Derive the solution for underground temperature oscillation without assuming that \(T_0 = 0\text{.}\). If you use Eulers formula to expand the complex exponentials, you will note that the second term will be unbounded (if \(B \neq 0\)), while the first term is always bounded. If you use Euler's formula to expand the complex exponentials, note that the second term is unbounded (if \(B \not = 0\)), while the first term is always bounded. Let us assume say air vibrations (noise), for example a second string. \frac{1+i}{\sqrt{2}}\), \(\alpha = \pm (1+i)\sqrt{\frac{\omega}{2k}}\text{. ]{#1 \,\, #2} We know how to find a general solution to this equation (it is a nonhomogeneous constant coefficient equation). Therefore, we pull that term out and multiply it by \(t\). i \sin \left(\omega t - \sqrt{\frac{\omega}{2k}}\, x\right) \right) . 0000003261 00000 n with the same boundary conditions of course. 2.6: Forced Oscillations and Resonance - Mathematics LibreTexts 0000005787 00000 n and what am I solving for, how do I get to the transient and steady state solutions? y_p(x,t) = For \(k=0.01\text{,}\) \(\omega = 1.991 \times {10}^{-7}\text{,}\) \(A_0 = 25\text{. To find the Ampllitude use the formula: Amplitude = (maximum - minimum)/2. It is very important to be able to study how sensitive the particular model is to small perturbations or changes of initial conditions and of various paramters. 0000004233 00000 n ODEs: Applications of Fourier series - University of Victoria PDF Vs - UH Thanks! When \(\omega = \frac{n \pi a}{L}\) for \(n\) even, then \(\cos (\frac{\omega L}{a}) = 1\) and hence we really get that \(B=0\text{. Write \(B = \frac{\cos (1) - 1}{\sin (1)}\) for simplicity. We now plug into the left hand side of the differential equation. That is, the amplitude does not keep increasing unless you tune to just the right frequency. We studied this setup in Section 4.7. }\), \(y(x,t) = \frac{F(x+t) + F(x-t)}{2} + \left( \cos (x) - We look at the equation and we make an educated guess, \[y_p(x,t)=X(x)\cos(\omega t). +1 , \end{equation}, \begin{equation*} a multiple of \( \frac{\pi a}{L}\). So I've done the problem essentially here? \nonumber \], \[ x_p''(t)= -6a_3 \pi \sin(3 \pi t) -9 \pi^2 a_3 t \cos(3 \pi t) + 6b_3 \pi \cos(3 \pi t) -9 \pi^2 b_3 t \sin(3 \pi t) +\sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-n^2 \pi^2 b_n) \sin(n \pi t). \nonumber \], We plug into the differential equation and obtain, \[\begin{align}\begin{aligned} x''+2x &= \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \left[ -b_n n^2 \pi^2 \sin(n \pi t) \right] +a_0+2 \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \left[ b_n \sin(n \pi t) \right] \\ &= a_0+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} b_n(2-n^2 \pi^2) \sin(n \pi t) \\ &= F(t)= \dfrac{1}{2}+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \dfrac{2}{\pi n} \sin(n \pi t).\end{aligned}\end{align} \nonumber \], So \(a_0= \dfrac{1}{2}\), \(b_n= 0\) for even \(n\), and for odd \(n\) we get, \[ b_n= \dfrac{2}{\pi n(2-n^2 \pi^2)}.

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