# Wavelength

by keemosabi
Tags: wavelength
 P: 109 1. The problem statement, all variables and given/known data 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? 2. Relevant equations v = f (lambda) 3. 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.
 HW Helper P: 4,430 If it's in the third harmonic, who many nodes and anti nodes are formed in the tube?
 P: 109 Doesn't it look like this? http://img101.imageshack.us/my.php?i...armonicoq0.png
P: 109

## Wavelength

Can someone tell me what I did wrong?
 HW Helper P: 4,430 In the tube closed at one end, third harmonic is the first overtone. So L = 3*lambda/4

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