A particle executes `SHM` on a straight line. At two positions, its velocities are `u` and `v` whle accelerations are `alpha` and `beta` respectively `[beta gt alpha gt ]`.The distance between these two positions is

Options

A. `(u^(2) + v^(2))/(a+ beta)`
B. `(u^(2) - v^(2))/(alpha + beta)`
C. `(u^(2) - v^(2))/(beta - alpha)`
D. `(u^(2) + v^(2))/(beta - alpha)`

Answer

B

Explanation

`mu = omega sqrt(A^(2) - x_(1)^(2))" "alpha =- omega^(2) x(1), beta =- omega^(2) x_(2)`
`v = omega sqrt(A^(2) - x_(2)^(2))`
`beta gt x_(0) rArr x_(1) and x_(2) "are -ve and" x_(2) lt x_(1)`
distance `= x_(1) - x_(2)`
`((v)/(omega))^(2) - ((u)/(omega))^(2) = x_(1)^(2) - x_(2)^(2) = (x_(1) - x_(2)) (x_(1) + x_(2))`
`(v^(2) - u^(2)) = (x_(1)^(2) -x_(2)^(2)) = (x_(1)-x_(2))(x_(1)+x_(2))`
`v^(2) - u^(2) = (x_(1) - x_(2)) - (alpha + beta)`
`x_(1) - x_(2) = (mu^(2) - v^(2))/(alpha + beta)`