Wavelength - distance between a node and adjacent antinode

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The discussion centers on calculating the distance between a node and an adjacent antinode in a tube open at one end, specifically at its third harmonic. The tube's length of 1.7 m corresponds to 1.25 wavelengths, leading to a calculated distance of 0.34 m between a node and an antinode. Clarification is sought regarding the number of nodes and antinodes present in the tube, with a diagram referenced for visual aid. It is noted that in a closed-end tube, the third harmonic represents the first overtone, implying that the length of the tube is equal to 3/4 of a wavelength. The conversation emphasizes the importance of understanding harmonic relationships in closed tubes.
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Homework Statement


A tube of air is open at only one end and has a length of 1.7 m. This tube sustains a standing wave at its third harmonic. What is the distance between one node and the adjacent antinode?


Homework Equations


v = f (lambda)


The Attempt at a Solution


If it's in the third harmonic, that means the length of the tube is equal to 1.25 wavelengths. So, 1.7 / 1.25 = 1.36. The distance between an antinode and the adjacent node is .25 lambda, so 1.36 * .25 = .34m. So .34m is the answer.
 
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If it's in the third harmonic, who many nodes and anti nodes are formed in the tube?
 


Can someone tell me what I did wrong?
 


In the tube closed at one end, third harmonic is the first overtone. So L = 3*lambda/4
 

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