Thread: Simple Harmonic Motion
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Old October 11th, 2007, 05:42 AM
ticbol ticbol is offline
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Quote:
Originally Posted by steve@thecostins.co.uk View Post
In a small harbour, the depth of the water at high tide is 10m, and at low tide is exactly zero. The depth of water follows approximately simple harmonic motion with amplitude (A), 5m and time period (T), 12 hours. On one day, high tide is noon.

If d is the depth of the water above the mean depth (5m), write an equation of the form d = A \ {\cos {\omega}t}, substituting the appropriate values for A and {\omega}.

Where, t is time.

I already know A = 5 and {\omega} = 0.5 \ rad \ h^{-1} or {\omega} = 1.45 \ rad \ s^{-1} but i am not sure what the question is asking me to do. Does it want me to make a function for time so that when you input a time of day it will return an appropriate value? I am not sure, any ideas anyone?

Thanks
"one day, high tide is noon"
So if the curve starts at a maximum, the sinusoidal function to use is cosine.

The general cosine equation is
y = A*cos(Bx -C) +D
where
A = amplitude----------------------------5m here
B = frequency; such that Period = 2Pi /B
C = horizontal shift
D = vertical shift
shift = relocation of the neutral axis from the x- or y-axis.

Since depth at high tide is given as = 10m and at low tide 0m, then the x-axis is at 0m. The mean depth, the neutral axis is then 5m above the x-axis. So the vertical shift is +5m.

If we start the curve at 12noon, at its maximum, then the horizontal shift is zero, and x (t in this Problem) starts at 12noon. Or t(0) is 12noon.

So the equation is
d = 5cos(B*t -0) +5 ----------(i)

What is B? --------------------your "w"
period = 2pi/B
12hrs = 2pi/B
B = 2pi/12 = pi/6 cycle/hr

Hence,
d = 5cos*((pi/6)*t) +5
d = 5cos(pi*t / 6) +5 ------------------------the equation.

---------------------------------------------------
Check:

If the period is 12hrs, and the start is at 12noon where there is a high tide, then the nexy high tide is 12 hours from 12noon. That will be at 12midnight.
Then the low tide is at half of one period (in a cosine curve, the lowest point is at half of one period) which is at 6pm.

d = 5cos(pi*t /6) + 5

At 12noon, or t = 0,
d = 5cos(0) +5 = 5(1) +5 = 10m -----checks.

At 6pm, or t = 6 hrs,
d = 5cos(pi*6 /6) +5 = 5cos(pi) +5 = 5(-1) +5 = 0 -----checks.

At 12midnight, or t = 12 hrs,
d = 5cos(pi*12 /6) +5 = 5cos(2pi) +5 = 5(1) +5 = 10m -----checks.
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