A-particle-moves-in-a-straight-line-along-x-axis-At-t-0-it-passes-origin-with-some-velocity-towards-positive-x-axis-and-with-an-acceleration-a-which-is-given-as-a-Kx-where-x-is-in-metre-and-K

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Question Number 24520 by Tinkutara last updated on 19/Nov/17

$$\mathrm{A}\mathrm{particle}\mathrm{moves}\mathrm{in}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{along}$$$$x-\mathrm{axis}.\mathrm{At}t=0\mathrm{it}\mathrm{passes}\mathrm{origin}\mathrm{with}$$$$\mathrm{some}\mathrm{velocity}\mathrm{towards}\mathrm{positive}x-\mathrm{axis}$$$$\mathrm{and}\mathrm{with}\mathrm{an}\mathrm{acceleration}a\mathrm{which}\mathrm{is}$$$$\mathrm{given}\mathrm{as},a=-Kx,\mathrm{where}x\mathrm{is}\mathrm{in}\mathrm{metre}$$$$\mathrm{and}K\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{constant}.\mathrm{The}\mathrm{time}$$$$\mathrm{at}\mathrm{which}\mathrm{its}\mathrm{velocity}\mathrm{becomes}\mathrm{half}\mathrm{of}\mathrm{its}$$$$\mathrm{value}\mathrm{at}t=0\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{time},\mathrm{is}$$

Answered by mrW1 last updated on 19/Nov/17

$$a=\frac{dv}{dt}=v\frac{dv}{dx}=-Kx$$$$vdv=-Kxdx$$$$\frac{v^2}{2}=-\frac{{Kx}^2}{2}+\frac{C}{2}$$$$atx=0,t=0,v=v_0$$$$\Rightarrow C=v_0^2$$$$\Rightarrow v^2=v_0^2-{Kx}^2$$$$\Rightarrow v=\sqrt{v_0^2-{Kx}^2}$$$$\Rightarrow \frac{dx}{dt}=\sqrt{v_0^2-{Kx}^2}$$$$\Rightarrow \frac{dx}{\sqrt{v_0^2-{Kx}^2}}=dt$$$$\Rightarrow \frac{d\sqrt{K}x}{\sqrt{v_0^2-{\left(\sqrt{K}x\right)}^2}}=\sqrt{K}dt$$$${\sin }^{-1}\frac{\sqrt{K}x}{v_0}+C=\sqrt{K}t$$$$att=0,x=0$$$$\Rightarrow C=0$$$$\Rightarrow t=\frac{1}{\sqrt{K}}{\sin }^{-1}\frac{\sqrt{K}x}{v_0}$$$$\Rightarrow x=\frac{v_0}{\sqrt{K}}\sin \left(\sqrt{K}t\right)$$$$\Rightarrow v=v_0\cos \left(\sqrt{K}t\right)$$$$$$$$forv=\frac{v_0}{2}$$$$\frac{v_0^2}{4}=v_0^2-{Kx}^2$$$$\Rightarrow x=\frac{\sqrt{3}{v}_0}{2\sqrt{K}}$$$$\Rightarrow t=\frac{1}{\sqrt{K}}{\sin }^{-1}\left(\frac{\sqrt{K}}{v_0}\times \frac{\sqrt{3}{v}_0}{2\sqrt{K}}\right)$$$$\Rightarrow t=\frac{1}{\sqrt{K}}{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3\sqrt{K}}$$$$$$$$or$$$$\frac{v_0}{2}=v_0\cos \left(\sqrt{K}t\right)$$$$\Rightarrow \sqrt{K}t=\frac{\pi }{3}$$$$\Rightarrow t=\frac{\pi }{3\sqrt{K}}$$

Commented by Tinkutara last updated on 20/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{I}\mathrm{want}\mathrm{because}\mathrm{I}\mathrm{had}\mathrm{not}$$$$\mathrm{studied}\mathrm{SHM}.$$

Answered by ajfour last updated on 20/Nov/17

$$a=-{\omega }^{2}xforS.H.M.alongx.$$$$letstakex=A\sin (\omega t+\varphi )$$$$asx=0att=0$$$$\Rightarrow \varphi =0,\pi$$$$v=\omega A\cos (\omega t+\varphi )$$$$asv>0att=0\Rightarrow \varphi =0and\ne \pi$$$$so\mathit{x}=\mathit{A}\sin \mathit{\omega }\mathit{t}$$$$andv=\omega A\cos \omega t$$$$v_0=\omega Aso\frac{v_0}{2}=\frac{\omega A}{2}$$$$v_1=\frac{\omega A}{2}=\omega A\cos \omega t_1$$$$\Rightarrow \cos \omega t_1=\frac{1}{2}$$$$or{t}_1=\frac{\pi }{3\omega }=\frac{\pi }{3\sqrt{K}}.$$

Commented by Tinkutara last updated on 20/Nov/17

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

Commented by Tinkutara last updated on 20/Nov/17

$$\mathrm{Using}\mathrm{SHM},\mathrm{I}\mathrm{will}\mathrm{need}\mathrm{it}\mathrm{later}.$$

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