3rd harmonic of a column of air with one end enclosed

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SUMMARY

The discussion centers on the calculation of the third harmonic frequency of a column of air with one end enclosed, where the expected frequency is 450Hz. However, the calculations presented yield a frequency of 750Hz, indicating a discrepancy. The participants agree that for the frequency to be 450Hz, it must correspond to the second harmonic rather than the third. The wave equation and boundary conditions are correctly applied, leading to the conclusion that the original solution may be incorrect.

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Miles123K
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Homework Statement


lGas78X

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The solution to this question says 450Hz. However, when I attempted to compute the frequency using the wave equation and find the normal mode solutions, I get 750Hz

2. Homework Equations

I suspect that the solution could be wrong, is that the case?

3. The Attempt at a Solution

## v^2 \frac {\partial^2 \psi} {\partial x^2} = \frac {\partial^2 \psi} {\partial t^2}##
##\omega = v k##
Boundary conditions:
##\psi (0, t) = 0## and ## \frac {\partial \psi} {\partial x} (57 \times 10^{-2}, t) = 0##
Normal mode solution should look like:
##\psi (x, t) = A sin(kx)cos(\omega t - \phi)##
## \frac {\partial \psi} {\partial x} (57 \times 10^{-2}, t)
= A cos(57 \times 10^{-2} k ) cos(\omega t - \phi) = 0##
## 57 \times 10^{-2} k = \frac {2n-1} {2} \pi ##
## k = \frac {2n-1} {2 \times 57 \times 10^{-2}} \pi##
Plug in ##n=3## and using ##\omega = 2 \pi f##
##\omega = \frac {5} {2 \times 57 \times 10^{-2}} \pi \times 342##
##f = \frac {\omega} {2 \pi} = 750 Hz ##
From my working, for the frequency to be 450, it had to be the 2nd harmonic, not the third.
 

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Miles123K said:
for the frequency to be 450, it had to be the 2nd harmonic, not the third.
I agree.
 
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