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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 = − 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
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{along} \\ $$$${x}-\mathrm{axis}.\:\mathrm{At}\:{t}\:=\:\mathrm{0}\:\mathrm{it}\:\mathrm{passes}\:\mathrm{origin}\:\mathrm{with} \\ $$$$\mathrm{some}\:\mathrm{velocity}\:\mathrm{towards}\:\mathrm{positive}\:{x}-\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}\:{a}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{as},\:{a}\:=\:−\:{Kx},\:\mathrm{where}\:{x}\:\mathrm{is}\:\mathrm{in}\:\mathrm{metre} \\ $$$$\mathrm{and}\:{K}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}.\:\mathrm{The}\:\mathrm{time} \\ $$$$\mathrm{at}\:\mathrm{which}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{becomes}\:\mathrm{half}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{value}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{time},\:\mathrm{is} \\ $$
Answered by mrW1 last updated on 19/Nov/17
a=(dv/dt)=v(dv/dx)=−Kx vdv=−Kxdx (v^2 /2)=−((Kx^2 )/2)+(C/2) at x=0, t=0, v=v_0 ⇒C=v_0 ^2 ⇒v^2 =v_0 ^2 −Kx^2 ⇒v=(√(v_0 ^2 −Kx^2 )) ⇒(dx/dt)=(√(v_0 ^2 −Kx^2 )) ⇒(dx/( (√(v_0 ^2 −Kx^2 ))))=dt ⇒((d(√K)x)/( (√(v_0 ^2 −((√K)x)^2 ))))=(√K)dt sin^(−1) (((√K)x)/v_0 )+C=(√K)t at t=0, x=0 ⇒C=0 ⇒t=(1/( (√K))) sin^(−1) (((√K)x)/v_0 ) ⇒x=(v_0 /( (√K))) sin ((√K)t) ⇒v=v_0 cos ((√K)t) for v=(v_0 /2) (v_0 ^2 /4)=v_0 ^2 −Kx^2 ⇒x=(((√3)v_0 )/(2(√K))) ⇒t=(1/( (√K))) sin^(−1) (((√K)/v_0 )×(((√3)v_0 )/(2(√K)))) ⇒t=(1/( (√K))) sin^(−1) (((√3)/2))=(π/(3(√K))) or (v_0 /2)=v_0 cos ((√K)t) ⇒(√K)t=(π/3) ⇒t=(π/(3(√K)))
$${a}=\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dx}}=−{Kx} \\ $$$${vdv}=−{Kxdx} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=−\frac{{Kx}^{\mathrm{2}} }{\mathrm{2}}+\frac{{C}}{\mathrm{2}} \\ $$$${at}\:{x}=\mathrm{0},\:{t}=\mathrm{0},\:{v}={v}_{\mathrm{0}} \\ $$$$\Rightarrow{C}={v}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={v}_{\mathrm{0}} ^{\mathrm{2}} −{Kx}^{\mathrm{2}} \\ $$$$\Rightarrow{v}=\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} −{Kx}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dx}}{{dt}}=\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} −{Kx}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dx}}{\:\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} −{Kx}^{\mathrm{2}} }}={dt} \\ $$$$\Rightarrow\frac{{d}\sqrt{{K}}{x}}{\:\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} −\left(\sqrt{{K}}{x}\right)^{\mathrm{2}} }}=\sqrt{{K}}{dt} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{{K}}{x}}{{v}_{\mathrm{0}} }+{C}=\sqrt{{K}}{t} \\ $$$${at}\:{t}=\mathrm{0},\:{x}=\mathrm{0} \\ $$$$\Rightarrow{C}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\:\sqrt{{K}}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{{K}}{x}}{{v}_{\mathrm{0}} } \\ $$$$\Rightarrow{x}=\frac{{v}_{\mathrm{0}} }{\:\sqrt{{K}}}\:\mathrm{sin}\:\left(\sqrt{{K}}{t}\right) \\ $$$$\Rightarrow{v}={v}_{\mathrm{0}} \mathrm{cos}\:\left(\sqrt{{K}}{t}\right) \\ $$$$ \\ $$$${for}\:{v}=\frac{{v}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}={v}_{\mathrm{0}} ^{\mathrm{2}} −{Kx}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{3}}{v}_{\mathrm{0}} }{\mathrm{2}\sqrt{{K}}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\:\sqrt{{K}}}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{{K}}}{{v}_{\mathrm{0}} }×\frac{\sqrt{\mathrm{3}}{v}_{\mathrm{0}} }{\mathrm{2}\sqrt{{K}}}\right) \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\:\sqrt{{K}}}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{3}\sqrt{{K}}} \\ $$$$ \\ $$$${or} \\ $$$$\frac{{v}_{\mathrm{0}} }{\mathrm{2}}={v}_{\mathrm{0}} \mathrm{cos}\:\left(\sqrt{{K}}{t}\right) \\ $$$$\Rightarrow\sqrt{{K}}{t}=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{t}=\frac{\pi}{\mathrm{3}\sqrt{{K}}} \\ $$
Commented by Tinkutara last updated on 20/Nov/17
Thank you very much Sir! That′s I want because I had not studied SHM.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$$$\mathrm{That}'\mathrm{s}\:\mathrm{I}\:\mathrm{want}\:\mathrm{because}\:\mathrm{I}\:\mathrm{had}\:\mathrm{not} \\ $$$$\mathrm{studied}\:\mathrm{SHM}. \\ $$
Answered by ajfour last updated on 20/Nov/17
a=−ω^2 x for S.H.M. along x. lets take x=Asin (ωt+φ) as x=0 at t=0 ⇒ φ = 0, π v=ωAcos (ωt+φ) as v>0 at t=0 ⇒ φ=0 and ≠π so x=Asin 𝛚t and v=ωAcos ωt v_0 =ωA so (v_0 /2)=((ωA)/2) v_1 = ((ωA)/2) =ωAcos ωt_1 ⇒ cos ωt_1 =(1/2) or t_1 =(π/(3ω)) = (π/(3(√K))) .
$${a}=−\omega^{\mathrm{2}} {x}\:\:\:{for}\:\:{S}.{H}.{M}.\:{along}\:{x}. \\ $$$${lets}\:{take}\:\:\:\:\:\:{x}={A}\mathrm{sin}\:\left(\omega{t}+\phi\right) \\ $$$${as}\:{x}=\mathrm{0}\:\:{at}\:\:{t}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\phi\:=\:\mathrm{0},\:\pi \\ $$$${v}=\omega{A}\mathrm{cos}\:\left(\omega{t}+\phi\right) \\ $$$${as}\:{v}>\mathrm{0}\:{at}\:{t}=\mathrm{0}\:\:\:\Rightarrow\:\:\:\phi=\mathrm{0}\:{and}\:\neq\pi \\ $$$${so}\:\:\:\:\:\boldsymbol{{x}}=\boldsymbol{{A}}\mathrm{sin}\:\boldsymbol{\omega{t}} \\ $$$${and}\:{v}=\omega{A}\mathrm{cos}\:\omega{t} \\ $$$${v}_{\mathrm{0}} =\omega{A}\:\:\:\:\:{so}\:\:\frac{{v}_{\mathrm{0}} }{\mathrm{2}}=\frac{\omega{A}}{\mathrm{2}} \\ $$$$\:\:\:{v}_{\mathrm{1}} =\:\frac{\omega{A}}{\mathrm{2}}\:=\omega{A}\mathrm{cos}\:\omega{t}_{\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{cos}\:\omega{t}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:\:\:{t}_{\mathrm{1}} =\frac{\pi}{\mathrm{3}\omega}\:=\:\frac{\pi}{\mathrm{3}\sqrt{{K}}}\:. \\ $$
Commented by Tinkutara last updated on 20/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by Tinkutara last updated on 20/Nov/17
Using SHM, I will need it later.
$$\mathrm{Using}\:\mathrm{SHM},\:\mathrm{I}\:\mathrm{will}\:\mathrm{need}\:\mathrm{it}\:\mathrm{later}. \\ $$

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