What is the correct frequency for a 20 cm pipe with open and closed ends?

AI Thread Summary
The discussion focuses on calculating the base frequency of a 20 cm pipe with different end conditions. For a pipe open at both ends, the first wavelength is determined to be 40 cm, leading to a calculated frequency of 870 Hz, while the expected answer is 1140 Hz. In the case of a pipe closed at one end, the wavelength is 80 cm, resulting in a frequency of 435 Hz, but the expected frequency is 570 Hz. The discrepancy in both cases suggests a consistent factor affecting the calculations. The participants are puzzled by the difference between their results and the expected values.
Karol
Messages
1,380
Reaction score
22

Homework Statement


What is the base frequency of a 20 cm pipe which is open on both ends.
Again but closed on one end.

Homework Equations


u=velocity, \lambda=wave length: \lambda=\frac{u}{f}
Velocity in air=348[m/sec]

The Attempt at a Solution


I understand these are standing waves in air.
In an open end there is a peak, so the first wavelength is \lambda=2\cdot 20=40[cm]
##0.4=\frac{348}{f_1}\rightarrow f_1=870[Hz]##
The answer should be 1140[Hz]
In the closed on one end condition, in the open end there is a peak and in the closed end a node, so the wavelength is \lambda=4\cdot 20=80[cm]
##0.8=\frac{348}{f_1}\rightarrow f_1=435[Hz]##
The answer should be 570[Hz]
Both answers differ from the right ones by the same factor
 
Physics news on Phys.org
Puzzling.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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