A-particle-moves-in-a-straight-line-along-x-axis-At-t-0-it-is-released-from-rest-at-x-a-Acceleration-of-the-particle-varies-as-d-2-x-dt-2-k-x-2-where-k-is-a-positive-constant-Time-r

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Question Number 22304 by Tinkutara last updated on 15/Oct/17

$$\mathrm{A}\mathrm{particle}\mathrm{moves}\mathrm{in}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{along}$$$$x-\mathrm{axis}.\mathrm{At}t=0,\mathrm{it}\mathrm{is}\mathrm{released}\mathrm{from}$$$$\mathrm{rest}\mathrm{at}x=a.\mathrm{Acceleration}\mathrm{of}\mathrm{the}$$$$\mathrm{particle}\mathrm{varies}\mathrm{as}\frac{d^{2}x}{{dt}^2}=-\frac{k}{x^2},\mathrm{where}k$$$$\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{constant}.$$$$\mathrm{Time}\mathrm{required}\mathrm{by}\mathrm{particle}\mathrm{to}\mathrm{reach}\mathrm{the}$$$$\mathrm{origin}\mathrm{will}\mathrm{be}$$

Answered by ajfour last updated on 16/Oct/17

$$\frac{d^{2}x}{{dt}^2}=\frac{vdv}{dx}=-\frac{k}{x^2}$$$$\Rightarrow {\int }_0^{v}vdv=-{\int }_a^x\frac{kdx}{x^2}$$$$\Rightarrow \frac{v^2}{2}=+k(\frac{1}{x}-\frac{1}{a})$$$$orv=-\sqrt{2k}\sqrt{\frac{1}{x}-\frac{1}{a}}$$$$\Rightarrow {\int }_a^x\frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}}=-\sqrt{2k}{\int }_0^{t}dt$$$$timetakentoreachx=0beT$$$$\Rightarrow T=\frac{1}{\sqrt{2k}}{\int }_0^a\sqrt{\frac{ax}{a-x}}dx$$$$leta-x=y^2\Rightarrow dx=-2ydy$$$$x=a-y^2;x=0\Rightarrow y=\sqrt{a}$$$$andx=a\Rightarrow y=0;then$$$$T=\frac{1}{\sqrt{2k}}{\int }_{\sqrt{a}}^0\frac{\sqrt{a(a-y^2)}}{y}(-2ydy)$$$$=\sqrt{\frac{2a}{k}}{\int }_0^{\sqrt{a}}\sqrt{{\left(\sqrt{a}\right)}^2-y^2}dy$$$$=\sqrt{\frac{2a}{k}}[\frac{y}{2}\sqrt{a-y^2}{\mid }_0^{\sqrt{a}}+\frac{a}{2}{\sin }^{-1}\left(\frac{y}{\sqrt{a}}\right){\mid }_0^{\sqrt{a}}]$$$$\mathit{T}=\sqrt{\frac{2\mathit{a}}{\mathit{k}}}\left(\frac{\mathit{a}}{2}\times \frac{\mathit{\pi }}{2}\right)=\frac{\mathit{\pi }\mathit{a}}{2}\sqrt{\frac{\mathit{a}}{2\mathit{k}}}.$$

Commented by squidward last updated on 16/Oct/17

$$4thlineisincorrecttoo$$

Commented by Tinkutara last updated on 15/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by squidward last updated on 16/Oct/17

$$your3rdlineiswrongandhowthehell$$$$didyougofrom3rdlineto4th?$$

Commented by ajfour last updated on 16/Oct/17

$$isitfinenow?$$

Commented by ajfour last updated on 16/Oct/17

$$No.afterbeingreleasedatx=a$$$$particledevelopsnegativevelocity.$$$$v=\pm \sqrt{2k}\sqrt{\frac{1}{x}-\frac{1}{a}}$$$$sowechoosethe-veone.$$

Commented by Tinkutara last updated on 25/Nov/17

$$Whywechose-veone?$$$$Whywechose-veone?$$

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