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QUESTION: I have no idea what to do for this problem. Any help would be appreciated.

A piece of cardboard tubing, closed at one end, is just the right length so that when it is cut into two pieces,

the lowest resonant frequency is 256 Hz for the piece with the closed end and 440 Hz for the other.

a) What is the lowest resonant frequency that would have been produced by the original cardboard tubing?

b) How long was the original tubing?

ANSWER: Hello Nick,

What's the wavelength of a 256 Hz sound? Using 331 m/s for the speed of sound, calculate the wavelength. Remembering that the period T is the time travel one wavelength
velocity = lambda/T
And since T = 1/frequency
velocity = 331 m/s = lambda*f = lambda*256 Hz
lambda = (331 m/s) / (256 cycles/s) = 1.29 m/cycle
(I've included the "cycle" above hoping to help you understand. That sound has to travel 1.29 meters before the compression/rarefaction conditions repeat -- in other words, before it has completed one cycle. Actually, the cycle is not a proper unit, so you should just say that the wavelength is 1.29 m.)

The lowest resonance the closed piece will support is the piece's fundamental frequency. If you look at your sketch for the waveform in a closed organ pipe when it is resonating at its fundamental, you should see that 1/4th of the full wavelength fits in the pipe. Since the wavelength is 1.29 m and 1/4th of the wavelength fits in the closed piece of tubing, that piece of tubing is 1.29/4 m long.

Now I'm going to ask you to repeat the method in the first paragraph to find the wavelength of the 440 Hz sound.

Since the original piece was closed at one end, after cutting in 2, the 2nd piece is open at both ends. If you look at your sketch for the waveform in an open organ pipe when it is resonating at its fundamental, you should see that 1/2 of the full wavelength fits in the pipe. So the length of the 2nd piece is 1/2 of your result for the wavelength of the 440 Hz sound.

So the answer to part b is the sum of the length of the 2 pieces.

a. The original pipe was closed at one end, so the fundamental resonant frequency of the original would have a wavelength 4 times as long as the pipe. So you know lambda of the fundamental. You can get the frequency, in Hertz, of the fundamental from
velocity = lambda*freq or freq = velocity/lambda

I hope this helps,

---------- FOLLOW-UP ----------

QUESTION: Hi, would the answer for part b) change if the question for that part included "Assume that on that day speed of sound is 343 m/s"

Hello Nick,

Yes it would. (I used 331 because it is given on the inside of the front cover of my college Physics text that the speed of sound for air at STP is 331 m/s. Your value is the speed of sound at 20 degrees C.)

Since the speed of sound is used when you know the frequency to calculate the wavelength,
and since the pipe must be some fraction of the wavelength, (one fourth in one case and one half in the other)
using a different value for the speed of sound will result in a different result for the pipe length.

I hope this helps,


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Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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