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Tides and Simple Harmonic Motion

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 

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


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 

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!   

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

Dr. Rick,

Thank you for your explanation.

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)?


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 

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!   
Associated Topics:
High School Physics/Chemistry

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