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Wavelength

by keemosabi
Tags: wavelength
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keemosabi
#1
Dec14-08, 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.
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rl.bhat
#2
Dec14-08, 07:57 PM
HW Helper
P: 4,437
If it's in the third harmonic, who many nodes and anti nodes are formed in the tube?
keemosabi
#3
Dec14-08, 08:10 PM
P: 109
Doesn't it look like this?



http://img101.imageshack.us/my.php?i...armonicoq0.png



keemosabi
#4
Dec14-08, 09:24 PM
P: 109
Wavelength

Can someone tell me what I did wrong?
rl.bhat
#5
Dec14-08, 09:59 PM
HW Helper
P: 4,437
In the tube closed at one end, third harmonic is the first overtone. So L = 3*lambda/4


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