A particle is moving with shm in a straight line. When the distance of the particle from the equilibrium position has the value x_1 and x_2the corresponding values of velocities are v_1and v_2show that period is T=2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)

A particle is moving with shm in a straight line. When the distance of

1.	A particle is moving with shm in a straight line. When the distance of the particle from the equilibrium position has the value x_1 and x_2the corresponding values of velocities are v_1and v_2show that period is T=2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)
Answer» Solution :VELOCITY at a DISTANCE `x_1` is `v_1=omegasqrt(a^2-x_1^2)` and `v_2=OMEGA SQRT(a^2-x_2^2)` SQUARING and subtracting `v_1^2-v_2^2=omega^2[a^2-x_1^2-a^2+x_2^2]` `omega^2=(v_1^2-v_2^2)/(x_2^2-x_1^2)` `omega=[(v_1^2-v_2^2)/(x_2^2-x_1^2)]^(1//2)` `(2pi)/T=omega"or " T=(2pi)/omega` `T = 2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)`