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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.

$$\mathrm{A}\:\mathrm{spring}\:\mathrm{with}\:\mathrm{one}\:\mathrm{end}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{mass}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{to}\:\mathrm{a}\:\mathrm{rigid}\:\mathrm{support}\:\mathrm{is} \\ $$$$\mathrm{stretched}\:\mathrm{and}\:\mathrm{released}. \\ $$$$\left({a}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{just}\:\mathrm{released}\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({b}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{at}\:\mathrm{equilibrium}\:\mathrm{position},\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({c}\right)\:\mathrm{Speed}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{equilibrium}\:\mathrm{position}. \\ $$$$\left({d}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{displacement}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{maximum}\:\mathrm{whenever}\:\mathrm{speed}\:\mathrm{is} \\ $$$$\mathrm{minimum}. \\ $$

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.

$${Displacement}\:{if}\:{it}\:{is}\:{change}\:{from} \\ $$$${initial}\:{position},\:{then}\:{initially}\:{it} \\ $$$${is}\:{zero}\:{but}\:{speed}\:{is}\:{also}\:{zero}. \\ $$

Commented by Tinkutara last updated on 10/Sep/17

Yes, thanks. But can you explaind why  (d) is wrong?

$$\mathrm{Yes},\:\mathrm{thanks}.\:\mathrm{But}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explaind}\:\mathrm{why} \\ $$$$\left({d}\right)\:\mathrm{is}\:\mathrm{wrong}? \\ $$

Commented by Tinkutara last updated on 10/Sep/17

Why wouldn′t be it is maximum there?

$$\mathrm{Why}\:\mathrm{wouldn}'\mathrm{t}\:\mathrm{be}\:\mathrm{it}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{there}? \\ $$

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)

