PhysicsEasy100×since 2002Q8160A particle is making simple harmonic motion along the X-axis. If at a distances x₁ and x₂ from the mean position the velocities of the particle are v₁ and v₂ respectively. The time period of its oscillation is given as :AT=2πx22+x12v12−v22T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}}T=2πv12−v22x22+x12BT=2πx22+x12v12+v22T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}}T=2πv12+v22x22+x12CT=2πx22−x12v12+v22T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}}T=2πv12+v22x22−x12DT=2πx22−x12v12−v22T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}}T=2πv12−v22x22−x12Check answerSkip