ps05_soln - 18.03 PS5 Solutions-Spring 09 1a(first solution...

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18.03 PS5 Solutions—Spring 09 1a. (first solution) Given mx 00 + kx = 0, the characteristic polynomial p ( r ) = mr 2 + k has roots r = ± ( p k/m ) i . Let ω = p k/m . Then z = ce iωt is a complex-valued solution for any constant c C . The real part has the form x = A cos( ω n t - φ ) with ω n = ω = p k/m Indeed, write c = Ae , so that z = Ae i ( ωt + θ ) is a complex-valued solution, and x = Re z = A cos( ωt + θ ) is a real-valued solution. This is equivalent to x = A cos( ωt - φ ) and φ = - θ . Starting with the other complex-valued solution, z = ce - iωt , yields the same family of real solutions. 1a. (second solution) We start by seeing under what conditions x = A cos( ωt - φ ) is a solution to the ODE mx 00 + kx = 0: mx 00 + kx = - mAω 2 cos( ωt - φ ) + kA cos( ωt - φ ) = A cos( ωt - φ )( - 2 + k ) which is zero exactly when ω = ± p k/m . Since frequency is positive by convention, ω = ω n = p k/m . To see that every solution is of the form x = A cos( ω n t - φ ) with ω n = p k/m , one should check that the form is flexible enough to support all possible val- ues of x (0) and x 0 (0). This leads to two simultaneous equations for the two parameters A and φ . But students are not required to carry this out to confirm that there is always a solution. The fact that there are no other solutions follows from the fact that a solution is uniquely specified by the initial conditions x (0) and x 0 (0). 1b. The initial conditions x (0) = x 0 , x 0 (0) = 0 for x = A cos( ω n t - φ ). lead to A cos φ = x 0 ω n A sin φ = 0 which means that φ = 0 and A = x 0 , so x = x 0 cos( ω n t ). The first time that x = 0, we have ω n t = π/ 2. In general, x 0 = - ω n x 0 sin( ω n t ), so at the time when ω n t = π/ 2, then x 0 = - ω n x 0 . The kinetic energy is (1 / 2) m ( x 0 ) 2 = (1 / 2) 2 n x 2 0 = (1 / 2) m ( k/m ) x 2 0 = (1 / 2) kx 2 0 . 1c. Given E = (1 / 2) m ( x 0 ) 2 + (1 / 2) kx 2 , we have E 0 = mx 0 x 00 + kxx 0 = x 0 ( mx 00 + kx ) = 0. The phase plane trajectories when b = 0 are shaped like ellipses. In the case m = 2, k = 1, b = 0, the ellipses are similar to each other, of the form ( x 0 ) 2 /a 2 + x 2 / ( λa ) 2 = 1 where λ 1 . 4 is the fixed ratio of the semiaxes and a is allowed to vary. The formula for
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