1.

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 equalto that of its acceleration. Then its time periodin second is :

Answer»

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

SOLUTION :`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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