Need some expert help on doppler's effect

[andy_un]

Homework Statement

There are in total 4 tubes. The 3rd harmonic is set up in each tube, and some of the sound is detected by detector D, which is moving directly away from the tubes. What is the speed of the detector if the detected frequency is equal to each tube's fundamental frequency?

Tube 1: length 1.0m, one open end
Tube 2: length 1.0m, two open end
Tube 3: length 2.0m, one open end
Tube 4: length 2.0m, two open end
speed of sound = 343 m/s

Homework Equations

f = n*v/(4L) for one end tubes
f = n*v/(2L) for two end tubes
f = f₀ [(v+/-v_D)/(v+/-v_S]

The Attempt at a Solution

So i just plug in the known values into the equation, but I got the same speed of the detector for each tube. Is that supposed to be correct?
For more clarification, I will provide an example of how I calculate using data for tube 1:

f₀ = n*v/(4L) = 3*(343m/s)/[4(1.0m)] = 257.25s^-1
f = n*v/(4L) = 1*(343m/s)/[4(1.0m)] = 85.75s^-1

f = f₀ [(v+/-v_D)/(v+/-v_S]
f = f₀ [(v-v_D)/v]
85.75s^-1 = 257.25s^-1 * [(343m/s-v_D)/343m/s]
v_D = 228.67 m/s

 

[Stonebridge]

Yes. Irrespective of the length of the tube, or whether it is open or closed, the 3rd harmonic is always 3 times the frequency of the fundamental.
So in the Doppler formula, you always have the same ratio of the two frequencies,and therefore the same speed for the detector.