# Train whistle frequency difference in approaching and receding

1. Apr 28, 2007

### euphtone06

1. The problem statement, all variables and given/known data
Whistle frequency is 595 Hz for the train and sounds higher or lower depending on whether it approaches it or recedes. Calculate the difference in frequency between approaching and receding train whistle for a train moving at a speed of 105 km/h. Speed of sound is 340 m/s

2. Relevant equations
change in f = (2frequency(u/v))/(1-u^2/v^2)

3. The attempt at a solution
105 km/h = 29.16 m/s
u/v = 29.16/340 = .08578

change if f = (2frequency(.08578))/(1-.08578^2)
Ok this may sound really stupid but I saw an example of this problem worked and it used frequency as 105 km/h the speed of the train so inputted looked like this
change if f = (2(105)(.08578))/(1-.08578^2)
This makes no sense to me but I just want to make sure it SHOULDNT be this (105) for frequency since its not actually in Hz. I assume frequency would be the whistle frequency 595 Hz correct?

2. Apr 28, 2007

### denverdoc

No problem, the shift in frequency, call if f*=f/(340+/-v) f in this case is the stationary frequency of 595 and v is the speed of the train/whatever. You'll notice that by using the eqn above, units don't change.

3. Apr 28, 2007

### euphtone06

Sorry Im still confused at what which frequency should be inputed in my equation 595/(340 + 29) 595(340 - 29) just 595? or the average of those?

4. Apr 28, 2007

### HallsofIvy

Staff Emeritus
I'm glad to hear it makes no sense to you! I suspect you have misread. Obviously, 105 km/h is the speed of the train, NOT a frequency at all.

Yes, the frequncy is 595 Hz, u is the speed of the train and v is the speed of sound (relative to the air). You use + or - depending on whether the train is coming toward you or going away from you.