J. Peter Campbell
To simplify the system, we assume that the motion is characterized by a linear restoring force in each direction, z and q. With the addition of a coupling factor, c, the equations of motion for the system become:
where m is the mass of the system, k is the spring constant for the z axis, I is the moment of inertia, delta is the rotational spring constant.
The first step in solving this system is to assume oscillatory motion, so that the solutions for z(t) and q(t) are:
Plugging these solutions into equations 3 and 4, respectively, yields:
Since the Wilberforce pendulum contains two degrees of freedom, the solution has two normal modes. Equation 7 can be solved to give the frequencies of the normal modes:
w1-w2 is defined as wb, the beat frequency, which is the frequency of the transfer of energy between the two modes.
Further, equations 8 & 9 can be used to determine the coupling as a function of the beat frequency, mass, and moment of inertia:
From this equation, we can calculate c for various I values to determine the I at which resonance occurs, or energy is most completely transferred between longitudinal and rotational motion.
Assuming sinusoidal oscillations, one can solve for z(t) and (t), given the initial conditions, zo and qo. This system is exactly solvable for the initial conditions: w'(0)=0 & z'(0) = 0. The solutions are:
where w1, w2 and w were determined experimentally.
This can be further simplified to provide the relationship between zo and qo for the normal modes w1 and w2 (equations 8 & 9).
The constant is defined as , which is called the radius of gyration, simply the radius of the spring. Therefore, the normal modes of the pendulum are to a close approximation: