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((x_(2)^(2) - x_(1)^(2))/(V_(1)^(2) - V_(2)^(2)))`
B. `2pi sqrt((V_(2)^(2) + V_(1)^(2))/(x_(1)^(2) + x_(2)^(2)))`
C. `2 pi sqrt((V_(1)^(2)-V_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`
D. `2pi sqrt((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))`
Substracting `(v_(1)^(2))/(omega^(2)) + x_(1)^(2) = (v_(2)^(2))/(omega^(2)) + x_(2)^(2)`
`rArr (v_(1)^(2) - v_(2)^(2))/(omega^(2)) = x_(2)^(2) - x_(1)^(2)`
`rArr omega = sqrt((v_(1)^(2) - v_(2)^(2))/(x_(2)^(2) - x_(1)^(2)))`
`rArr T = 2pi sqrt((x_(2)^(2) - x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))`