k1 and k2 attached to a mass m as shown in figure. The results can be generalized to any number of springs in parallel.
Let the force F be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement.
Therefore, net force for the displacement of mass m is
F = -Kp X ......(1)
where kp is called effective spring constant. Let the first spring be elongated by a displacement x due to force F1 and second spring be elongated by the same displacement x due to force F2, then the net force
F = – k1 x – k2 x …..(2)
Equating equations (2) and (1), we get kp = k1 + k2 …..(2)
Generalizing, for n springs connected in parallel,
If all spring constants are identical i.e., k1 = kk2 = … = kn = k then
This implies that the effective spring constant increases by a factor n. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.