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Unformatted text preview: the results from Chapter 4, gives A = g/(ω 2 − ω0 ) =
g/(2ω 2 − k/m), and B = 0. So the entire solution (including the homogeneous
solution from Exercise 6.25, under the assumption that ω < k/m) is
r(t) = g
2ω 2 − k/m cos ω t + C cos(ω0 t + φ). We see that the motion goes to inﬁnity if ω =
undamped driven oscillator at resonance. (288) k/2m. In this case, we have an
6.27. Coﬀee cup and mass
The Lagrangian is (up to an additive constant)
M r2 + mr2 + mr2 θ2 − M gr + mgr sin θ.
2 (289) The equations of motion are then
(M + m)¨
mrθ2 − M g + mg sin θ,
−2rθ + g cos θ.
˙˙ (290) A Maple program that determines the smallest value of r is the following. It includes
a counter, i, that yields the total time of the process.
while r1<.000001 do
# initial r value
initial r speed
initial angular speed
small time interval
value of g
value of m/M
initial value of counter
do process while r1 is negative
count steps (to get final time)
the first E-L equation
the second E-L equation
how r changes
how r1 changes
how q changes
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