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A particle is executing SHM along a stra...

A particle is executing SHM along a straight line. Its velocities at distances `x_(1)` and `x_(2)` from the mean position are `v_(1)` and `v_(2)`, respectively. Its time period is

A

`2pisqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2))`

B

`2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2))`

C

`2pisqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2))`

D

`2pisqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2))`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) Let A be the amplitude of oscillation then
`v_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))" "…(i)`
`v_(2)^(2)=omega^(2)(A^(2)-x^(2))" "…(ii)`
On subtracting Eq. (ii) from Eq. (i), we get
`V_(1)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))`
`impliesomega=sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))implies(2pi)/(T)=sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))impliesT=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`
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