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