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
-k1 x1 = k2 x2 = 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