With the assumption of no slipping, determine the mass m of the block which must be placed on the top of a `6 kg` cart in order that the system period is `0.75s`. What is the minimum coefficient of static friction `mu_(s)` for which the block will not slip relative to the cart is displaced `50 mm` from the equilibrium position and released? Take `(g = 9.8 m//s^(2))`.

Explanation

We know.
`T = 2 pi sqrt((m//k))`
`= 2 pi sqrt(((m + 6)/(600)))`
`implies 0.75 = 2 pi sqrt(((m + 6)/(600)))`
`m = 2.56 kg`
Maximum acceleration of SHM is `a_(max) = omega^(2)A (A = "amplitude")`
To avoid slipping . Maximum force of mass m is `m omega^(2)A` which is being proveded by the force of friction between blocks A and B
`mu_(s) mg ge m omega^(2)A`
`implies mu_(s) ge (omega^(2)A)//g`
`ge (2 pi//T)^(2) (A//g)`
`ge ((2 pi)/(0.75))^(2) ((0.05)/(9.8))` `(A = 50 nm)`
`= 0.358`