Question

A simple harmonic oscillator consists of a block of mass 3.80 kg attached to a spring of spring constant 350 N/m. When t = 1.20 s, the position and velocity of the block are x = 0.137 m and v = 4.450 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] let the amplitude be A,

then maximum spring energy = energy at any instant

0.5kA^2 = 0.5kx^2+0.5mv^2

0.5*350*A^2 = 0.5*350*0.137^2+0.5*3.80*4.45^2

A = 0.4835m

b] We have position x = A sin (wt+phi)

0.137 = 0.4835*sin( sqrt(350/3.8)*1.2+phi)

  ( sqrt(350/3.8)*1.2+phi) = arcsin(0.137/0.4835) = 0.287286

phi = 0.287286 -sqrt(350/3.8)*1.2 = -11.229

position x = Asin (w*0+phi)

= 0.4835*sin -11.229 rad

= 0.4704m

c) velocity v = Aw cos (w*0+phi) = 0.4835*sqrt(350/3.8)*cos(-11.229) = 1.073 m/s