Both the springs shown in figure are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration ?

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Solution

Let the block m towards left through displacement x

$∴ F_{1}$ = k . x (compressed)

$F_{2} = k_{2} x$ (expanded)

Since they are in same direction,

$∴$ Resultant $F = F_{1} + F_{2} = k_{1} x + k_{2} x$

$= x (k_{1} + k_{2})$

So, acceleration

$a = \frac{F}{m} = \frac{x (k_{1} + k_{2})}{m}$ opposite to the displacement

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Both the springs shown in figure are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration ?

Let the block m towards left through displacement x ∴F1 = k . x (compressed) F2=k2x(expanded) Since they are in same direction, ∴ Resultant F=F1+F2=k1x+k2x =x(k1+k2) So, acceleration a=Fm=x(k1+k2)m opposite to the displacement