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Physics LibreTexts

11.5: Exercises

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Exercise 11.5.1

Find the time-domain Green’s function of the critically-damped harmonic oscillator (γ=ω0).

Exercise 11.5.2

Consider an overdamped harmonic oscillator (γ>ω0) subjected to a random driving force f(t), which fluctuates between random values, which can be either positive or negative, at each time t. The random force satisfies f(t)=0andf(t)f(t)=Aδ(tt), where denotes an average taken over many realizations of the random force and A is some constant. Using the causal Green’s function, find the correlation function x(t1)x(t2) and the mean squared deviation [x(t+Δt)x(t)]2.

Answer

For the over-damped oscillator, the Green’s function is G(t,t)=Θ(tt)eγ(tt)Γsinh[Γ(tt)],whereΓ=γ2ω20. Hence, the response to the force f is x(t)=1mΓtdteγ(tt)sinh[Γ(tt)]f(t). From this, we get the following expression for the desired correlation function: x(t1)x(t2)=1m2Γ2t1dtt2dt Note that the \langle\cdots\rangle can be shifted inside the integrals, because it represents taking the mean over independent sample trajectories. Now, without loss of generality, let us take t_1 \ge t_2. Since \langle f(t') f(t'')\rangle = A \delta(t'-t'') which vanishes for t' \ne t'', the double integral only receives contributions from values of t' not exceeding t_2 (which is the upper limit of the range for t''). Thus, we revise \int^{t_1} dt' into \int^{t_2} dt'. The delta function then reduces the double integral into a single integral, which can be solved and simplified with a bit of tedious algebra: \begin{align} \langle x(t_1)\, x(t_2)\rangle &= \frac{A}{m^2\Gamma^2} e^{-\gamma(t_1+t_2)} \int^{t_2}_{-\infty} dt' e^{2\gamma t'} \sinh\big[\Gamma(t'-t_1)\big] \, \sinh\big[\Gamma(t'-t_2)\big] \\ &= \frac{A}{8m^2\Gamma^2} e^{-\gamma(t_1+t_2)}\Bigg[\frac{e^{-\Gamma t_1} e^{(2\gamma+\Gamma)t_2}}{\gamma+\Gamma} + \frac{e^{\Gamma t_1}e^{(2\gamma-\Gamma)t_2}}{\gamma-\Gamma} \nonumber \\ &\qquad\qquad\qquad\qquad\qquad - \frac{e^{-\Gamma t_1} e^{(\Gamma+2\gamma)t_2} + e^{\Gamma t_1} e^{(-\Gamma+2\gamma)t_2}}{\gamma}\Bigg] \\ &= \frac{A}{8m^2\Gamma\gamma} \left[\frac{e^{-(\gamma-\Gamma)(t_1-t_2)}}{\gamma-\Gamma} - \frac{e^{-(\gamma+\Gamma)(t_1-t_2)}}{\gamma+\Gamma} \right].\end{align} Hence, \begin{align} \left\langle [x(t+\Delta t) - x(t)]^2\right\rangle &= 2\Big[\left\langle x(t)^2\right\rangle - \left\langle x(t+\Delta t) x(t)\right\rangle\Big] \\ &= \frac{A}{4m^2\Gamma\gamma} \left[\frac{1-e^{-(\gamma-\Gamma)\Delta t}}{\gamma-\Gamma} - \frac{1-e^{-(\gamma+\Gamma)\Delta t}}{\gamma+\Gamma} \right].\end{align}


This page titled 11.5: Exercises is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform.

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