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

Options

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

Answer

A

Explanation

`v_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))`
`v_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))`
`implies v_(1)^(2)-v_(2)^(2)=omega^(2)[A^(2)-x_(1)^(2)-A^(2)+x_(2)^(2)]`
`implies omega=sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
`implies T=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`.