Recent content by Emily_20

Ok Thanks for the help! I appreciate it :smile:

So I understand that the wavelength of the string is length of the can (L) divided by 15 is that right? I got 15 from plugging L/10 for Ls in the equation λs=2/3Ls

Is L/10 considered as the wavelength of the string too?

Ok so the equation is going to be λs=2/3Ls

λs=2/3L=2/3 L/10 Its going to be 2L/30 which is L/15

I thought the the wavelength of the string is the diameter of the can. 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.

Unfortunately, I am very confused. Physics is not my strength. This problem is very important for me

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

Simon Bridge said that I can sketch in the pressure wave for the can what does he mean by pressure wave?

4L? because the can has a frequency of f1 and the string is going to have a frequency of f3

Thanks for the help I appreciate it. If I understand its right so the can is actually going to have a frequency of Fn = nv/4L and the string is going to have a frequency of f=V/λ

I'm not even sure if my original equation is correct. I though at the beginning that the string will act with a frequency for a tube with one open end and one closed end that's why I had a 4. For the wavelength of the can I used L=λc

I think I made a mistake in the calculation in Vt/L = 3/2 Va/ L/15

Hello sir, thank you for your help I appreciate it. I wrote out the wavelength in terms of L for both can and string and I tried to write the mathematical relationship using it, is that correct?

Here is an attempt for a sketch but I am not sure if its right..