A block is kept on a rough horizontal plank. The coefficient of friction between block and plank is `1//2`. Plank is undergoing `SHM` is angular frequency `10 rad//s`. Find the maximum amplitude of plank in which the block does slip over plank `(g = 10 m//s^(2))`.

Explanation

Maximum acceleration in `SHM` is `a_(max) = •^(2)A` this will be provided to the block by friction. Hence,
`a_(max) = •g or •^(2)A = •g`
or `A = (•g)/(•^(2)) = (((1)/(2))(10))/((10)^(2)) =0.05m = 5cm`