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A block resting on the horizontal surfac...

A block resting on the horizontal surface executes S.H.M. in horizontal plane with amplitude 'A'. The frequency of oscillation for which the block just starts to slip is (`mu` = coefficient of friction, g = gravitational acceleration)

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
`F=mumg`
`kA=mumg`
`momega^(2)A=mumg`
`4pi^(2)n^(2)A=mug`
`n^(2)=(1)/(4pi^(2))(mug)/(A)`
`n=(1)/(2pi)sqrt((mug)/(A))`
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