Answer:
1) You are in the presence of an undamped oscillation, therefore you can simulate this system as a simple harmonic motion (SHM) system. The equations used in this case are:
X= X₀ + A·cos(ω·t+Ф)
V=-Aω·sin(ω·t+k·2π)
a=-Aω²·cos(ω·t+k·2π) ⇒ = m·a
ΔU=m·g·ΔX ⇒ W = -ΔU (if is an SHM system, all forces are conservative).
2) If t₀ = 0s is 12:00 am (maximum value of X) then:
A=(Xmax - X min)/2 = 0.5m and X₀=Xmax - A = 0.8m
But Max( cos(α)) is obtained with α = 0 or 2π, Therefore 0 = ω·0s+Ф ⇒Ф=0
3) You can obtain ω as ω = 2π/T remember (T=43,200s or 12 hours which is a completed cycle of 2π).
ω =1.454· s⁻¹
4) In SHM, a=-Xω² and = m·a = -m·Xω² = - K·X Therefore:
K= m·ω² = 8.4616 N/m
5) Maximum velocity is Vmax = Aω = 7.272· m/s
Maximum acceleration of the buoy is a = Aω² = 1.058· m/s²
6) The total energy of the buoy due to tidal displacement is constant (due this is an undamped oscillation), therefore:
E = K + U = (calculated in the maximum tidal point, where V = 0m/s) Umax = m·g·Xmax=510.12J
7) The work W done by the tidal force from low tide to high tide:
W = -ΔU = -m·g·ΔX = 392.4J
The Power P for this work is:
P=W/t=392.4J/6h=65.4J/h=0.018W
the buoy requires 3577 times less power for its movement than a light bulb at the same time of usage.