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 is a positive constant. The time at which its velocity becomes half of its value at t = 0 for the first time, is


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

$A particle moves in a straight line along$ $x - axis . At t = 0 it passes origin with$ $some velocity towards positive x - axis$ $and with an acceleration a which is$ $given as, a = - K x, where x is in metre$ $and K is a positive constant . The time$ $at which its velocity becomes half of its$ $value at t = 0 for the first time, is$
	Answered by mrW1 last updated on 19/Nov/17

	$a = \frac{d v}{d t} = v \frac{d v}{d x} = - K x$ $v d v = - K x d x$ $\frac{v^{2}}{2} = - \frac{{K x}^{2}}{2} + \frac{C}{2}$ $a t x = 0, t = 0, v = v_{0}$ $\Rightarrow C = v_{0}^{2}$ $\Rightarrow v^{2} = v_{0}^{2} - {K x}^{2}$ $\Rightarrow v = \sqrt{v_{0}^{2} - {K x}^{2}}$ $\Rightarrow \frac{d x}{d t} = \sqrt{v_{0}^{2} - {K x}^{2}}$ $\Rightarrow \frac{d x}{\sqrt{v_{0}^{2} - {K x}^{2}}} = d t$ $\Rightarrow \frac{d \sqrt{K} x}{\sqrt{v_{0}^{2} - {(\sqrt{K} x)}^{2}}} = \sqrt{K} d t$ $\sin^{- 1} \frac{\sqrt{K} x}{v_{0}} + C = \sqrt{K} t$ $a t t = 0, x = 0$ $\Rightarrow C = 0$ $\Rightarrow t = \frac{1}{\sqrt{K}} \sin^{- 1} \frac{\sqrt{K} x}{v_{0}}$ $\Rightarrow x = \frac{v_{0}}{\sqrt{K}} \sin (\sqrt{K} t)$ $\Rightarrow v = v_{0} \cos (\sqrt{K} t)$ $f o r v = \frac{v_{0}}{2}$ $\frac{v_{0}^{2}}{4} = v_{0}^{2} - {K x}^{2}$ $\Rightarrow x = \frac{\sqrt{3} v_{0}}{2 \sqrt{K}}$ $\Rightarrow t = \frac{1}{\sqrt{K}} \sin^{- 1} (\frac{\sqrt{K}}{v_{0}} \times \frac{\sqrt{3} v_{0}}{2 \sqrt{K}})$ $\Rightarrow t = \frac{1}{\sqrt{K}} \sin^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3 \sqrt{K}}$ $o r$ $\frac{v_{0}}{2} = v_{0} \cos (\sqrt{K} t)$ $\Rightarrow \sqrt{K} t = \frac{π}{3}$ $\Rightarrow t = \frac{π}{3 \sqrt{K}}$
	Commented by Tinkutara last updated on 20/Nov/17

	$Thank you very much Sir!$ ${That}^{'} s I want because I had not$ $studied SHM .$
	Answered by ajfour last updated on 20/Nov/17

	$a = - ω^{2} x f o r S . H . M . a l o n g x .$ $l e t s t a k e x = A \sin (ω t + ϕ)$ $a s x = 0 a t t = 0$ $\Rightarrow ϕ = 0, π$ $v = ω A \cos (ω t + ϕ)$ $a s v > 0 a t t = 0 \Rightarrow ϕ = 0 a n d \neq π$ $s o x = A \sin ω t$ $a n d v = ω A \cos ω t$ $v_{0} = ω A s o \frac{v_{0}}{2} = \frac{ω A}{2}$ $v_{1} = \frac{ω A}{2} = ω A \cos ω t_{1}$ $\Rightarrow \cos ω t_{1} = \frac{1}{2}$ $o r t_{1} = \frac{π}{3 ω} = \frac{π}{3 \sqrt{K}} .$
	Commented by Tinkutara last updated on 20/Nov/17

	$Thank you very much Sir!$
	Commented by Tinkutara last updated on 20/Nov/17

	$Using SHM, I will need it later .$
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