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 dis"tan"ce 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

B

Explanation

When th S.H.M. particle is at positions `x_(1) and x_(2)`, its velocities and aceelerations at these positions are given by
`u^(2)=omega^(2)(A^(2)-x_(1)^(2))[because v^(2)=omega^(2)(A^(2)-x^(2))]`
and `v^(2)=omega^(2)(A^(2)-x_(2)^(2))`
and `alpha=omega^(2)x_(1)`
and `beta=omega^(2)x_(2)`
Subtracting (2) from (1), we get
`u^(2)-v^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))`
and adding (3) and (4), we get
`alpha +beta=omega^(2)(x_(1)+x_(2))`
Dividing (5) by (6), we get
`(u^(2)-v^(2))/(alpha+beta)=(x_(2)^(2)-x_(1)^(2))/(x_(2)+x_(1))=x_(2)-x_(1)`
Thus the dis"tan"ce between the positions is
`x_(2)-x_(1)=(u^(2)-v^(2))/(alpha+beta)`