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))/(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

A

Explanation

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)`