A 27.0 kg block at rest on a horizontal frictionless air trackis connected to th
ID: 1727698 • Letter: A
Question
A 27.0 kg block at rest on a horizontal frictionless air trackis connected to the wall via a spring. The equilibrium position ofthe mass is defined to be at x=0. Somebody pushes the mass to theposition x= 0.350 m, then lets go. The mass undergoes simpleharmonic motion with a period of 4.30 s. What is the position ofthe mass 3.526 s after the mass is released?I tried using the equationx(t)=xmcos(t+) however I keepgetting the answer to be 0.3486 m and according to the homeworkthat is incorrect. Any guidance would beappreciated.
Explanation / Answer
if you assume the solution is of the form x(t) = xm cos(t + ) then, amplitude xm is .35 m and =0 since at t=0the displacement is maximum. x(t) = .35 cos(t) = 2f = 2/T = 2/4.3 x(t) = .35 cos(2t/4.3) x(3.526) = .35*cos(2*3.526/4.3) = .149 m