1.Find the amplitude of SHM of a particle if, at distances x1x_1x1 and x2x_2x2 from the equilibrium position, its velocities are v1v_1v1 and v2v_2v2 respectively.a.v12x22−v22x12v12−v22\sqrt{\dfrac{v_1^2 x_2^2 - v_2^2 x_1^2}{v_1^2 - v_2^2}}v12−v22v12x22−v22x12b.v22x12−v12x22v22−v12\sqrt{\dfrac{v_2^2 x_1^2 - v_1^2 x_2^2}{v_2^2 - v_1^2}}v22−v12v22x12−v12x22c.x22−x12v1x2−v2x1\dfrac{x_2^2 - x_1^2}{v_1 x_2 - v_2 x_1}v1x2−v2x1x22−x12d.x12−x22v1x2−v2x1\dfrac{x_1^2 - x_2^2}{v_1 x_2 - v_2 x_1}v1x2−v2x1x12−x22Login to continueOnly logged in users canattempt or see the solution.
