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Applying a sinusoidal function to tidal patterns on the earth.

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data

    At the time of a full moon, the tides repeat with a period of about 12 hours, and the depth of water in a certain channel varies between 2 meters and 6 meters in a way that can be modeled by an equation of the form D(t) = A + Bsin(ct + d), where A is the average depth and B is the amplitude of the tide (difference from average to high).

    So;
    D(t) = depth of water in channel
    A = average depth = 4 meters
    B = amplitude = 2 meters
    c = unknown (at least to myself)
    d = (solved in #3, see below)

    The question then goes on to ask us to sketch a rough graph of this function, for t = 0 to 24 hours, and then says "suppose we know that the water level at 6:00 AM is 3 meters and rising". We are then asked to estimate, using the graph, when a barge or boat with a draft of 5 meters will be able to pass through a channel. The draft of a boat is the absolute distance from the water line to the bottom of the boat (how deep the boat goes in the water). THEN it asks what are the values of A, B, c and d?

    I managed to answer these three questions without any problems (see "attempt" below), except for finding "c".

    After we went over this in class, we were asked one last question: "Using the results from this question, determine EXACTLY when the barge is able to pass through.

    2. Relevant equations

    D(t) = A + Bsin(ct + d)

    3. The attempt at a solution

    After sketching my rough graph, I determined that the barge will be able to pass at approx. 9 AM till 12 PM (noon) and then again at 9 PM till 12 AM (midnight)

    Now, start out by finding the unknown values that I'm clearly given...A and B. We can solve d by plugging in what we have...you'll see that we don't need to solve for "c":

    D(t) = A + Bsin(ct + d)
    3 meters = 4 meters + (2 meters)(sin c(0) + d) << we know A and B, and t = 0, because this is the only point on the graph we are given. We let t = 0, which is basically time = 6:00 AM. So t = 0 = 6:00 AM.

    3 = 4 + 2(sin c(0) + d)
    -1/2 = sin( 0 + d)
    -1/2 = 0 + sin d
    d = sin-1 (-1/2)

    d = -.52398776...

    Today during my class we confirmed that the above calculations were true.

    Then the final question (the one on which I am stumped), I couldn't really get anywhere. I need to find EXACTLY when this barge can pass through the channel of water.

    What I did was:

    D(t) = A + Bsin(ct + d)
    plug in the depth and values of A, B, and d. Inequality because the barge will go through when the depth is greater than 5 meters.
    5 < 4 + 2sin(ct + sin-1 (-1/2))
    1/2 = sin(ct + sin-1 (-1/2))
    sin-1 (1/2) = ct + sin-1 (-1/2)

    But now we have two variables, c and t...but as I said earlier, I couldn't manage to find the variable "c". This was because we were given only one point on the graph, and this one point is the point where t = 0, so because t = 0, ct = 0 as well.

    Now, because I have two unknown variables, we are given only one complete point (at 6 AM, the depth is 3 meters), and the period of the function was given to us as "approximately 12 hours", it would seem like this question is impossible.

    Perhaps I've missed something. Sorry, my post was rather long, and perhaps confusing. I hope you guys are able to read it through and understand it.

    Thank you so much in advance.
     
  2. jcsd
  3. Jan 7, 2010 #2

    D H

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    The only information you have is that the period is about 12 hours. So assume it is exactly 12 hours.
     
  4. Jan 7, 2010 #3
    But even then, we don't know what the depth is at time = 12 hours, the period will only give us when the same depth occurs, and the only depth we know is at 6:00 AM, depth = 3 meters...
     
  5. Jan 7, 2010 #4

    D H

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    The period is twelve hours. That gives you the coefficient 'c'. The coefficient 'd' is determined by the fact that at 6:00 AM the level is 3 meters and rising. (You would get a different phase angle if the level at 6:00 AM was 3 meters and falling.)
     
  6. Jan 7, 2010 #5
    OH I didn't even see that, there ya go.

    Thanks so much for your help!
     
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