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A block rests on a horizontal table whic...

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`.

A

`(1)/(2pi)sqrt((mug)/(A))`

B

`(1)/(4pi)sqrt((mug)/(A))`

C

`2pisqrt((A)/(mug))`

D

`4pisqrt((A)/(mug))`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) When restoring force will become equal to the frictional force, block will start to slip.
`therefore` Restoring force=Friction force
`implies kA=mumg" "[because F_("rest"=-kx]...(i)`
Now, frequency , `f=(1)/(2pi)sqrt((k)/(m))`
From Eq. , we get `f=(1)/(2pi)sqrt((mug)/(A))`
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