An arrangement of three identical spring and a block of mass 10 kg is shown in the adjoining figure. When the platform on which block is placed is slightly pressed down and released. It performs S.H.M. with a period of 2 s.
(i) Calculate the spring constant of the spring.
(ii) If a block of mass m is further placed on the platform, the new period of S.H.M. becomes 3.5 s. Find the value of m. (Assume platform to be massless)

Explanation

(i) given, spring constant of each spring = K
The three springs are connected in parallel.
`therefore` Spring constant of combination, K = 3K
`M = 10 kg`
now, time period is
`T = 2pi sqrt((M)/(k.))`
`rArr K = (4pi^(2)M)/(T^(2)) = (4 xx 9.86 xx 10)/(2^(2))`
= 98.6 N/m
`therefore K = K.//3 = 32.87` N/m
= 33 N/m
(ii) When a block of mass m is also placed on the platform, the period of oscillation, will be
`T. = 2pisqrt((m + M)/(K.))`
`therefore (T.)/(T) = sqrt((m + M)/(M))`
Given, `T. = 3.5s`
`therefore (3.5)/(2) = sqrt((m + 10)/(10))`
`(1.75)^(2) = (m + 10)/(10)`
`rArr 3.0625 xx 10 = m + 10 `
`or m = 20.625 ~~ 21` kg