For a linear SHM, when the distance of the oscillator from the equilibrium position has values y_(1)" and "y_(2) the velocities are v_(1)" and "v_(2). Show that the time period of oscillation is T= 2pi [(y_(2)^(2) -y_(1)^(2))/(v_(1)^(2)-v_(2)^(2))]^(1/2).

For a linear SHM, when the distance of the oscillator from the equilib

1.	For a linear SHM, when the distance of the oscillator from the equilibrium position has values y_(1)" and "y_(2) the velocities are v_(1)" and "v_(2). Show that the time period of oscillation is T= 2pi [(y_(2)^(2) -y_(1)^(2))/(v_(1)^(2)-v_(2)^(2))]^(1/2).
Answer» Solution :Velocity `v_(1)` at displacement `y_(1)` is `v_(1) = OMEGA SQRT(A^(2) - y_(1)^(2))"""……."(1)` and velocity `v_(2)` at displacement `y_(2)` is `v_(2) = omegasqrt(A^(2) - y_(2)^(2))""".........."(2)` `therefore v_(1)^(2) -v_(2)^(2) = omega^(2)[A^(2) -y_(1)^(2) - A^(2)+ y_(2)^(2)]` `= omega^(2) [ y_(2)^(2)- y_(1)^(2)]` `therefore omega^(2)= (v_(1)^(2)- v_(2)^(2))/(y_(2)^(2)- y_(1)^(2))` `therefore omega = [(v_(1)^(2)- v_(2)^(2))/(y_(2)^(2)- y_(1)^(2))]^(1/2)` `therefore (2pi)/(T) = [(v_(1)^(2)- v_(2)^(2))/(y_(2)^(2)- y_(1)^(2))]^(1/2)` `therefore T =2pi[(v_(1)^(2)- v_(2)^(2))/(y_(2)^(2)- y_(1)^(2))]^(1/2)` is PROVED.