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    This is supposed to be a physics question, But maybe the Mathematicians could help out.

    The height of the water on a beach can be approximated as simple harmonic motion with a period of 12hours. If the mean water height is 3.5m, the amplitude of the tide is 1.6m, and `high water' occurs at 7am one day, what would you predict the height of the water to be at 11am?

    I used the equation, x=Acos(w)t But couldnt get the Answer
    w is angular frequency
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    Your equation should include a constant offset for the (given) mean water height. If that doesn’t fix it please post your working.
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    Can u post ur working out? Its the easiest way to see where u made a mistake so we can work to fix it.
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    Is t not supposed to be inside the bracket next to w?
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    (Original post by Shaanv)
    Can u post ur working out? Its the easiest way to see where u made a mistake so we can work to fix it.
    Sorry for late reply, i hadnt seen a notification so i thought no one replied to the post.

    my working:

    Amplitude =1.6
    Time Period=12hours*3600(to convert to seconds)
    time elapsed=4 hours

    x=Acos(2pi/T)t
    x=1.6cos(2pi/(12*3600)* (4*3600)
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    (Original post by H3n)
    Is t not supposed to be inside the bracket next to w?
    I dont think so, but ill try that
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    (Original post by old_engineer)
    Your equation should include a constant offset for the (given) mean water height. If that doesn’t fix it please post your working.
    im not sure what you mean by constant offset
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    (Original post by joyoustele)
    I dont think so, but ill try that
    It is supposed to be inside the bracket

    Edit: This definitely belongs in Physics, not Maths! I’ve moved it across.
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    Omg lol OCR A question? I did that one yesterday 😂😂😂
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    (Original post by Protostar)
    It is supposed to be inside the bracket

    Edit: This definitely belongs in Physics, not Maths! I’ve moved it across.
    I tried it, it didnt work
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    (Original post by joyoustele)
    I tried it, it didnt work
    Your calculator is in radians right? I haven’t actually tried the question myself but surely it must work...
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    (Original post by joyoustele)
    I tried it, it didnt work
    U can keep everything in hours as thats what all the relevant measurements are going to be in. The equilibrium position is at 3.5m.

    So if u imagine a graph of Acos(wt) in this situation it would not suffice as the equilibrium point is 0 not 3.5, so ur equation should be along the lines of Acos(wt) + 3.5
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    (Original post by Protostar)
    Your calculator is in radians right? I haven’t actually tried the question myself but surely it must work...
    It is in radians. Have you seen my working? is that correct?
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    (Original post by Shaanv)
    U can keep everything in hours as thats what all the relevant measurements are going to be in. The equilibrium position is at 3.5m.

    So if u imagine a graph of Acos(wt) in this situation it would not suffice as the equilibrium point is 0 not 3.5, so ur equation should be along the lines of Acos(wt) + 3.5
    Oh wow, It worked Thanks so much.
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    (Original post by joyoustele)
    It is in radians. Have you seen my working? is that correct?
    I think that you need to take the 3.5m into account - this is effectively the equilibrium position and so the amplitude will be 1.6m above the 3.5m. So you’d have to add this on somewhere?
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    (Original post by joyoustele)
    Sorry for late reply, i hadnt seen a notification so i thought no one replied to the post.

    my working:

    Amplitude =1.6
    Time Period=12hours*3600(to convert to seconds)
    time elapsed=4 hours

    x=Acos(2pi/T)t
    x=1.6cos(2pi/(12*3600)* (4*3600)
    I know this sounds dumb but wondering how you got t=4 (i can see it is from 11-7) but dont know why it isnt just 11?
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    (Original post by Kyrpt)
    I know this sounds dumb but wondering how you got t=4 (i can see it is from 11-7) but dont know why it isnt just 11?
    BTW im no expert, but the simple harmonic motion graph is displacement (Y axis) against time (x axis). Because I started when the tide had maximum amplitude at 7am (cos graph) t=0 at 7 am. And therefore at 11am, t has to = 4hours
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    (Original post by joyoustele)
    BTW im no expert, but the simple harmonic motion graph is displacement (Y axis) against time (x axis). Because I started when the tide had maximum amplitude at 7am (cos graph) t=0 at 7 am. And therefore at 11am, t has to = 4hours
    ahhh thankyou
 
 
 
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