A mass M attached to a spring oscillation with a period of `2 s`. If the mass is increased by `2 kg` , the period increases by 1s, find the initial mass m assuming that Hooke's low is obeyed.

Explanation

The period of oscillation of a mass M attached to a spring of force constant K is
`T = 2 pi sqrt(((m)/(k)))`
IN the 1st case, `T = 2s`
`2 = 2 pi sqrt(((M)/(K)))`
IN the 2 nd case, `T = 3s`
`3 = 2 pi sqrt(((M +2)/(K)))`
Dividing Eq. (ii) by Eq (i) , we get
`(3)/(2) = sqrt(((M +2)/(K)))`
Squaring,
`(3)/(2) = sqrt(((M +2)/(K)))`
`(2)/(M) = (9)/(4) - 1 = (5)/(4)`
`:. M = (4 xx 2)/(5) = (8)/(5) = 1.6 kg`