A particle is executing SHM along a straight line. Its velocities at distance `x_(1)" and "x_(2)` form the mean position are `v_(1)" and "v_(2)`, respectively. Its time period is………..

Options

A. `2pi sqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`
B. `2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))`
C. `2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`
D. `2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`

Answer

B

Explanation

For SHM particle,
`v= omega sqrt(A^(2)-x^(2))`
`therefore v_(1)= omega sqrt(A^(2)-x_(1)^(2))" and "v_(2)= omega sqrt((A^(2)-x_(2)^(2))`
`therefore v_(1)^(2)= omega^(2)(A^(2)-x_(1)^(2))" and "v_(2)^(2)= omega^(2) (A^(2)-x_(2)^(2))`
`therefore v_(1)^(2) -v_(2)^(2) = omega^(2)[A^(2)-x_(1)^(2)-A^(2)+x_(2)^(2)]`
`= omega^(2) [x_(2)^(2)-x_(2)^(2)]`
`therefore omega^(2)= (v_(1)^(2)-v_(2)^(2))/((x_(2)^(2)-x_(1)^(2)))`
`therefore omega = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
`therefore (2pi)/(T) = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
`therefore T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`.