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

A particle is executing SHM along a straight line. Its velocities at d

1.	A particle is executing SHM along a straight line. Its velocities at distance x_(1)" and "x_(2) form the mean position are v_(1)" and "v_(2), respectively. Its time period is………..
Answer» `2PI SQRT((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))` `2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))` `2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))` `2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))` Solution :For SHM particle, `V= omega sqrt(A^(2)-x^(2))` `therefore v_(1)= omega sqrt(A^(2)-x_(1)^(2))" and "v_(2)= omega sqrt((A^(2)-x_(2)^(2))` `therefore v_(1)^(2)= omega^(2)(A^(2)-x_(1)^(2))" and "v_(2)^(2)= omega^(2) (A^(2)-x_(2)^(2))` `therefore v_(1)^(2) -v_(2)^(2) = omega^(2)[A^(2)-x_(1)^(2)-A^(2)+x_(2)^(2)]` `= omega^(2) [x_(2)^(2)-x_(2)^(2)]` `therefore omega^(2)= (v_(1)^(2)-v_(2)^(2))/((x_(2)^(2)-x_(1)^(2)))` `therefore omega = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` `therefore (2pi)/(T) = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` `therefore T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`.