A cubical block of mass M vibrates horizontally with amplitude of 4.0 cm and a frequency of 2.0 Hz. A small block of mass is placed on the bigger block. In order that the smaller block does not side on the bigger block, the minimum value of the coefficient of static friction between the two blocks is (0.16x). Find the valule of x(Take `pi^(2)`=10 and g=10m/`s^(2)`)

Answer

4

Explanation

`mug>`maximum acceleration in SHM
`therefore mu>omega^(2)A therefore mu>(omega^(2)A)/(g)` or `mu_(min)=((2piv)^(2)A)/(g)`
`mu_(min)=(4xx10xx(2)^(2)xx0.04)/(10)=0.64` 0.16x=0.64 `Rightarrow`x=4