How Does String Tension Affect the Sound of a New Musical Instrument?

[Simon Bridge]

No worries, just be careful when you write down an equation... make sure it says what you mean.
Its OK to invent new variable names. You don't have to, and often shouldnt, use the variables in the "standard form" of a physics equation.
Also... get used to drawing pictures of everything. You'll notice your teachers do that all the time.

 

[BvU]

I got that the wavelength of the string is 2/3 (L/10) and the wavelength of the can is 4L is that correct?

Yes, it's correct -- in a way. But the fact that you have to ask worries me.

The wavelength of the standing wave in the string is 2/3 (L/10). L/10 is the length of the string. L is the length of the can. You'll do well not to use L otherwise -- it'll only cause confusion. See for example your post #20. And earlier on, Simon's reaction in post #17 to your post #6 where it really says $L = {L\over 10}$ (top right). I understand what you mean but it really isn't right. And it's costing marks needlessly.
All this isn't physics, and it barely is math; just common sense.

You can see this 2/3 in your drawing.

If the string has a 3rd harmonic frequency and its an open pipe I though using the equation λs=2/3L. For L I used the diameter of the can because its the length.

At least write something like "λ_s=2/3L_s. L_s = the diameter of the can". And yes, that is the length of the string.

On to your post #18. "confusion, not my strength" Nothing wrong with that. That's what being a student of something is about. Was that a reaction to post #17 or did it cross post #17 and come as an afterthought after post #16 ? Because that was bulls eye, spot on, just right, and what have you.

Then: I hope #13 was an answer to my first ... in post #12. 4L is right. A pipe with one end open and one end closed has its lowest resonance at a frequency corresponding with a wavelength of four times its length.

At a closed end the air can't move (zero amplitude of the back-and-forth motions) so the amplitude of the pressure variations is maximum. At an open end the pressure is equal to the pressure in the outside world, i.e. (almost) zero variations in pressure and maximum amplitude in the collective motion of the air molecules (sigh... I find it difficult to formulate it all correctly and still keep it legible ).​

There is a second set of ... in my post # 12 and I want to pick up on that. The problem statement is clear:

The string is tensioned such that the 3rd harmonic frequency of the vibrating string matches the fundamental frequency of the can

Is it clear to you that the tone is equal, i.e. the frequency ? Your post #13 is a little bit puzzling: you don't say explicitly that this f₁ of the can is equal to the f₃ of the string, which worried me.
$\$
[edit]Don't understand. I got to see posts #21 and higher only after posting this. Thought I should reply to post #20, which looks silly when you two were at #31 already and getting along just fine. Sorry, Simon (not the first time either, today!). And if I didn't make a mess of it, perhaps Emmy doesn't have to suffer from my barging in.

At least I didn't spoil anything. Did I miss the ultimate answer, or did I see it come by at some point