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

Options

A. `2 pi sqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`
B. `2 pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`
C. `2 pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`
D. `2 pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`

Answer

B

Explanation

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)`
Subtracting Eq. (ii) from Eq. (i), we get
`v_(1)^(2)-v_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))`
`rArr" "omega=sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
`rArr" "(2pi)/(T)=sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
`rArr" "T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`