Question

A simple harmonic oscillator consists of a block of mass 4.50 kg attached to a spring of spring constant 210 N/m. When t = 1.90 s, the position and velocity of the block are x = 0.143 m and v = 3.870 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

(a)

w = sqrt(k/m) = sqrt(210/4.5) = 6.83

From SHM we have x = A*cos(wt + phi) and v = -wA*sin(wt + phi )

So dividing the 2nd eqn by the first gives

v/x = -w*sin(wt + phi)/cos(wt + phi) = -w*tan(wt + phi)

so (wt + phi) = -arctan(v/(x*w)) = - arctan(3.870/(0.143*6.83)) = -1.3235

So phi = -1.3235 – (6.83 *1.9) = -14.3 rad

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

=> A = x/cos(wt + phi) = 0.143/cos(6.83 *1.9 – 14.3) = 0.584m

b) At t = 0 then x = 0.584*cos(0 – 14.3) = -0.09467m

c) v = -6.83 *0.584*sin(0 - 14.3) = 3.94m/s