for the large solution, and
for the small solution.
In the general case, we must allow θ to vary. Recall that θ is the angle between the axis of symmetry and the z-axis, vertical in this case. As the axis of symmetry precesses in φ a variation of θ produces the nodding motion of the axis. This is called nutation, latin for nodding.
It turns out that for this situation, the energy can be expressed as a function of the angle θ alone. This simplifies the analysis of the problem. Begin with E = T + U, Start with the kinetic energy for a symmetric top as given in Equation~(10.105),
and
to eliminate ψdot and φdot in the expression for the total energy
where
We are treating an idealized problem of a frictionless top where the energy is constant, or at least working in an approximation where the friction is small and the energy is approximately constant. Therefore, we can write
In principal, this can be integrated to find θ as a function of t. It is a little easier if we use the clever substitution u = cosθ which yields udot = -θdot sinθ, or θdot = -udot/sinθ = -udot/√(1 - u²). Upon substituting this into the expression for θdot² we find
Turning points of the motion occur for udot=0, or therefore at the roots of the equation f(u) = 0. The equation in udot² can be integrated to yield
Since θ lies between 0 and π, and u = cosθ, we see that u must lie in the range -1 to +1. In fact, for a top spinning on a table, θ is restricted to the range from 0 to π/2, meaning that u lies in the range from 0 to 1. The derivative udot is real, and udot² is positive. Therefore, the turning points of the motion occur when θdot is zero, corresponding to udot = 0, or udot² = f(u) = 0. Clearly the turning points of the motion correspond to the solutions for f(u) = 0, under the restriction that u lies in the range from 0 to 1.
The function f(u) is a cubic polynomial in u with either 1, 2, or 3 real roots.
from which we see that if Lz is larger than L3, then φdot can never vanish, meaning that the top always precesses forward. If Lz is less than L3, then φdot can vanish, and the top can precess backward during some parts of the motion.
In chapter 5 we studied systems with a single oscillator, subject to damping and driving forces, learning about the behavior of resonance. In this chapter we will consider systems of coupled oscillators and examine the behavior possible in such systems.
Our model system for this study consists of two masses, m1 and m2, each connected to a fixed support by springs of spring constants k1 and k3, and coupled to each other by a third spring of spring constant k2. There is some equilibrium position where the net force on each mass is zero. Measure the locations of each mass from their corresponding equilibrium position, call them x1 and x2. Then the kinetic energy of the system is
T = ½m1x1dot² + ½m2x2dot²The potential energy of the system comes from stretching or compressing the springs. The change in length of k1 is x1 and the change in length of k3 is x2. The change in length of k2 is (x1 - x2); if both masses are displaced by an equal amount in the same direction, then k2 doesn't change length. The potential energy is
U = ½k1x1² + ½k2(x1 - x2)² + ½k3x2²The Lagrangian for the system is
L = T - U = ½m1x1dot² + ½m2x2dot² - ½k1x1² - ½k2(x1 - x2)² - ½k3x2²Lagrange's equations yield the following equations of motion
mx1ddot = -k1x1 - k2(x1 - x2) = -(k1 + k2)x1 + k2x2 mx2ddot = k2(x1 - x2) -k3x2 = k2x1 - (k2 + k3)x2This can be written more compactly (and usefully) by introducing matrix notation. Let x be the column vector (x1, x2), then xdot is the column vector (x1dot, x2dot), and xddot is the column vector (x1ddot, x2ddot). The equations of motion can be written in the form
Mxddot = -Kxwhere M is the diagonal matrix [(m1, 0), (0, m2)] and K is the matrix [((k1 + k2), -k2), (-k2, (k2 + k3))].