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A-spring-with-one-end-attached-to-a-mass-and-the-other-to-a-rigid-support-is-stretched-and-released-a-Magnitude-of-acceleration-when-just-released-is-maximum-b-Magnitude-of-acceleration-when-a

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Question Number 21012 by Tinkutara last updated on 10/Sep/17
A spring with one end attached to a mass and the other to a rigid support is stretched and released. (a) Magnitude of acceleration, when just released is maximum. (b) Magnitude of acceleration, when at equilibrium position, is maximum. (c) Speed is maximum when mass is at equilibrium position. (d) Magnitude of displacement is always maximum whenever speed is minimum.
Aspringwithoneendattachedtoamassandtheothertoarigidsupportisstretchedandreleased.(a)Magnitudeofacceleration,whenjustreleasedismaximum.(b)Magnitudeofacceleration,whenatequilibriumposition,ismaximum.(c)Speedismaximumwhenmassisatequilibriumposition.(d)Magnitudeofdisplacementisalwaysmaximumwheneverspeedisminimum.
Commented by ajfour last updated on 11/Sep/17
Displacement if it is change from initial position, then initially it is zero but speed is also zero.
Displacementifitischangefrominitialposition,theninitiallyitiszerobutspeedisalsozero.
Commented by Tinkutara last updated on 10/Sep/17
Yes, thanks. But can you explaind why (d) is wrong?
Yes,thanks.Butcanyouexplaindwhy(d)iswrong?
Commented by Tinkutara last updated on 10/Sep/17
Why wouldn′t be it is maximum there?
Whywouldn′tbeitismaximumthere?
Answered by alex041103 last updated on 11/Sep/17
We use F_(spring) ^→ =−kΔx^→ , where k is constant. But also F_(sp) ^→ =ma^→ ⇒ a^→ =−(k/m)Δx^→ =−cΔx^→ Also Δx^→ =x^→ −x_(eq) ^→ , x_(eq) ^→ =const.^(→) And a^→ =(d^2 x^→ /dt^2 ). Now if we make the substitution u^→ =x^→ −x_(eq) ^→ ⇒ (d^2 u^→ /dt^2 )=(d^2 x^→ /dt^2 )=a^→ and (du^→ /dt)=(dx^→ /dt)=v^→ . Or let′s just relabel x as x^→ =Δx^→ . We know that a^→ =(dv^→ /dt)=(dx^→ /dx^→ ) (dv^→ /dt)=(dx^→ /dt) (dv^→ /dx^→ ) ⇒a^→ =v^→ (dv^→ /dx^→ ) ⇒ a^→ dx^→ =v^→ dv^→ We now integrate: ∫_( x_0 ) ^( x) (−cx)dx=∫_( 0) ^v vdv c∫_x ^x_0 xdx=∫_( 0) ^v vdv c(x_0 ^2 −x^2 )=v^2 ⇒v=(√c)(√(x_0 ^2 −x^2 )) So in order for a to be max x has to be max too (a^→ =−cx^→ ) ⇒At maximum displacement a=a_(max) =cx_0 In order for v = v_(max) x=x_(min) =0 (0 displacepment or at equilibrium) And in order for v=v_(min ) , x=x_(max) =x_0 ⇒Ans. (a),(c),(d)
WeuseF→spring=−kΔx→,wherekisconstant.ButalsoF→sp=ma→⇒a→=−kmΔx→=−cΔx→AlsoΔx→=x→−x→eq,x→eq=const.→Anda→=d2x→dt2.Nowifwemakethesubstitutionu→=x→−x→eq⇒d2u→dt2=d2x→dt2=a→anddu→dt=dx→dt=v→.Orlet′sjustrelabelxasx→=Δx→.Weknowthata→=dv→dt=dx→dx→dv→dt=dx→dtdv→dx→⇒a→=v→dv→dx→⇒ad→x→=vd→v→Wenowintegrate:∫xx0(−cx)dx=∫v0vdvc∫x0xxdx=∫v0vdvc(x02−x2)=v2⇒v=cx02−x2Soinorderforatobemaxxhastobemaxtoo(a→=−cx→)⇒Atmaximumdisplacementa=amax=cx0Inorderforv=vmaxx=xmin=0(0displacepmentoratequilibrium)Andinorderforv=vmin,x=xmax=x0⇒Ans.(a),(c),(d)
Commented by Tinkutara last updated on 11/Sep/17
But answer given in book is only (a) and (c).
Butanswergiveninbookisonly(a)and(c).
Commented by alex041103 last updated on 11/Sep/17
I think I saw the misconseption. You can even imagine it. When you are at max displacement the directon of speed is changing so there the magnitude of the speed is 0. And because 0 is the minimum possible value for a magnitude, (d) is ok. But we measure speed like a vector so minimum speed will be negative. It′s obvious that the minimum speed then will be at the equilibrium point. Then (d) is wrong.
IthinkIsawthemisconseption.Youcanevenimagineit.Whenyouareatmaxdisplacementthedirectonofspeedischangingsotherethemagnitudeofthespeedis0.Andbecause0istheminimumpossiblevalueforamagnitude,(d)isok.Butwemeasurespeedlikeavectorsominimumspeedwillbenegative.It′sobviousthattheminimumspeedthenwillbeattheequilibriumpoint.Then(d)iswrong.

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