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 isA. `(u^(2) - v^(2))/(alpha + beta)`B. `(u^(2) + v^(2))/(alpha + beta)`C. `(u^(2)0 - v^(2))/(alpha - beta)`D. `(u^(2) - v^(2))/(alpha - beta)`

A particle executes `SHM` on a straight line. At two positions, its ve

1.	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 isA. `(u^(2) - v^(2))/(alpha + beta)`B. `(u^(2) + v^(2))/(alpha + beta)`C. `(u^(2)0 - v^(2))/(alpha - beta)`D. `(u^(2) - v^(2))/(alpha - beta)`
Answer» Correct Answer - A Let the distance be p when velocity is u and acceleration `alpha`. Let the distance q when velocity is v and acceleration `beta`. If `omega` is the angular frequency. then `alpha =omega^(2)p` and `beta = omega^(2) q` `:. alpha + beta = omega^(2) (p + q) "..."(i)` Also, `u^(2) = omega^(2)A^(2) - omega^(2) p^(2)` and `v^(2) = omega^(2)A^(2) - omega^(2)q^(2)` `rArr v^(2) - u^(2) = omega^(2) (p^(2)- q^(2))` `v^(2) - u^(2) = omega^(2) (p - q) (p+q) "...."(ii)` By Eqs. (i) and (ii), we get `v^(2) - u^(2) = (p-q) (alpha + beta)` `:. p -q = (v^(2) - u^(2))/(alpha + beta)`or `q- p = (u^(2) - v^(2))/(alpha + beta)`