A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude A. What will be the frequency of oscillation, the block will just start to slip? Coefficient of friction`=mu`.

Options

A. `(1)/(2pi ) sqrt((mu g)/(A))`
B. `(1)/(4pi ) sqrt((mu g)/(A))`
C. `2pi sqrt((A)/(mu g))`
D. `4pi sqrt((A)/(mu g))`

Answer

A

Explanation

When restoring force will become equal to the frictional force , block will start to slip .
`therefore` Restoring force = Friction force
`implies kA = mu mg " " … (i)`
Now , frequency f `= (1)/(2pi) sqrt((k)/(m))`
From Eq.(i)
`f = (1)/(2pi) sqrt((mu g)/(A))`