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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

	$w h e n t = 0 p o s i t i o n i s (- A)$ $v e l o c i t y = 0$ $f o r c e = F \vec{i} \vec{i} = u n i t v e c t o r a l o n g x a x i s$ $a c c e l a r a t i o n = \frac{F}{m}$ $s = u t + \frac{1}{2} {a t}^{2}$ $d i s p l a c e m e n t f r o m - \frac{A}{2} t o 0$ $= 0 \vec{i} - (- \frac{A}{2} \overset{\to)}{i})$ $= \frac{A}{2} \vec{i}$ $w h e n p a r t i c l e r e a c h a t (\frac{- A}{2} \vec{i}) i t h a s i n i t i a l$ $v e l o c i t y . . . t o c a l c u l a t e t h a t v a l u e . . . t o f i n d t i m e$ $t o r e a c h f r o m (- \frac{A}{2} \vec{i}) t o (0 \vec{i})$ $v^{2} = 0^{2} + 2 \frac{F}{m} \times \frac{A}{2}$ $v = \sqrt{\frac{F A}{m}}$ $n o w p u t t i n g t h i s \sqrt{\frac{F A}{m}} i n s = u t + \frac{1}{2} {a t}^{2}$ $\frac{A}{2} = \sqrt{\frac{F A}{m}} \times t + \frac{1}{2} \times \frac{F}{m} t^{2}$ $\frac{F}{2 m} t^{2} + \sqrt{\frac{F A}{m}} \times t - \frac{A}{2} = 0$ $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $v a l u e o f b^{2} - 4 a c$ $= \frac{F A}{m} - 4 \times \frac{F}{2 m} \times - \frac{A}{2}$ $= 2 \frac{F A}{m}$ $t = \frac{\sqrt{b^{2} - 4 a c} - b}{2 a}$ $= \frac{\sqrt{\frac{2 F A}{m}} - \sqrt{\frac{F A}{m}}}{2 \times \frac{F}{2 m}}$ $= \frac{m}{F} \times \sqrt{\frac{F A}{m}} \times (\sqrt{2} - 1)$ $= \sqrt{\frac{m A}{F}} \times (\sqrt{2} - 1)$
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