.31 m/s for the moving train may seem surprising, but it is important to remember that sound travels much slower than visual objects. The speed of sound is affected by temperature, humidity, and other factors, so it is possible that the speed of sound in this scenario is lower than the standard 343 m/s. Additionally, the beat frequency may not be very noticeable at higher speeds, so the conductor may not be able to detect it as easily. Overall, your solution appears to be correct and within the realm of possibility.