Standing waves in tube closed tube

AI Thread Summary
A tone with a frequency of 2,000 Hz produces sound maxima in a closed tube filled with water, where the first maximum occurs at 5.7 cm and the fifth at 39.3 cm below the opening. The difference in water levels corresponds to four wavelengths, leading to confusion over the calculated wavelength, which should be 16.8 cm instead of 8.4 cm. The misunderstanding arises from the definition of sound maxima as points of resonance rather than wave peaks. The water level acts as the closed end of the tube, while the opening remains open. Understanding the relationship between resonance and standing waves is crucial for solving the problem accurately.
Arcthor
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Homework Statement


A tone with frequency 2,00 * 10^3 Hz is held above a tube filled with water. The water level is increased and decreased.

When the water level is 5.7cm below the opening of the tube, the first sound maxima is heard.

When the water level is 39.3cm below the opening of the tube, the fifth sound maxima is heard.

Questions:
1. what is the wavelength?
2. What is the speed of sound in this specific experiment?

Homework Equations


v=wavelength*frequency

The Attempt at a Solution


My attempt is as follows:

(39.3 - 5.7) = delta distance, and there are four wavelengths between the first and the fifth maxima, therefore the answer should be:

(39.3-5.7) / 4 = 8.4

But the answer is 16.8, double my answer. Why is this?
 
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When is there a sound resonance ? Your title says closed tube. Are both sides closed ?
 
..there are four wavelengths between the first and the fifth maxima..

That's not correct. Try drawing the 5 cases.
 
BvU said:
When is there a sound resonance ? Your title says closed tube. Are both sides closed ?
Sorry, the water makes up the closed part. The opening is open.
 
You have to look at the amplitudes of standing waves (fig 3).
 
+1 to that.

"the first sound maxima" is a reference to resonance not to wave peaks. See the link BvU posted.
 
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