posted by Andrea on .
Find the instantaneous velocity of a mass on a spring oscillating on a horizontal frictionless surface at the instant when its displacement is half of its maximum displacement x=(x_max/2). Assume the max velocity of that mass during each oscillation is v_max= 2m/s.
I know that I will have this: v = (2*pi)/T *(x_max/2) I'm having trouble because I don't have the period(T). I also don't know x_max
You don't need to know the period. The sum of the kinetic (1/2) M V^2 and potential energy (1/2) kX^2 is constant. When X is half the maximum value, the P.E. is 1/4 of the maximum value. That means the kinetic energy is 3/4 of its maximum value, since energy shifts from all-kinetic to all-potential.
The maximum KE is (1/2)MV-max^2 . When it is 3/4 of that, V^2 = 3/4 V-max^2
V = sqrt(3/2) V_max = sqrt 3 m/s