An easy way to derive the above equation is to draw an analogy between a rod and a spring. The analogy works because both objects are continuously distributed elements, in that their stiffness and mass are spread uniformly throughout their interiors.
Both the spring and the rod obey Hooke's Law when used in static applications,
where DL is the change in length of the spring or rod. The stiffness k for the rod is given by,
where E is the Youngs modulus of the material, and A and L are defined in the picture above.
The change in length of the rod for dynamic applications is given by,
where the wavenumber n is given by,
and f is the driving frequency (in Hz). To check that we have the right equation, we note 2 things: The dynamic equation for DL satisfies the governing partial differential equation for the rod, and DL becomes Hooke's Law in the static limit (f goes to zero),
To find the natural frequencies for the rod, we look at where the change in length of the rod blows up. This occurs when nL in the denominator equals one of the following: {p, 2p, 3p, ...}. The first natural frequency occurs when,
We solve for f_{res} and substitute in k_{rod} and the volume of the rod (A*L),
We recognize that the density times the volume equals the mass of the rod. We can therefore simplify the resonant frequency formula to,
By analogy, the spring's first natural frequency will have the same equation,
where k is now the spring stiffness, and M is the spring mass (which can be found by weighing the spring).
Note that this equation is similar to that for a lumped springmass oscillator. However, the frequency here is a factor of p higher due to the distributed mass in the spring (and rod).
