Two identical springs (springs constant K) are connected together in three different ways. In which case will the spring factor of the oscillation of the body be least ?

Explanation

(a) If `k_(1)` is the spring factor of the combined system and R be the restoring force in each spring, then
`2F = k_(1)y`
[where y is the displacement in each spring]
`rArr F = (-k_(1))/(2)y" "…(i)`
Also, for each spring
F = -Ky ...(ii)
comparing (i) and (ii), we get
`(K_(1))/(2) = K or K_(1) = 2k`
(b) In this case, the length of the spring is doubled, So, extension produced by the mass m will also be doubled, i.e, 2y. If `k_(2)` is the spring constant of the combined system, then
`F = K_(2)(2y) = -2k_(2)y" "...(iii)`
Comparing (ii) and (iii)
`2k_(2) = K or K_(2) = k//2`
(c ) In this case, mass m produces an extension in the upper spring and a compression in the lower spring. So, each of the spring will give rise to the restoring force F in the same direction.
If `K_(3)` is the spring factor of the combined system in this case, then
`2F = k_(3)y`
`rArr F = (k_(3))/(2)y`
Comparing (ii) and (iv),
`(K_(3))/(2) = K or K_(3) = 2K`
So, in case (b) spring constant will be least.