In the figure a block A of mass m is attached at one end of a light spring and the other end of the spring is connected to another block B of $$mass2m$$ through a light string. A is held and B has obtaiwwned equilibrium. Now A is released. The acceleration of Ajust after that instant is a. The same thing is repeated for B.In that case the acceleration of B is b, then the value of $$a/b$$ is $$__________$$.

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In Ist CaseIn IIst $$CaseT=2mg$$&$$a=2mg$$−mgm⇒$$a=g$$↑$$kx$$$$=mg$$&$$b=2mg$$−$$mg2m$$⇒$$b=g/2$$↓∴ $$ab=2$$

Newton's laws of motion

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In I^st Case
In II^st Case
$\begin{array}{l} T = 2 m g & a = \frac{2 m g - m g}{m} \Rightarrow a = g ↑ \\ k x = m g & b = \frac{2 m g - m g}{2 m} \Rightarrow b = g / 2 ↓ \\ ∴ \frac{a}{b} = 2 \end{array}$

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Newton's laws of motion

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In Ist CaseIn IIst CaseT=2mg&a=2mg−mgm⇒a=g↑kx=mg&b=2mg−mg2m⇒b=g/2↓∴ ab=2

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In I^st Case
In II^st Case
$\begin{array}{l} T = 2 m g & a = \frac{2 m g - m g}{m} \Rightarrow a = g ↑ \\ k x = m g & b = \frac{2 m g - m g}{2 m} \Rightarrow b = g / 2 ↓ \\ ∴ \frac{a}{b} = 2 \end{array}$