
#1
Dec1408, 07:51 PM

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. 



#2
Dec1408, 07:57 PM

HW Helper
P: 4,442

If it's in the third harmonic, who many nodes and anti nodes are formed in the tube?




#3
Dec1408, 08:10 PM

P: 109





#4
Dec1408, 09:24 PM

P: 109

Wavelength
Can someone tell me what I did wrong?




#5
Dec1408, 09:59 PM

HW Helper
P: 4,442

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



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