Simple Harmonic Question (new New Seniors Mathematics 3UNIT Fitzpatrick) (1 Viewer)

StudyHard123 · Feb 19, 2017

(new New Seniors Mathematics 3UNIT Fitzpatrick) Ex13.4 pg.264

Q.10 Solve the differential equation d^2x/dt^2 +16x = 0 subject to the conditions x = 3 and dx/dt =16 when t = 0. Find the maximum displacement and the maximum speed if x metres is the displacement of the particle moving in a straight line at time t seconds.

InteGrand · Feb 19, 2017

StudyHard123 said:
(new New Seniors Mathematics 3UNIT Fitzpatrick) Ex13.4 pg.264

Q.10 Solve the differential equation d^2x/dt^2 +16x = 0 subject to the conditions x = 3 and dx/dt =16 when t = 0. Find the maximum displacement and the maximum speed if x metres is the displacement of the particle moving in a straight line at time t seconds.

$\noindent The differential equation is $\ddot{x} = -4^{2}x$, so the solution is $x = A\cos \left(4t - \phi\right)$ for some constants $A,\phi$. Since $x(0) = 3$, we have $A \cos \phi = 3$. Also, $\dot{x} = -4A\sin \left(4t - \phi\right)$. So $\dot{x}(0) = 16 \Rightarrow 4A\sin \phi = 16 \Rightarrow A \sin \phi = 4$. In summary, $A\cos \phi = 3$ and $A \sin \phi = 4$, so by drawing a right-angled triangle, we see we may take $A = 5$, $\phi = \cos^{-1}\left(\frac{3}{5}\right)$.$

$\noindent Now, the maximum displacement is just $A$, i.e. $5$ metres. The maximum speed is $4A$ (amplitude of the $\dot{x}(t)$ equation), so it is $20$ metres per second.$

Simple Harmonic Question (new New Seniors Mathematics 3UNIT Fitzpatrick) (1 Viewer)

StudyHard123

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InteGrand

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