A particle executes a linear SHM. In two of its positions the velocities are u and v and the corresponding acceleration are `alpha " and " beta` respectively `(0 lt alpha lt beta)`. What is the distance between the positions ?

Options

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

Answer

A

Explanation

Let A be the amplitude of S.H.M.
Let `x_(1)` be the dis"tan"ce (displacement) of the particle when the velocity is u and acceleration is `alpha`.
and Let `x_(2)` be the dis"tan"ce (displacement ) of the particle when the velocity is v and the acceleration is `beta`.
Let `omega` be the angular frequency.
then `alpha=omega^(2)x_(1)` (numerically)
and `beta=omega^(2)x_(2)` (numerically)
`therefore alpha+beta=omega^(2)(x_(1)+x_(2))`
and `u^(2)=omega^(2)(A^(2)-x_(1)^(2))`
and `v^(2)=omega^(2)(A^(2)-x_(2)^(2))`
`therefore v^(2)-u^(2)=omega^(2)[A^(2)-x_(2)^(2)-A^(2)+x_(1)^(2)]`
`=omega^(2)(x_(1)^(2)-x_(2)^(2))`
`=omega^(2)(x_(1)+x_(2))(x_(1)-x_(2))`
From (1), `alpha+beta=omega^(2)(x_(1)+x_(2))`
`therefore v^(2)-u^(2)=(alpha+beta)(x_(1)-x_(2))`
`therefore (x_(1)-x_(2))=(v^(2)-u^(2))/(alpha+beta)`
`therefore` The dis"tan"ce between the two positions is given by
`x_(2)-x_(1)=(u^(2)-v^(2))/(alpha+beta)`