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Unformatted text preview: Examine equations for type of solution • Notice that the x and y directions are cou pled together, and the z direction is not coupled to either. • If you look at the z equation first, you’ll see that it can be solved by a function whose second derivative is the negative of itself, i.e., sines and cosines. Really a linear com bination. • Then we just need to solve for the coeffi cients in terms of initial conditions. L. Healy – ENAE404 – Spring 2007 – Lecture 19 (Apr. 5) 1 Solve for z motion As noted, z is decoupled from the other direc tions. It is a simple harmonic oscillator. Set the coefficients correctly to match initial con ditions, z ( t ) = z cos nt + ˙ z n sin nt. L. Healy – ENAE404 – Spring 2007 – Lecture 19 (Apr. 5) 2 Solve for x motion Take the x equation ¨ x = 2 n ˙ y + 3 n 2 x and differentiate once x (3) = 2 n ¨ y + 3 n 2 ˙ x, then substitute the y equation ¨ y = 2 n ˙ x x (3) = 4 n 2 ˙ x + 3 n 2 ˙ x, or x (3) + n 2 ˙ x = 0 . This should be recognizable as the simple har monic oscillator in v = ˙ x : ¨ v + n 2 v = 0 . L. Healy – ENAE404 – Spring 2007 – Lecture 19 (Apr. 5) 3 Solve for v Harmonic oscillator solution v ( t ) = v x cos nt + v y sin nt is the most general, which means that the so lution for x is the integral x ( t ) = Z v ( t ) dt = x + v x n sin nt v y n (cos nt 1) . L. Healy – ENAE404 – Spring 2007 – Lecture 19 (Apr. 5) 4 x and derivatives initial conditions All that remains is to find the constants, which we shall do by looking at x and its derivatives at t = 0, x (0) = x ˙ x ( t ) = v x cos nt + v y sin nt ˙ x (0) =˙ x = v x ¨ x ( t ) = nv x sin nt + nv y cos nt ¨ x (0) =¨ x = nv y so we have solved for v x and v y in terms of the initial x and derivatives, v x = ˙ x , v y = ¨ x n = 2˙ y + 3 nx where we substitute ¨ x = 2 n ˙ y + 3 n 2 x from the original differential equation. Thus x ( t ) = 4 x + 2˙ y n + ˙ x n sin nt 2˙ y n + 3 x cos nt. L. Healy – ENAE404 – Spring 2007 – Lecture 19 (Apr. 5) 5 y solution Note that y equation depends on x and vice versa, so we can substitute our solution for x ( t ) ¨ y = 2 n ˙ x = 2 n ˙ x cos nt + ¨ x n sin nt = 2 n ˙ x cos nt (4 n ˙ y + 6 n 2 x )sin nt....
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This document was uploaded on 04/27/2011.
 Spring '11

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