We have produced white noise using LabView program described earlier in the class. (Soundwhitenoisegenerator.vi) the computer sent the signal to the speakers.
Microphone from Kunds' tube set recorded the  sound and sent it to the oscilloscope and then through LabView  interface to the  computer file.

read=ReadList["C:\\Users\\intelliome\\Desktop\\depottex\\depottex25msbox\\c25c.txt",Number];

ListPlot[read,PlotJoined->True]

Length[read]

1893/2.

FOURIER TRANSFORM - transform of physical , captured , image from time space to the conjugate frequency space T --> 1/T

af=Abs[Fourier[read]];

ListPlot[af,PlotJoined->True,PlotRange->All]

Mathematical artifact is the double , so we show 1893/2 = 956.5 points in frequency space

ListPlot[Abs[Fourier[read]],PlotJoined->True,PlotRange->{{0,947},{0.,200}}]

NOTE: 
-the results show only certain modes are active in the tube (resonance, see quantum cavities such as electrons in atoms and molecules (particle in the box, lasers,   NMR etc) )
-different modes have different amplitude  -populations are different 
-modes are at the n * fundamental frequency
- we have seen that modes depend on open or closed cavity (tube)
-we have done experiment in a very basic form, did not use any filtering , 
-there is much room for improvement (next group)

af[[215]]

af[[415]]

We can take 14 or 15  peaks ( highest number less error) and fit the values or just check the total /number of peaks     (Fit from the linear fit program)

ESTIMATED number of points for n * fundamental

USE: RIGHT click over Figure and choose GET COORDINATES

500/14.

data={{1,0},{2,34},{3,64},{4,100},{5,134},{6,170},{7,200},{8,240},{9,277},{10,309},{11,343},{12,382},{ 13,418},{14,452},{15,497}}

ListPlot[data]

ListPlot[data,PlotJoined->True]

Fit[data,{1,x},x]

the fundamental is 35.29   on the FT scale
What is that in frequency scale?

Firstly the number of points

Length[af]/2.

"Secondly transform of time to frequency

"time  unit  -i- from the oscilloscope scale    100microseconds =0.1  millisecond
100* 10^-6 = 0.000001 *100  =0.0001
for 1/0.0001   
  frequecy FFT scale is

1/0.0001

that is 10kHz/2   =5khz

so our scale is 5khz   corresponds to length in points/2     946

and for the fit result that is

that is

5000*35.3/946

vs accurate 190 Hz   (check effective L,  the cavity length moduleate by piston and calculate the fundamental

l/2 is the first resonant mode in the cavity as measured by microfon scanning the whole length of the tube

OPEN versus closed  cavity

should be

Solve[5000. *x/946 ==190,x]

Look in moe detaILS AND FILTER NOISE