Wavelength - distance between a node and adjacent antinode

[keemosabi]

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.

 

[rl.bhat]

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

 

[keemosabi]

Doesn't it look like this?

http://img101.imageshack.us/my.php?image=thirdharmonicoq0.png

http://img101.imageshack.us/my.php?image=thirdharmonicoq0.png"

http://img101.imageshack.us/img101/5733/thirdharmonicoq0.png
http://g.imageshack.us/img101/thirdharmonicoq0.png/1/

 

[keemosabi]

Can someone tell me what I did wrong?

 

[rl.bhat]

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