The Helmholtz Resonator
The longest wavelength standing wave in a long thin tube of length L that is closed on one end is 4L (node at the closed end, antinode at the open end, so L = l /4). Thus the fundamental (lowest) frequency is v/4L, where v is the speed of sound. But if you blow over the top of an empty beer bottle, you will hear a lower sound. It doesn't behave like a thin tube. I have seen text problems involving bottles of a given length with no other dimensions given, implying that you should treat it like a thin tube. If you run into this, go ahead and do the calculation, but when you try it experimentally you will see that the calculation is way off.
We can model bottle oscillations after the mass on a spring:
w = (k/m)1/2, where m is the
mass of air in the neck, and k is due to the springiness (springyness?) of the
volume V in the wide part of the bottle.
Consider a short tube of length L and cross-sectional area A connected to a
large volume V. Our mass m is
r
D P = -BD
V/V, where B is the bulk modulus. So imagine a piston moving distance
D x in the tube. Write
D F = -kD
x with D F = AD P=-ABD V/V. But
D V=AD
x, so plug this in. Now the equivalent k is -D
F/D x = A2B/V
Hence w = (k/m)1/2 = (BA2/r ALV)1/2. Cancel one factor of A
and recall that the speed of sound is
v = (B/r )1/2, so we find
that w = v(A/LV)1/2.
I know I have seen this referred to as a Helmholtz resonator, but I looked in several books and didn't find it. Now resonate back to the main waves page.
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mechanics
fluids, heat, electricity and magnetism
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