Posted by URGENT on Saturday, December 15, 2007 at 12:11pm.
The "first nodal line" is a line along which the path difference is 1/2 wave, a point of destructive interference. In this case, 9 cm is the wavelength, since there would be one extra wavelength in path difference along the second nodal line.
The frequency is then f = (35 cm/s)/9 cm = 3.9 Hz. Those are very slow waves, by the way. Are you sure you got the units right?
It does not matter whether the two sources are in phase with one another or not. It affects where the nodes are, but not the separation between the nodes.
For more background about this, see
http://www.glenbrook.k12.il.us/gbssci/phys/Class/light/u12l3a.html
mmmh are you sure it is 9 cm? It is destructive interference which means a minimum, therefore PmS1-PmS2= (m+1/2) lambda...I get 9 cm when I don't account for the m + 1/2
You are correct. 9 cm is (3/2) wavelength along a second nodal line. The wavelength is therefore 6 cm.
Thanks for all of your help, I talked to my other friend who had originally posted, and we both appreciate this, thanks!
Sorry about the error. I'm glad you and BobPursley were on the ball. I was thinking of the path length difference between the first and second nodal line.
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