# Resonance at low and high frequencies of 2 closed end tubes

1. Apr 13, 2017

### Taniaz

1. The problem statement, all variables and given/known data
The figure (attached) shows two tubes that are identical except for their slightly different lengths. Both tubes have one open end and one closed end. A speaker connected to a variable frequency generator is placed in front of the tubes, as shown. Te speaker is set to produce a note of very low frequency and then turned on. Te frequency is then slowly increased to produce resonances in the tubes. Students observe that at first only one of the tubes resonates at a time. Later, as the frequency gets very high, there are times when both tubes resonate. In a clear, coherent, paragraph-length answer, explain why there are some high frequencies, but no low frequencies, at which both tubes resonate. You may include diagrams and/or equations as part of your explanation

2. Relevant equations
For resonance of closed end tubes, the length of the tube must be an odd multiple of a quarter wavelength of the sound.

3. The attempt at a solution
I don't understand their solution: In order to resonate, the length of a tube must be an odd multiple of a quarter wavelength of the sound, as shown below. For resonance at low frequencies, the wavelength of the sound is of the order of the length of the tubes. So the match can occur for only one tube at a time — the difference in tube lengths is much smaller than a half wavelength. As the frequency increases, the wavelength decreases and many more wavelengths ft inside a tube. When half the wavelength becomes of the order of the difference in tube lengths, the tubes can contain an odd multiple of quarter wavelengths for the same wavelength at the same time — for instance, one tube might contain 17 quarter wavelengths while the other contains 19 quarter wavelengths.

Why are they comparing the differences in length of the tubes to the wavelength?

Thank you

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2. Apr 13, 2017

### BvU

Hello Tania,
Your first sentence leads to the conclusion almost automatically: if one tube is 19/4 $\lambda$ long and the other 17/4 then the length difference is 1/2 $\lambda$. As you say,

3. Apr 13, 2017

### Taniaz

That's the solution they provided, why must the difference be λ/2? And they gave these values as an example.

I don't understand what they mean by "at low frequencies, the wavelength of the sound is of the "order" of the length of the tubes-so the match can occur for only one tube at a time-the difference in tube lengths is much smaller than a half wavelength"

4. Apr 13, 2017

### Taniaz

Oh so when the difference between them is λ/2 then the tubes contain an odd multiple of a quarter wavelength of the sound. This part I get now after drawing the diagram but why is that at low frequencies the match can only occur for one tube at a time?

5. Apr 13, 2017

### BvU

Al lower frequencies the length of the tubes is a small odd number times $\lambda/4$ and the difference in lengths is $\ \ <\lambda/2$ , so no common resonances.

6. Apr 13, 2017

### Taniaz

The difference has to be λ/2 for resonance to occur in both tubes? Or is it the condition for resonance to occur simultaneously?

And even if it is a small odd number times λ/4 wouldn't resonance still occur?

7. Apr 13, 2017

### Taniaz

What do you mean by a small odd number? Can you give an example please?

8. Apr 13, 2017

### haruspex

Like 1, 3, 5...
Suppose both resonate. Each has a length which is an odd number multiplied by λ/4. Say, (2m+1)λ and (2n+1)λ.
What is the difference between those two lengths? Keeping λ fixed, what is the smallest possible nonzero value of that difference?

If the difference in their lengths is very small, what does that tell you about λ?