A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is…….

Answer

B

Explanation

Velocity in SHM `v= pm omega sqrt(A^(2)-x^(2))` and
Acceleration `a= -omega^(2)x`
but `|v|=|a|`
`therefore omega sqrt(A^(2)-x^(2))= omega^(2)x`
`sqrt(A^(2)-x^(2))= omega x`
`therefore A^(2)- x^(2) = omega^(2) x^(2)`
`therefore omega^(2) = (A^(2)-x^(2))/(x^(2))`
`= ((3)^(2)-(2)^(2))/((2)^(2)`
`= (9-4)/(4)`
`=(5)/(4)`
`therefore omega = (sqrt(5))/(2)`
Now `omega = (2pi)/(T)" " therefore T= (2pi)/(omega)`
`therefore T= (2pi)/((sqrt(5))/(2))= (4pi)/(sqrt(5))`.