A particle executing linear SHM has velocities `v_(1) " and " v_(2)` at dis"tan"ce `x_(1) " and " x_(2)`, respectively from the mean position. The angular velocity of the particle is

Options

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

Answer

B

Explanation

`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_(2)^(2)-v_(1)^(2)=omega^(2)[A^(2)-x_(2)^(2)-A^(2)+x_(1)^(2)]`
`=omega^(2)(x_(1)^(2)-x_(2)^(2))`
`therefore omega =sqrt((v_(2)^(2)-v_(1)^(2))/(x_(1)^(2)-x_(2)^(2)))`