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Question Number 35739 by rahul 19 last updated on 22/May/18

Commented by rahul 19 last updated on 22/May/18

Also find period of motion ?

Commented by rahul 19 last updated on 22/May/18

ans. are 13- A , 14-B

Answered by tanmay.chaudhury50@gmail.com last updated on 23/May/18

when t=0 position is (−A)  velocity=0  force=Fi^→   i^→ =unit vector along x axis  accelaration=(F/m)  s=ut+(1/2)at^2   displacement from −(A/2)to 0  =0i^→ −(−(A/2)i^(→)) )  =(A/2)i^→   when particle reach at (((−A)/2)i^→ )it has initial  velocity...to calculate that value ...to find time  to reach from (−(A/2)i^→ )to(0i^→ )    v^2 =0^2 +2(F/m)×(A/2)  v=(√(((FA)/m) ))  now putting this (√(((FA)/m) ))  in s=ut+(1/2)at^2     (A/2)=(√(((FA)/m) )) ×t+(1/2)×(F/m)t^2   (F/(2m))t^2 +(√((FA)/m)) ×t−(A/2)=0  t=((−b±(√(b^2 −4ac)) )/(2a))  value of b^2 −4ac  =((FA)/m)−4×(F/(2m))×−(A/2)  =2((FA)/m)  t=(((√(b^2 −4ac))  −b)/(2a))  =(((√((2FA)/m)) −(√((FA)/m)))/(2×(F/(2m))))  =(m/F)×(√((FA)/m)) ×((√2) −1)  =(√(((mA)/F) )) ×((√2) −1)

$${when}\:{t}=\mathrm{0}\:{position}\:{is}\:\left(−{A}\right) \\ $$$${velocity}=\mathrm{0} \\ $$$${force}={F}\overset{\rightarrow} {{i}}\:\:\overset{\rightarrow} {{i}}={unit}\:{vector}\:{along}\:{x}\:{axis} \\ $$$${accelaration}=\frac{{F}}{{m}} \\ $$$${s}={ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$${displacement}\:{from}\:−\frac{{A}}{\mathrm{2}}{to}\:\mathrm{0} \\ $$$$=\mathrm{0}\overset{\rightarrow} {{i}}−\left(−\frac{{A}}{\mathrm{2}}\overset{\left.\rightarrow\right)} {{i}}\right) \\ $$$$=\frac{{A}}{\mathrm{2}}\overset{\rightarrow} {{i}} \\ $$$${when}\:{particle}\:{reach}\:{at}\:\left(\frac{−{A}}{\mathrm{2}}\overset{\rightarrow} {{i}}\right){it}\:{has}\:{initial} \\ $$$${velocity}...{to}\:{calculate}\:{that}\:{value}\:...{to}\:{find}\:{time} \\ $$$${to}\:{reach}\:{from}\:\left(−\frac{{A}}{\mathrm{2}}\overset{\rightarrow} {{i}}\right){to}\left(\mathrm{0}\overset{\rightarrow} {{i}}\right) \\ $$$$ \\ $$$${v}^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} +\mathrm{2}\frac{{F}}{{m}}×\frac{{A}}{\mathrm{2}} \\ $$$${v}=\sqrt{\frac{{FA}}{{m}}\:} \\ $$$${now}\:{putting}\:{this}\:\sqrt{\frac{{FA}}{{m}}\:}\:\:{in}\:{s}={ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$$\:\:\frac{{A}}{\mathrm{2}}=\sqrt{\frac{{FA}}{{m}}\:}\:×{t}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{F}}{{m}}{t}^{\mathrm{2}} \\ $$$$\frac{{F}}{\mathrm{2}{m}}{t}^{\mathrm{2}} +\sqrt{\frac{{FA}}{{m}}}\:×{t}−\frac{{A}}{\mathrm{2}}=\mathrm{0} \\ $$$${t}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}} \\ $$$${value}\:{of}\:{b}^{\mathrm{2}} −\mathrm{4}{ac} \\ $$$$=\frac{{FA}}{{m}}−\mathrm{4}×\frac{{F}}{\mathrm{2}{m}}×−\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{2}\frac{{FA}}{{m}} \\ $$$${t}=\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:\:−{b}}{\mathrm{2}{a}} \\ $$$$=\frac{\sqrt{\frac{\mathrm{2}{FA}}{{m}}}\:−\sqrt{\frac{{FA}}{{m}}}}{\mathrm{2}×\frac{{F}}{\mathrm{2}{m}}} \\ $$$$=\frac{{m}}{{F}}×\sqrt{\frac{{FA}}{{m}}}\:×\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right) \\ $$$$=\sqrt{\frac{{mA}}{{F}}\:}\:×\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right) \\ $$

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