Number of Different resonances in a closed Box

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SUMMARY

The discussion focuses on determining the number of different resonance frequencies in a closed 3D box with side length 'a', filled with gas at constant sound speed 'c'. The resonance frequencies are expressed as constant multiples of the term ##(l^2+m^2+n^2)^{1/2}##, where l, m, and n are integers. The final conclusion derived is that the number of distinct resonance frequencies below a certain frequency 'v' is approximately ##\frac{4\pi a^3 v^3}{9\sqrt{3}c^3}##. The analysis involves solving the wave equation under boundary conditions and calculating the volume of a sphere in the context of integer solutions in l, m, n space.

PREREQUISITES
  • Understanding of wave equations and boundary conditions
  • Familiarity with resonance frequencies in acoustics
  • Knowledge of spherical volume calculations
  • Basic concepts of integer solutions in three-dimensional space
NEXT STEPS
  • Study the derivation of wave equations in three dimensions
  • Explore the relationship between resonance frequencies and integer solutions
  • Learn about the implications of boundary conditions on wave functions
  • Investigate the geometric interpretation of resonance frequencies in l, m, n space
USEFUL FOR

Students and researchers in physics, particularly those focusing on acoustics, wave mechanics, and mathematical physics, will benefit from this discussion.

Marcus95
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Homework Statement


Show that the possible resonance frequencies in a 3D box with side a are constant multiples of ##(l^2+m^2+n^2)^{1/2}##, where l, m and n are integers. Assume that the box with sides a is filled with a gas in which the speed of sound is constant. Hence show that the number of different resonancec frequencies less than v is ##\approx \frac{4\pi a^3 v^3}{9\sqrt{3}c^3}##.

Homework Equations


##c = \frac{\omega}{k}##
##f = \frac{\omega}{2\pi}##

The Attempt at a Solution


I solved the first part by assuming that the variables could be decoupled and that the waveequation is on the form: ##\Psi(x,y,z,t) = X(x) Y(y) Z(z) cos(\omega t)## where ##X(x) = A_xcos(k_xx) + B_xsin(k_xx)## and equally for Y and Z. Using the boundary conditions that no particle motion should occur at the walls, I ended up with:
##\Psi(x,y,z,t) = C sin(k_xx) sin(k_yy) sin(k_zz) cos(\omega t) ##
where ##k_x = \frac{\pi n}{a}##, ##k_y = \frac{\pi m}{a}##, ##k_z = \frac{\pi l}{a}##
This ultimately lead to:
##f= \frac{c}{2a} (l^2+m^2+n^2)^{1/2} ## as was to be shown.

However, I am completely stuck on the last part. I esentially end up with the inequality:
##(l^2+m^2+n^2)^{1/2} < \frac{2av}{c} ##
but from here I have no idea how to progress to find the number of solutions.
 
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What is the volume of a sphere with radius ##2af\over c## ?
 
BvU said:
What is the volume of a sphere with radius ##2af\over c## ?
The volume is ##\frac{32\pi a^3f^3}{3c^3}## which is close to what we want to prove, but I don't see how a sphere comes into the number of integer solutions in a cubic box...
 
BvU said:
What is the volume of a sphere with radius ##2af\over c## ?
Is it related to somekind of point density?
##V=\frac{4\pi r^3}{3}=\frac{32\pi a^3f^3}{3c^3}=16\cdot3\sqrt{3}\frac{4\pi a^3f^3}{9\sqrt{3}c^3}=\frac{1}{\rho}N## ?
I still don't see why we get a sphere in a cubic box though and how we are supposed to know ##\rho##.
 
Not in the box with physical dimensions a, but in the space of l, m, n... where a radius is ##2af\over c##
 
BvU said:
Not in the box with physical dimensions a, but in the space of l, m, n... where a radius is ##2af\over c##
I am sorry, but I still don't have any idea how to solve this.
 
You are after the number of
Marcus95 said:
resonance frequencies less than v
which is roughly the number of grid points in ## l,m,n ## space for which ## l^2+m^2+n^2 \le ## a certain number, the (radius of a sphere)2
 

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