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A particle executes SHM on a straight li...

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

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)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let distance by `x_(1)` when velocity if u and accleration `alpha`.
Let distance by `x_(2)` when velocity is v and acceleration `beta`.
If `omega` is the angular frequencey then,
`alpha=omega^(2)x_(2)`
`therefore alpha+beta=omega^(2)(x_(1)+x_(2)),.......(1)`
`"Also," u^(2)=omega^(2)A^(2)-omega^(2)x_(1)^(2)`
`and v^(2)=omega^(2)A^(2)-omega^(2)x_(2)^(2`
`v^(2)-u^(2)=omega^(2)(x_(1)^(2)-x_(2)^(2))`
`v^(2)-u^(2)=omega^(2)(x_(1)-x_(2))(x_(1)+x_(2)).....(ii)`
By. eq (i) we get
`v^(2)-u^(2)=(x_(1)-x_(2))(alpha+beta)`
`therefore x_(1)-x_(2)=(u^(2)-v^(2))/(alpha+beta) or x_(2)-x_(1)=(u^(2)-v^(2))/(alpha+beta)`
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