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

Answer

D

Explanation

`v_(1)^(2) = omega^(2) (A^(2) - x_(1)^(2))" "v_(2)^(2) = omega^(2) (A^(2) - x_(2)^(2))`
`v_(1)^(2) - v_(2)^(2) = omega^(2) (A^(2) - x_(1)^(2)) - omega^(2) (A^(2) - x_(2)^(2))`
`v_(1)^(2) - v_(2)^(2) = omega^(2) (x_(2)^(2) - x_(1)^(2))`
So, `T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))`