Question

Consider the ODE given by: x''(t) + c x'(t) + k x(t) = r(t). For c = 1 and k = 4, plot the position and velocity over the time interval [0, 20] seconds for the following cases. Analyze and discuss each result. Compare the two results. 1) r(t) is zero, x(0) = 1, x'(0) = 0. 2) r(t) is zero, x(0) = 0, x'(0) = 1. 

          Consider the ODE given by: x''(t) + c x'(t) + k x(t) = r(t). For c = 1 and k = 4, plot the position and velocity over the time interval [0, 20] seconds for the following cases. Analyze and discuss each result. Compare the two results.
1) r(t) is zero, x(0) = 1, x'(0) = 0.
2) r(t) is zero, x(0) = 0, x'(0) = 1.
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider the ODE given by: x''(t) + c x'(t) + k x(t) = r(t). For c = 1 and k = 4, plot the position and velocity over the time interval [0, 20] seconds for the following cases. Analyze and discuss each result. Compare the two results. 1) r(t) is zero, x(0) = 1, x'(0) = 0. 2) r(t) is zero, x(0) = 0, x'(0) = 1.
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Transcript

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00:01 For this problem on the topic of second order linear ode's, we want to solve the initial value problem y double prime plus 2ky prime plus k squared plus omega squared y is equal to 0, where y at 0 is 1 and y prime at 0 is minus k.
00:17 Now the characteristic equation from above is lambda squared plus 2k lambda plus k squared plus omega squared, which we can write as lambda plus k, all squared plus omega squared, is equal to zero.
00:39 And so the roots here are lambda equal to minus k plus or minus i omega...
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