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Question Number 95679 by Rio Michael last updated on 26/May/20

$$\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{motion}\mathrm{for}\mathrm{a}\mathrm{particle}\mathrm{moving}\mathrm{in}\mathrm{a}\mathrm{straight}\mathrm{line}$$$$\mathrm{along}\mathrm{the}\mathrm{O}X\mathrm{axes}\mathrm{is}\mathrm{given}\mathrm{by}\frac{d^{2}x}{{dt}^2}+\sqrt{7}\frac{dt}{dx}+4x=0.$$$$\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{motion}\mathrm{is}\mathrm{an}\mathrm{oscilatory}\mathrm{motion}\mathrm{hence}\mathrm{find}$$$$\mathrm{its}\mathrm{period}.$$

Commented by Tony Lin last updated on 27/May/20

$${\lambda }^2+\sqrt{7}\lambda +4=0$$$$\lambda =\frac{-\sqrt{7}\pm 3i}{2}$$$$x=e^{-\frac{\sqrt{7}}{2}t}(c_{1}cos\frac{3}{2}t+c_{2}sin\frac{3}{2}t)$$$$=e^{-\frac{\sqrt{7}}{2}t}\sqrt{c_1^2+c_2^2}(\frac{c_1}{\sqrt{c_1^2+c_2^2}}cos\frac{3}{2}t+\frac{c_2}{\sqrt{c_1^2+c_2^2}}sin\frac{3}{2}t)$$$$let\frac{c_1}{\sqrt{c_1^2+c_2^2}}=sin\theta$$$$\theta ={sin}^{-1}\frac{c_1}{\sqrt{c_1^2+c_2^2}}$$$$\Rightarrow x=e^{-\frac{\sqrt{7}}{2}t}\sqrt{c_1^2+c_2^2}\left[sin(\frac{3}{2}t+\theta )\right]$$$$\frac{dx}{dt}=-\frac{\sqrt{7}}{2}{e}^{-\frac{\sqrt{7}}{2}t}\sqrt{c_1^2+c_2^2}\frac{3}{2}\left[cos(\frac{3}{2}t+\theta )\right]$$$$\frac{d^{2}x}{{dt}^2}=-\frac{7}{4}{e}^{-\frac{\sqrt{7}}{2}t}\sqrt{c_1^2+c_2^2}\frac{9}{4}\left[sin(\frac{3}{2}t+\theta )\right]$$$$=-\frac{63}{16}x$$$$F=-\frac{63m}{16}x$$$$\to oscillatorymotion$$$$periodT=\frac{2\pi }{\omega }=\frac{2\pi }{\frac{3}{2}}=\frac{4}{3}\pi$$$$butwhent\to \mathrm{\infty },e^{-\frac{\sqrt{7}}{2}t}\to 0$$$$sotheamplitudewoulddecreasegradually$$

Commented by Rio Michael last updated on 27/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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