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### Tides and Simple Harmonic Motion

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Date: 08/11/98 at 22:08:27
From: William Liao
Subject: Simple Harmonic Motion

Dr. Maths,

The question is as follows:

It is accepted that the rise and fall of the tide at a particular inlet
is simple harmonic, with the time difference between successive high
tides being 10 hours. The entrance to the inlet has a depth of 20m at
high tide and 8m at low tide. If the low tide occurs at 10:00 a.m. on a
certain day, find the earliest that a vessel requiring a minimum water
depth of 15 m can pass through the entrance.

I have identified that since T = 2pi/omega, and T = 10 hours = 36000
seconds in this case, then omega must equal to pi/18000. Also I have
found a, the amplitude, to be 6m. This is obtained by 20m - 8m = 12m,
but since this is its total displacement, the amplitude must be 6m as
it oscillates 6m up and down, which makes up the 12m. I also found
the desired displacement in this case to be -1, since low tide +
medium tide is 8 + 6 = 14m, but 15m means it has to oscillate backwards
for 1m, which is -1. I then used the equation for displacement,
x = a cos(omega*t), substituting all the value except for t (which is
what I have to find), I got -1 = 6 cos ((pi/18000)t).

However, the answer I have obtained is just simply not correct, and I
don't know why, or which step I have done is incorrect. The answer
given on the back of my textbook is 12:46 p.m., and I have tried to
work backwards to see what I did wrong, but that didn't seem to help
either.

Can you please point out what I did wrong?  Thank you.

Sincerely,
Will
```

```
Date: 08/12/98 at 09:05:43
From: Doctor Rick
Subject: Re: Simple Harmonic Motion

Hi, Will. Thanks for going through your thinking so clearly.

You haven't done anything wrong as far as you have stated it. I think
that all you need is to be a little more careful in understanding the
meaning of quantities. In particular, it would help to write an
equation.

We need to decide how to define the time. You chose to express time in
seconds, but I don't see anything about when time starts. Let's say
that time t is 0 at 10:00 A.M., the time of low tide.

Then the equation for water depth, h, at the entrance to the inlet
(measured in meters), is:

h = 14 - 6*cos(2pi * t / T)

where T is the period, 10 hours, or 36000 seconds. Do you see how I
got this? As you said, the mean tide is 14m, and the amplitude is 6m.
I chose a negative cosine so that the sinusoid will have its minimum
at t = 0 -- the low tide. And you understand the term inside the
cosine, omega * t.

Now if I set h = 15m and solve for t, I get:

t = 36000/(2pi) * arccos(-1/6)
= 5729.6 * 1.73824
= 9959.4 sec
= 2.7665 hours
= 2 hours + (0.7665 * 60) min
= 2 hours + 45.99 min

Finally, remember that t is the time after 10:00 A.M., so the time
you're after is 12:46 P.M.

In other words, I think the only thing you forgot is to add your
number of seconds to 10:00 A.M. By writing the equation carefully, and
spelling out at the beginning what t means, you can be sure you're
doing it right. I hope this helps.

- Doctor Rick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 08/12/98 at 09:32:14
From: William Liao
Subject: Re: Simple Harmonic Motion

Dr. Rick,

Your calculation seems perfectly correct to me. However, there is
still one thing I don't quite understand. You stated that
t = 36000/(2pi) * arccos(-1/6).  36000/(2pi) would be omega of course.
According to my textbook, T = 2pi/omega, and T equals to 36000 in this
case, so wouldn't 36000 = 2pi/omega, which means omega = 2pi/36000 =
pi/18000?  Why is it 36000/(2pi)?

Sincerely
Will
```

```
Date: 08/12/98 at 10:32:05
From: Doctor Rick
Subject: Re: Simple Harmonic Motion

Hi again. You are correct that omega = 2pi/T. This is why the equation
contains cos((2pi/T) * t). The coefficient of t inside the cosine is
omega.

Let's examine in more detail how I solved the equation for t:

h = 14 - 6*cos((2pi/T) * t)
h - 14 = -6*cos((2pi/T) * t)
14 - h = 6*cos((2pi/T) * t)
(14 - h)/6 = cos((2pi/T) * t)
arccos((14 - h)/6) = (2pi/T) * t
(T/2pi) * arccos((14 - h)/6) = t

You see that I had to _divide_ the inverse cosine by omega (2pi/T) in
the process of solving for t. I hope this makes it clearer.

- Doctor Rick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Physics/Chemistry

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