Question

A simple harmonic oscillator consists of a block of mass 3.50 kg attached to a spring of spring constant 400 N/m. When t = 1.70 s, the position and velocity of the block are x = 0.121 m and v = 4.020 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

ω = sqrt(k/m) = sqrt(400/3.5) = 10.69

From SHM we have x = A*cos(ωt + φ) and v = -ωA*sin(ωt + φ )

So dividing the 2nd eqn by the first gives

v/x = -ω*sin(ωt + φ)/cos(ωt + φ) = -ω*tan(ωt + φ)

so (ωt + φ) = -arctan(v/(x*ω)) = - arctan(4.020/(0.121*10.69)) = -1.264

So φ = -1.264- 10.69*1.7= -19.44

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

=> A = x/cos(ωt + φ) = 0.121/cos(10.69*1.7- 19.44) = 0.4044
b) At t = 0 then x = 0.4044*cos(0 - 19.44) = 0.3359

c) v = -10.69*0.4044sin(0 - 19.44) = 2.41m/s