How Low Can a Pop Bottle Go in Pitch?

AI Thread Summary
The discussion centers on the theoretical limits of pitch achievable in a closed tube, specifically a pop bottle. It highlights that the human ear can only perceive a limited range of frequencies, suggesting a practical limit to the lowest note. Additionally, it notes the inverse relationship between frequency and wavelength, indicating that at very low frequencies, the wavelength may exceed the tube length, making it impossible to produce those notes. These factors contribute to the conclusion that there is indeed a limit to the lowest pitch achievable in such a setup. Understanding these principles is essential for exploring sound production in closed tubes.
Numzie
Messages
15
Reaction score
0
Is there a theoretical limit to the lowest note achievable in a closed tube(Pop bottle)?

Heres the question in context:
I've got what I believe to be the correct:
http://img141.imageshack.us/img141/9301/1azw1.jpg
^^ May want to check that, if you don't mind ^^
 
Last edited by a moderator:
Physics news on Phys.org
I've searched all over for something but can't find anything. From what I can reason there would be a limit for two reasons:
1) Because the human ear can only hear a limited about of frequencies.(I assume this is correct but doesn't really relate to the given situation)
2) Frequency and wavelength are inversely proportional so eventually at a low enough frequency the wavelength will be so great that 1/4 of its wavelength is greater than the length of the tube.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top