End correction of the pipe mouth(standing wave)

AI Thread Summary
Resonance in a pipe closed at one end occurs at specific wavelengths, but the open end's air isn't completely free, necessitating an end correction to account for this. The commonly accepted value for this correction is approximately 0.6 times the pipe's radius, though some sources suggest it may be closer to 0.58r, depending on the tube's shape. To determine the end correction experimentally, one can measure two consecutive resonating lengths and use their difference to calculate the wavelength. It's important to note that the speed of sound in the pipe differs from that in free air. Understanding these nuances is crucial for accurate acoustic modeling.
EHT
Messages
13
Reaction score
1
Consider a pipe closed at one end and open at the other.a simple model tells that resonance will occurs when the sound wavelength of the resonator is 1/4,3/4,5/4.. of pipe length,by assumption that antinode occurs at the open end.But it comes out that the air at open end isn't completely "free"(the pipe wall makes it can't expand freely) so it's not a perfect antinode .We have to apply an end correction, the pipe appears to be acoustically somewhat longer than its physical length.I'm searching for a model that can be use to find this end correction and I've trouble doing the experiment because of the non uniform ity of speed of sound due to viscosity.Do anyone knows the model to compute this correction?

sorry,my english is bad
 
Physics news on Phys.org
If my memory serves me correctly the end correction is equal to 0.6 times the pipes radius.Try googling waves in pipes.Good luck.
 
Yes Dadface was correct. The end correction, c, is 0.6r where r is the pipe radius.

Note:
In your post you say
"when the sound wavelength of the resonator is 1/4,3/4,5/4.. of pipe length,"

It is the pipe length (plus end correction) that is 1/4, 3/4 etc of the sound wavelength.
 
@stonebridge:yeah there I'm explaining standing wave without correction in basic physics textbook so I put the wrong one.I've read about that 0.6 r rayleigh's correction,but what i want to know is where does it come from.how do we calculate that 0.6r?

thanks
 
  • Like
Likes Aman Poddar
I've never seen an analytical derivation of this and have always assumed it to be an experimentally determined value. I checked in some of my books and it seems the most accurate value is 0.58r (It is independent of the wavelength but does depend on the shape of the tube. Value quoted is for circular cross section)
Normally in these experiments, you eliminate (and can calculate) the end correction by finding two consecutive resonating lengths L1 and L2 at the same frequency, for example the two cases where the L1+c is 1/4 wavelength and L2+c is 3/4 wavelength.
Subtracting gives L2 - L1 = half wavelength.

These experiments, of course, calculate the speed of sound in the tube. It must be remembered that, this is not the same as the speed in free air.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top