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.


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Question Number 21012 by Tinkutara last updated on 10/Sep/17

$A spring with one end attached to a$ $mass and the other to a rigid support is$ $stretched and released .$ $(a) Magnitude of acceleration, when$ $just released is maximum .$ $(b) Magnitude of acceleration, when$ $at equilibrium position, is maximum .$ $(c) Speed is maximum when mass is at$ $equilibrium position .$ $(d) Magnitude of displacement is$ $always maximum whenever speed is$ $minimum .$
Commented by ajfour last updated on 11/Sep/17

$D i s p l a c e m e n t i f i t i s c h a n g e f r o m$ $i n i t i a l p o s i t i o n, t h e n i n i t i a l l y i t$ $i s z e r o b u t s p e e d i s a l s o z e r o .$
Commented by Tinkutara last updated on 10/Sep/17

$Yes, thanks . But can you explaind why$ $(d) is wrong ?$
Commented by Tinkutara last updated on 10/Sep/17

$Why {wouldn}^{'} t be it is maximum there ?$
	Answered by alex041103 last updated on 11/Sep/17

	$W e u s e {\vec{F}}_{s p r i n g} = - k Δ \vec{x}, w h e r e k i s$ $c o n s t a n t .$ $B u t a l s o {\vec{F}}_{s p} = m \vec{a} \Rightarrow \vec{a} = - \frac{k}{m} Δ \vec{x} = - c Δ \vec{x}$ $A l s o Δ \vec{x} = \vec{x} - {\vec{x}}_{e q}, {\vec{x}}_{e q} = \vec{c o n s t .}$ $A n d \vec{a} = \frac{d^{2} \vec{x}}{{d t}^{2}} .$ $N o w i f w e m a k e t h e s u b s t i t u t i o n$ $\vec{u} = \vec{x} - {\vec{x}}_{e q} \Rightarrow \frac{d^{2} \vec{u}}{{d t}^{2}} = \frac{d^{2} \vec{x}}{{d t}^{2}} = \vec{a}$ $a n d \frac{d \vec{u}}{d t} = \frac{d \vec{x}}{d t} = \vec{v} .$ $O r {l e t}^{'} s j u s t r e l a b e l x a s \vec{x} = Δ \vec{x} .$ $W e k n o w t h a t \vec{a} = \frac{d \vec{v}}{d t} = \frac{d \vec{x}}{d \vec{x}} \frac{d \vec{v}}{d t} = \frac{d \vec{x}}{d t} \frac{d \vec{v}}{d \vec{x}}$ $\Rightarrow \vec{a} = \vec{v} \frac{d \vec{v}}{d \vec{x}} \Rightarrow \vec{a d} \vec{x} = \vec{v d} \vec{v}$ $W e n o w i n t e g r a t e :$ $\underset{x_{0}}{\int^{x}} (- c x) d x = \underset{0}{\int^{v}} v d v$ $c \underset{x}{\int^{x_{0}}} x d x = \underset{0}{\int^{v}} v d v$ $c (x_{0}^{2} - x^{2}) = v^{2}$ $\Rightarrow v = \sqrt{c} \sqrt{x_{0}^{2} - x^{2}}$ $S o i n o r d e r f o r a t o b e m a x$ $x h a s t o b e m a x t o o (\vec{a} = - c \vec{x})$ $\Rightarrow A t m a x i m u m d i s p l a c e m e n t$ $a = a_{m a x} = {c x}_{0}$ $I n o r d e r f o r v = v_{m a x}$ $x = x_{m i n} = 0 (0 d i s p l a c e p m e n t o r a t e q u i l i b r i u m)$ $A n d i n o r d e r f o r v = v_{m i n}, x = x_{m a x} = x_{0}$ $\Rightarrow A n s . (a), (c), (d)$
	Commented by Tinkutara last updated on 11/Sep/17

	$But answer given in book is only (a)$ $and (c) .$
	Commented by alex041103 last updated on 11/Sep/17

	$I t h i n k I s a w t h e m i s c o n s e p t i o n .$ $Y o u c a n e v e n i m a g i n e i t .$ $W h e n y o u a r e a t m a x d i s p l a c e m e n t$ $t h e d i r e c t o n o f s p e e d i s c h a n g i n g$ $s o t h e r e t h e m a g n i t u d e o f t h e s p e e d$ $i s 0 . A n d b e c a u s e 0 i s t h e m i n i m u m$ $p o s s i b l e v a l u e f o r a m a g n i t u d e,$ $(d) i s o k .$ $B u t w e m e a s u r e s p e e d l i k e a v e c t o r$ $s o m i n i m u m s p e e d w i l l b e n e g a t i v e .$ ${I t}^{'} s o b v i o u s t h a t t h e m i n i m u m s p e e d t h e n$ $w i l l b e a t t h e e q u i l i b r i u m p o i n t .$ $T h e n (d) i s w r o n g .$
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