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A particle executes S.H.M. with an ampli...

A particle executes S.H.M. with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of tis velocity is equal to that of its acceleration. Then its time period in second is :

A

`(1)/(2pi sqrt(3))`

B

`2pi sqrt(3)`

C

`(2pi)/(sqrt(3))`

D

`(sqrt(3))/(2pi)`

Text Solution

Verified by Experts

The correct Answer is:
C
`omega sqrt(r^(2)-y^(2))=omega^(2)y`
`sqrt(r^(2)-y^(2))=omega y=(2pi)/(T)y`
`T=(2pi y)/(sqrt(r^(2)-y^(2)))=(2pi xx1)/(sqrt(2^(2)-1^(2)))=(2pi)/(sqrt(3))` second.
Correct choice is ( c ).
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