A particle is executing a linear SHM. `v_(1) " and " v_(2)` are its velocities at dis"tan"ce `x_(1) " and " x_(2)` from the equilibrium. What is its period of oscillation ?

Options

A. `T=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`
B. `T=2pisqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
C. `T=2pi((v_(1)-v_(2)))/((x_(2)-x_(1)))`
D. `T=2pisqrt(((x_(2)-x_(1))^(2))/((v_(1)-v_(2))^(2)))`

Answer

A

Explanation

`v_(1)=omegasqrt(A^(2)-x_(1)^(2))`
and `v_(2)=omegasqrt(A^(2)-x_(2)^(2))`
`therefore v_(1)^(2)-v_(2)^(2)=omega^(2)(A^(2)-x_(1)^(2))-omega^(2)(A^(2)-x_(2)^(2))`
`=omega^(2)[x_(2)^(2)-x_(1)^(2))`
`therefore omega =(2pi)/(T)=sqrt((v_(1)^(2)-v_(2)^(2))/((v_(1)^(2)-v_(2)^(2))))`
`therefore T=2pisqrt((x_(2)^(2)-x_(1)^(2))/((v_(1)^(2)-v_(2)^(2))))`