a 1 kg block oscillates on a spring with spring constant 20N/m at t=0s the block is 20 cm to the right of the equilibrium position and movin to the left at a speed of 100 cm/s determine the period of oscillation and draw a graph of position versus time

T = π/√5 ≈ 1.40s

Explanation:

ω = √(k/m) = √(20/1) = 2√5 rad/s

T = 2π/ω = 2π/2√5 = π/√5 ≈ 1.4049 s

x(t) = Acos(ωt + φ)

v(t) = -Aωsin(ωt + φ)

v(t)/x(t) = -ωtan(ωt + φ)

If we let RIGHT be the positive direction and evaluate at t = 0

-0.100 / 0.20 = -2√5tan(φ)

φ = arctan(-0.5/-2√5) = 0.111341...≈ 0.111 rad

0.20 = Acos(0.111)

A = 0.201246...≈ 0.201 m

x(t) = 0.201cos(2√5t + 0.111)

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over