# Wavelength - distance between a node and adjacent antinode

1. Dec 14, 2008

### keemosabi

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. Dec 14, 2008

### rl.bhat

Re: Wavelength

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

3. Dec 14, 2008

### keemosabi

Last edited: Dec 14, 2008
4. Dec 14, 2008

### keemosabi

Re: Wavelength

Can someone tell me what I did wrong?

5. Dec 14, 2008

### rl.bhat

Re: Wavelength

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