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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((mu g)/(A))`

B

`(1)/(4pi)sqrt((mu g)/(A))`

C

`2pi sqrt((A)/(mu g))`

D

`4pi sqrt((A)/(mu g))`

Text Solution

Verified by Experts

The correct Answer is:
A


`m omega^(2)A= mu mg therefore omega =sqrt((mu g)/(A))=2 pi n`
` therefore n =(1)/(2pi)sqrt((mu g)/(A))`
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