Question

A simple harmonic oscillator consists of a block of mass 3.70 kg attached to a spring of spring constant 410 N/m. When t = 1.60 s, the position and velocity of the block are x = 0.102 m and v = 3.050 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

Answer 1

? = sqrt(k/m) = sqrt(410/3.70) = 10.526

From SHM we have x = A*cos(?t + ?) and v = -?A*sin(?t + ? )

So dividing the 2nd eqn by the first gives

v/x = -?*sin(?t + ?)/cos(?t + ?) = -?*tan(?t + ?)

so (?t + ?) = -arctan(v/(x*?)) = - arctan(3.050/(0.102*10.526)) = -72.941

So ? = -72.941 - 10.526*1.60 = -89.7826 rad

Now plugging into the position eqn we have x = A*cos(?t + ?)

=> A = x/cos(?t + ?) = 0.102/cos(10.526*1.60 - 89.7826) = 0.3477m

b) At t = 0 then x = 0.3477*cos(0 - 89.7826) = 0.00131m

c) v = -10.526*0.3477*sin(0 - 89.7826) = 3.659m/s