Simple Harmonic Motion Question?
At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.9 h. How long does it take for the water to fall a distance 0.330d from its highest level?
When I solved this, I got 2.15 but that is wrong.
- Anonymous2 months ago
0.33*2 = 0.66
Θ = arccos 0.66 ≅ 48.7°
t = 12.9/4*48.7/90 = 1.745 h
- PhilomelLv 72 months ago
The period is12.9 h. half is rise and half is fall = 6.45*.33=2.13h.
- nyphdinmdLv 72 months ago
You are given a period T = 12.9 hr and the fact that this is simple harmonic motion. So chose to represent the height of the water, y by
y = d*cos(2*pi*t/T) where t = time in hours. Note that at t = 0, the tide is at its highest point
Now the water falls a distance 0.33d so the height of the water at that time is y = d - 0.33d
y = 0.67 d = d*cos(2*pi*t/T) now take the arccos and solve for t
arccos(0.67) = 2*pi*t/T --> t = T*arccos(0.67)/(2*pi) = 1.72 hours
Remember to get arccos back in radians, not degrees