Science:Math Exam Resources/Courses/MATH215/December 2013/Question 03 (b)

MATH215 December 2013

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Question 03 (b)
Suppose the displacement x(t) of a damped mass-spring system subject to sinusoidal forcing of amplitude $F_{0}$ is modelled by: $x''+2x'+5x=F_{0}\sin(2t),\quad x(0)=0,x'(0)=1$. (b) Now take $F_{0}=1$, and find the steady periodic solution (the part of the solution x(t) which remains as $t\rightarrow \infty$).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint
The general solution is a sum of a homogeneous solution and a particular solution. What happens to the homogeneous solution as $t\rightarrow \infty$? What solution to you need to find?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We seek a particular solution since the homogeneous solution will decay to 0 as $t\rightarrow \infty$. As $\sin(2t)={\text{Im}}(e^{2it})$, we will look for a particular solution of form $x(t)={\text{Im}}(\phi (t))$ where $\phi =Ae^{2it}$ satisfies the ODE in complex form: $\displaystyle \phi ''+2\phi '+5\phi =e^{2it}$ so that $\displaystyle -4Ae^{2it}+4iAe^{2it}+5Ae^{2it}=e^{2it}$ which upon cancelling the $e^{2it}$ terms and simplification becomes $\displaystyle A(1+4i)=1$ Solving for A we obtain $A={\frac {1}{1+4i}}={\frac {1-4i}{(1+4i)(1-4i)}}={\frac {1-4i}{17}}$. Now that we have A, we can find x by using A and Euler's identity: ${\begin{aligned}x(t)&={\text{Im}}\left({\frac {1-4i}{17}}e^{2it}\right)\\&={\frac {1}{17}}{\text{Im}}((1-4i)(\cos(2t)+i\sin(2t))\\&={\frac {1}{17}}{\text{Im}}(\cos(2t)+4\sin(2t)+i(\sin(2t)-4\cos(2t)))\\&={\frac {1}{17}}\sin(2t)-{\frac {4}{17}}\cos(2t)\end{aligned}}$ The steady solution is $x(t)={\frac {1}{17}}\sin(2t)-{\frac {4}{17}}\cos(2t)$.