Question

Due to tides mean sea level off of Newport Beach reaches a height of 1.3 meters during high tide and 0.3 meters during low tide. Successive high tides occur every 12 hours (43,200 seconds). A buoy with mass m = 40 kg is floating in the ocean off of Newport Beach.1) Relevant concepts/equations. (5 points.)2) Assume we begin to measure the buoy’s displacement at High tide which occurs exactly at 12:00 am (0 seconds). Also assume we can model the buoy’s displacement as a simple undamped oscillation. What is the amplitude and phase angle for the buoy’s displacement? (10 points)3) During one half cycle of six hours (21600 seconds), the buoy’s displacement passes through an angle of 180 degrees. From this information, what is the angular frequency ω of the buoy? (5 points)4) Using your previous answer, what is the force constant ‘k’ acting on the buoy? (5 points)5) What is the maximum velocity of the buoy? What is the maximum acceleration of the buoy? (10 points)6) What is the energy of the buoy due to tidal displacement? (5 points)7) How much work is done during one low tide to high tide cycle? How much Power per hour is required to accomplish this? (Assume g= 9.81m/s^2 , compare your answer to a 65W light bulb which uses 65 watts per hour.)

Answer 1

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) ⇒ $F_{tide}$ = 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·$10^{-4}$ s⁻¹

4) In SHM, a=-Xω² and  $F_{tide}$ = m·a = -m·Xω² = - K·X Therefore:

K= m·ω² = 8.4616 $10^{-7}$ N/m

5) Maximum velocity is Vmax = Aω = 7.272·$10^{-5}$ m/s

Maximum acceleration of the buoy is a = Aω² = 1.058·$10^{-8}$ 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.