Question about resonating in a closed tube.

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To determine the speed of sound in a closed tube resonating at 440 Hz with lengths of 0.195 m and 0.586 m, the relationship between wavelength and tube length must be utilized, specifically that wavelength equals four times the length of the tube. The speed of sound can be calculated using the formula v = f × wavelength, where f is the frequency. Two simultaneous equations can be established from the two lengths provided to solve for the speed of sound. The discussion emphasizes the importance of recognizing that the wavelength can vary based on the harmonic mode of resonance. Ultimately, applying these principles will yield the desired speed of sound calculation.
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Homework Statement


A closed tube resonates at 440 Hz when it is 0.195 m long and 0.586 m long. Determine the speed of sound. How do I do this?

Homework Equations



wavelength=4L
v=fw

The Attempt at a Solution


v= 440 (4L) I'm just not sure how to use both lengths given, can someone help me out?
 
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You do not know that wavelength = 4L.
The wavelength can be 2(L/1), 2(L/2), 2(L/3), 2(L/4), 2(L/5)...etc
You also know that v = f x wavelength. And v does not change
The values you have been given should enable you to write down 2 simultaneous equations from which you can get the answer.
 
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