The mass is pulled down by one unit of length beyond

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Unformatted text preview: ￿￿ t cos 4q − p2 − φ . 2 We now look at several examples. Please use JOde to plot solutions in each case! Summary: General solution and properties of (6.7) L[ y ] = d2 y dy +p + qy = 0. 2 dt dt √2 √2 Case 1. p2 > 4q. y (t) = Ce(−p+ p −4q)t/2 + De(−p− p −4q)t/2 with C and D real. For non-zero C, D; |y (t)| → ∞ if | y ( t) | → 0 if q < 0 or if p < 0, q > 0 p > 0, q > 0. p Case 2. p2 = 4q. y (t) = (C + Dt)e− 2 t , C, D real. Again for non-zero C, D; |y (t)| → ∞ if | y ( t) | → 0 if 12 p￿0 p > 0. Case 3. p2 < 4q . ￿ ￿ ￿ ￿ p y (t) = e− 2 t C cos 4q − p2 t/2 + D sin 4q − p2 t/2 ￿￿ ￿ ￿ p or C 2 + D2 e− 2 t cos 4q − p2 t/2 − φ √ √ where cos φ = C/ C 2 + D2 and sin φ = D/ C 2 + D2 . Again for non-zero C, D; | y ( t) | → ∞ | y ( t) | → 0 y ( t) if if remains finite and bounded if 13 p<0 p>0 p = 0. Application to Mass-Spring systems. Units: The units of mass M is kg., the unit of distance is meter (m), the unit of time is second (sec). A mass M kg. under gravity g exerts a force of M g newtons. Interpretation of results of example for damped mass spring system d2 x dx +p + qx = 0, dt2 dt where p is the friction divided by the mass and is the spring constant divided by the mass. Case 1: p = 4, q = 1, overdamped. The mass is pulled down by one unit of length beyond equilibrium Case 1. > > > > > > > > >equilibrium> ② - - - - - - - - - >- ✻> > x(0) > > ❄② and then let go. The damping is so strong that it dominates the spring force so that there are no oscillations. Instead x(t) simply decays exponentially in time from x(0) = 1 to zero. Case 2: p = 2, q = 1, critically damped. Similar behavior. However it is borderline in that the mass is very close to making small oscillations before damping to zero. Case 3: p = 1, q = 1, p2 < 4q underdamped. In this case, the friction is not enough to prevent oscillations, but merely dampens them. The motion is not periodic unless p = 0. Periodic means there exists T > 0 such that x(t + T ) = x(√ for all t. For p = 0, x(t) t) 2 is periodic with period √π . On the other hand, even for 4q > p > 0, the distance q between zeros t0 at which x￿ (t0 ) > 0 is constant and equal to ￿ 2π 2 q− p 4 . (Underdamped mass-spring systems are a very important paradigm for understanding many physical, engineering and chemical processes.) 14 Find the solution of the initial value problem d2 y dy +p + 4y = 0, dt2 dt y (0) = 0, dy (0) = 1, dt in the three cases (i) p = 5, (ii) p = 4, (iii) p = 2. In each case: a) find the behavior of the solutions near t = 0 and the dominant behavior for large t, (b) use JOde to draw the graphs of y (t), (c) find ymax and the value of t where it occurs. 15 Properties of solutions of linear, homogeneous odes L[y ] ≡ (D2 + p(t)D + q (t))y ≡ d2 y dy + p( t ) + q ( t) y = 0 dt2 dt (6.15) with initial conditions dy...
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