We present, without proof, the following fourth-order differential equation that describes the system of a vibrating rigid rod.
¶2/¶x2(E*I ¶2y/¶x2) = - r ¶2y/¶t2
E is one of the essential elastic properties of a solid that is known as the Young's Modulus, the ratio of stress to strain in the rod. I is called the geometrical moment of inertia. In this case, I is not the normal moment of inertia that one usually thinks about (i.e. mass times the square of the distance from the axis of rotation). Rather, it is the cross section weighted by the square of the distance from the clamped end (see appendix of ref. 2 for more information). Rho is density of the material. The solutions to this equation are of the form
y = y(x)*ei*w*t
Since we are mainly interested in the standing wave patterns, we may ignore the time dependent term and limit our discussion to the y(x) term. Making the substitution
k4 = rw2/E*I
We can obtain the solutions of y(x) that are of the form
y(x) = A cosh(kx) + B sinh(kx) + C cos(kx) +D sin(kx)
Since we used a fourth order differential equation, four boundary conditions are needed to determine a unique solution for each oscillation frequency and length. These boundary conditions are
Implementing the boundary conditions will yield a transcendental equation that can be solved graphically. A sample solution is shown below for a fixed rod length of 43.2 cm. The solutions are the x-intercepts of the graph which correspond to allowed values for k and thus of frequency.
(note: y-axis has no physical meaning)
Chladni's law, for two-dimensional circular objects, states that the resonant frequencies are proportional to (m + 2n)2, where m is the number of nodal diameters and n represents the number of concentric nodal circles. Since the problem we are considering is only one dimensional, we expect to see resonant frequencies that are proportional to the square of number of nodes if Chladni's law is applicable to this one-dimensional case.
see also references in conclusion section.
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