Explain the horizontal oscillations of a spring.

Explanation

(i) consider only two springs whose spring constant are `k_(1) andk_(2)` and which can be attached to a mass m . The results thus obtained can be generalized for any number of springs in series.
(ii) Let F be the applied force towards right as shown in figure.
(iii)Since the spring constants for different spring are different and the connection points between them is not rigidly fixed , the strings can stretch in different lengths.

(iv) 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)`
From Hooke's law, the net force
`F=-k_(s)(x_(1)+x_(2))impliesx_(1)+x_(2)=-(F)/(k_(s))" "....(1)`
For spring in series connection
`-k_(1)x_(1)=-K_(2)x_(2)=F" "...(A)`
`impliesx_(1)=-(F)/(K_(1))and x_(1) = - (F)/(k_(2)) " "...(2)`
(v) Therefore , substituting equation in (1) , 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)`
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))=sum_(i=n)^(n)(1)/(k_(i))`
If all spring constants are identical i.e. .,
`K_(1)=k_(2) =....k_(n)=k` then
`(1)/(k_(s))=n/kimpliesk_(s) =k/n`
(iv) This means that the effective spring constant reduce by the factor n. Hence , for spring in series connection, the individual spring constant is lesser than the individual spring constant.
(vii) From equation (A) , we have ,
`k_(1)x_(1) =k_(2)x_(2)`
Then the ratio of compressed distance or elongated distance `x_(1) and x_(2)` is
`(x_(2))/(x_(1))=(k_(1))/(k_(2))`
(vii) The elastic potential energy stored in first and second springs are `U_(1) =1/2 k_(1) x_(1)^(2)=U_(2) =1/2 k_(2) x_(2)^(2) ` = respectively . Then their ratio is
`(U_(1))/(U_(1)) =(1/2 k_(1) x_(1)^(2)) /(1/2 k_(2) x_(2)^(2))=(k_(1))/(k_(2))((x_(1))/(x_(2)))=(k_(2))/(k_(1))`