Let `x_(1) and x_(2)` be the elongation of springs from their equilibrium position (un-stretched position) due to the applied force F. Then, the net displacement of the mass point is
`x=x_(1)+x_(2)" "...(1)`
From Hooke.s law, the net force
`F=-k_(s)(x_(1)+x_(2))impliesx_(1)+x_(2)=-(F)/(k_(s))" "...(2)`
For springs in series connection
`-k_(1)x_(1)=-k_(2)x_(2)=F`
`implies x_(1)=-(F)/(k_(1))andx_(2)=-(F)/(k_(2))" "...(3)`
Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as
`-(F)/(k_(1))-(F)/(k_(2))=-(F)/(k_(s))`
`(1)/(k_(s))=(1)/(k_(1))+(1)/(k_(2))" (or) "k_(s)=(k_(1)k_(2))/(k_(1)+k_(2))Nm^(-1)" "...(4)`
Suppose we have n springs connected in series, the effective spring constant in series is
`(1)/(k_(s))=(1)/(k_(1))+(1)/(k_(2))+(1)/(k_(3))+...+(1)/(k_(n))=underset(i=l)overset(n)Sigma(1)/(k_(i))" "...(5)`
If all spring constants are identical i.e., `k_(1)=k_(2)=...=k_(n)=k` then
`(1)/(k_(s))=(n)/(k)impliesk_(s)=(k)/(n)" "...(6)`
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 constant.
From equation (3), we have,
`k_(1)x_(1)=k_(2)x_(2)`
Then the ratio of compressed distance or elongated distance `x_(1)andx_(2)` is
`(x_(2))/(x_(1))=(k_(1))/(k_(2))" "...(7)`
The elastic potential energy stored in first and second springs are `V_(1)=(1)/(2)k_(1)x_(1)^(2)andV_(2)=(1)/(2)k_(2)x_(2)^(2)` respectively. Then, their ratio is
`(V_(1))/(V_(2))=((1)/(2)k_(1)x_(1)^(2))/((1)/(2)k_(2)x_(2)^(2))=(k_(1))/(k_(2))((x_(1))/(x_(2)))^(2)=(k_(2))/(k_(1))" "...(8)`
