Applications of Systems of Equations 487A spring pendulum has two "natural frequencies." The first frequency WI =
VifLo corresponds to the natural frequency of the pendulum with the spring
replaced by a rod of fixed length, L 0. The second frequency w 2 = JkTiTi,
corresponds to the natural frequency of a spring-mass system which oscillates
vertically without swinging back and forth.EXERCISES 10. 7l. All solutions of system (4), the linearized simple pendulum model with
no damping and no forcing function, are periodic with period 21r. For initial
conditions YI (0) = 0 and -2 < Y2 (0) < 2 the solutions of system (6) are
periodic. Is the period 27r or does the period vary with y 2 (0)? Find out
by using SOLVESYS or your computer software to solve system (6) on the
interval [O, 10] for YI(O) = 0 and (a) Y2(0) = .1 and (b) Y2(0) = 1.9. Print
solution values for YI (t) on the monitor and determine the period or graph
YI (t) and estimate the period.
- a. Determine the critical points of system (7), the simple pendulum model
with damping but no forcing function. Are they the same as for system (6)?
b. Use SOLVESYS or your computer software to solve system (7) on the
interval [O, 10] for C = .1, YI(O) = 0, and (i) Y2(0) = 2, (ii) Y2(0) = 2.5,
and (iii) y 2 (0) = 3. In each case, display a phase-plane graph of Y2 versus
YI. Does the solution corresponding to initial conditions (i) approach the
unstable critical point (7r, O) as it did in the undamped case? Give a physical
interpretation for the phase-plane graph for the initial conditions (ii) and
(iii). What critical point does the solution approach? What does this mean
the pendulum has done?
c. Solve system (7) on the interval [O, 10] for C = .5, YI (0) = 0, and
(i) y 2 (0) = 3 and (ii) y 2 (0) = 3.5. In each case, display a phase-plane graph
of Y2 versus YI and interpret the results physically.
3. a. Show that for [A[ < 1 the critical points of system (8), the simple
pendulum with constant forcing function model, are (en, O) and (dn, 0) where
Cn = 2n7r+arcsin A, dn = (2n+ l)7r-arcsin A, and n is an any integer. So, for
[A[ < 1 and A =/=- 0 the critical points are not equ ally spaced along the YI -axis.
Show that as A --> 1-, Cn --> (2n7r + 7r /2)-, and dn --> (2n7r + 7r /2)+ - that
is, as A approaches 1 from below, the critical points (en, 0) move toward
(2n7r + 7r/2, 0) from the left and the critical points (dn,O) move toward
(2n7r + 7r /2, 0) from the right.
b. Show that for [A[ > 1 system (8) has no critical points.
- Numerically so lve system (8), the simple pendulum with constant forcing