Find the time period in each of the following cases, if the mass is pulled and left.

Explanation

If the mass is pulled down by y and released.
(a) `F_(res) =-Ky =- K_(eff) y =- m omega^(2)y`
`therefore" "T = 2pi sqrt((m)/(K))`

(b) Here if the mass is pulled down by .y.. The spring will be stretched by y/2.
`therefore` it will give a force of ky/2,
which gets distributed in each string
`therefore" "F(res) =- Ky//4 =- K_(eff)y =- m omega^(2) y` br> `therefore" "T = 2pi sqrt((4m)/(K))`
(c) Here, if the the mass is pulled through .y.. the spring gets stretched by 2y.
`therefore` it gives a force 2Ky

`therefore` Restoring tension in string connected to the mass .m. is 4 Ky
`therefore" "F_(res) =- 4Ky =-K_(eff) y =- m omega^(2) y`
`therefore" "T = 2pi sqrt((m)/(4K))`
(d) `F_(res) = (K_(1) y_(1) + K_(2) y_(2))`
`K_(1) y_(1) = K_(2) y_(2)`
now `(y_(1) + y_(2))/(2) = y`

`rArr y_(1) = 2y ((K_(2))/(K_(1) + K_(2))) and y_(2) = 2y ((K_(1))/(K_(1) + K_(2)))`
`rArr" "F_(res) = 2K_(1) y_(1) = 4y ((K_(2) K_(1))/(K_(1) + K_(2))) = - m omega^(2)y`
`therefore" "K_(eq) = 4 ((K_(2)K_(1))/(K_(1) + K_(2)))`
`therefore" "T = 2pi sqrt((m (K_(1) + K_(2)))/(4 K_(1) K_(2)))`