$$ \\ $$$${We}\:{use}\:\overset{\rightarrow} {{F}}_{{spring}} =−{k}\Delta\overset{\rightarrow} {{x}},\:{where}\:{k}\:{is} \\ $$$${constant}. \\ $$$${But}\:{also}\:\overset{\rightarrow} {{F}}_{{sp}} ={m}\overset{\rightarrow} {{a}}\:\Rightarrow\:\overset{\rightarrow} {{a}}=−\frac{{k}}{{m}}\Delta\overset{\rightarrow} {{x}}=−{c}\Delta\overset{\rightarrow} {{x}} \\ $$$${Also}\:\Delta\overset{\rightarrow} {{x}}=\overset{\rightarrow} {{x}}−\overset{\rightarrow} {{x}}_{{eq}} ,\:\overset{\rightarrow} {{x}}_{{eq}} =\overset{\rightarrow} {{const}.} \\ $$$${And}\:\overset{\rightarrow} {{a}}=\frac{{d}^{\mathrm{2}} \overset{\rightarrow} {{x}}}{{dt}^{\mathrm{2}} }. \\ $$$${Now}\:{if}\:{we}\:{make}\:{the}\:{substitution} \\ $$$$\overset{\rightarrow} {{u}}=\overset{\rightarrow} {{x}}−\overset{\rightarrow} {{x}}_{{eq}} \:\Rightarrow\:\frac{{d}^{\mathrm{2}} \overset{\rightarrow} {{u}}}{{dt}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} \overset{\rightarrow} {{x}}}{{dt}^{\mathrm{2}} }=\overset{\rightarrow} {{a}} \\ $$$${and}\:\:\frac{{d}\overset{\rightarrow} {{u}}}{{dt}}=\frac{{d}\overset{\rightarrow} {{x}}}{{dt}}=\overset{\rightarrow} {{v}}. \\ $$$${Or}\:{let}'{s}\:{just}\:{relabel}\:{x}\:{as}\:\overset{\rightarrow} {{x}}=\Delta\overset{\rightarrow} {{x}}. \\ $$$${We}\:{know}\:{that}\:\overset{\rightarrow} {{a}}=\frac{{d}\overset{\rightarrow} {{v}}}{{dt}}=\frac{{d}\overset{\rightarrow} {{x}}}{{d}\overset{\rightarrow} {{x}}}\:\frac{{d}\overset{\rightarrow} {{v}}}{{dt}}=\frac{{d}\overset{\rightarrow} {{x}}}{{dt}}\:\frac{{d}\overset{\rightarrow} {{v}}}{{d}\overset{\rightarrow} {{x}}} \\ $$$$\Rightarrow\overset{\rightarrow} {{a}}=\overset{\rightarrow} {{v}}\:\frac{{d}\overset{\rightarrow} {{v}}}{{d}\overset{\rightarrow} {{x}}}\:\Rightarrow\:\overset{\rightarrow} {{a}d}\overset{\rightarrow} {{x}}=\overset{\rightarrow} {{v}d}\overset{\rightarrow} {{v}} \\ $$$${We}\:{now}\:{integrate}: \\ $$$$\underset{\:{x}_{\mathrm{0}} } {\overset{\:\:{x}} {\int}}\left(−{cx}\right){dx}=\underset{\:\mathrm{0}} {\overset{{v}} {\int}}{vdv} \\ $$$${c}\underset{{x}} {\overset{{x}_{\mathrm{0}} } {\int}}{xdx}=\underset{\:\mathrm{0}} {\overset{{v}} {\int}}{vdv} \\ $$$${c}\left({x}_{\mathrm{0}} ^{\mathrm{2}} −{x}^{\mathrm{2}} \right)={v}^{\mathrm{2}} \\ $$$$\Rightarrow{v}=\sqrt{{c}}\sqrt{{x}_{\mathrm{0}} ^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${So}\:{in}\:{order}\:{for}\:{a}\:{to}\:{be}\:{max}\: \\ $$$${x}\:{has}\:{to}\:{be}\:{max}\:{too}\:\left(\overset{\rightarrow} {{a}}=−{c}\overset{\rightarrow} {{x}}\right) \\ $$$$\Rightarrow{At}\:{maximum}\:{displacement}\: \\ $$$${a}={a}_{{max}} ={cx}_{\mathrm{0}} \\ $$$${In}\:{order}\:{for}\:{v}\:=\:{v}_{{max}} \\ $$$${x}={x}_{{min}} =\mathrm{0}\:\left(\mathrm{0}\:{displacepment}\:{or}\:{at}\:{equilibrium}\right) \\ $$$${And}\:{in}\:{order}\:{for}\:{v}={v}_{{min}\:} ,\:{x}={x}_{{max}} ={x}_{\mathrm{0}} \\ $$$$\Rightarrow{Ans}.\:\left({a}\right),\left({c}\right),\left({d}\right) \\ $$

Commented by Tinkutara last updated on 11/Sep/17

But answer given in book is only (a)  and (c).

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{in}\:\mathrm{book}\:\mathrm{is}\:\mathrm{only}\:\left({a}\right) \\ $$$$\mathrm{and}\:\left({c}\right). \\ $$

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.

$${I}\:{think}\:{I}\:{saw}\:{the}\:{misconseption}. \\ $$$${You}\:{can}\:{even}\:{imagine}\:{it}. \\ $$$${When}\:{you}\:{are}\:{at}\:{max}\:{displacement} \\ $$$${the}\:{directon}\:{of}\:{speed}\:{is}\:{changing} \\ $$$${so}\:{there}\:{the}\:{magnitude}\:{of}\:{the}\:{speed} \\ $$$${is}\:\mathrm{0}.\:{And}\:{because}\:\mathrm{0}\:{is}\:{the}\:{minimum} \\ $$$${possible}\:{value}\:{for}\:{a}\:{magnitude}, \\ $$$$\left({d}\right)\:{is}\:{ok}. \\ $$$${But}\:{we}\:{measure}\:{speed}\:{like}\:{a}\:{vector} \\ $$$${so}\:{minimum}\:{speed}\:{will}\:{be}\:{negative}. \\ $$$${It}'{s}\:{obvious}\:{that}\:{the}\:{minimum}\:{speed}\:{then} \\ $$$${will}\:{be}\:{at}\:{the}\:{equilibrium}\:{point}. \\ $$$${Then}\:\left({d}\right)\:{is}\:{wrong}. \\ $$

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