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# So the entire solution including the homogeneous

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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) L= 1 1 1 ˙ M r2 + mr2 + mr2 θ2 − M gr + mgr sin θ. ˙ ˙ 2 2 2 (289) The equations of motion are then (M + m)¨ r ¨ rθ = = ˙ 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. r:=1: r1:=0: q:=0: q1:=0: e:=.00001: g:=10: k:=.1: i:=0: while r1<.000001 do i:=i+1: r2:=(k*r*q1^2-g+k*g*sin(q))/(1+k): q2:=(-2*r1*q1+g*cos(q))/r: r:=r+e*r1: r1:=r1+e*r2: q:=q+e*q1: q1:=q1+e*q2: end do: r; i*e; q; # # # # # # # # # # # # # # # # # # # # initial r value initial r speed initial angle 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 how...
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