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Derive the expression for resultant spring constant when two springs having constant k1 and k2 are connected in series.

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Let x1 and x2 be the elongation of springs from their equilibrium position (un-stretched position) due to applied force F. Then, the net displacement of the mass point is

x = x1 + x2 ........(1)

From hooke's law , the net force

F = -ks(x1 + x2) ⇒ x1 + x2 = \(-\frac{F}{k}\) .......(2)

Effective spring constant in series connection For springs in  series connection

-kx1 = kx2 = F

Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as

Suppose we have n springs connected in series, the effective spring constant in series is

If all spring constants are identical i.e., k1 = k2 = … = kn = k then

This means that the effective spring constant reduces by the factor n. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (3), we have,

Then the ratio of compressed distance or elongated distance x1 and x2 is

The elasticity potential energy stored in first and second springs are 

respectively. Then their ratio is

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