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Ferris Wheel - Trigonometric Function

  1. Oct 25, 2008 #1
    Pendulum Ride - Trigonometric Function

    1. The problem statement, all variables and given/known data
    At Canada's Wonderland, a thrill seeker can ride the Xtreme Skyflyer. This is essentially a large pendulum of which the rider is the bob. The height of the rider is given for various times:

    Time(s) 0 1 2 3 4 5 6 7 8 9
    Height(m) 55 53 46 36 25 14 7 5 8 15

    Find the amplitude, period, vertical translation, and phase shift for this function. [Note: that the table does not follow the bob through one complete cycle, s

    2. Relevant equations
    y=a sin [b(x-c)] + d

    3. The attempt at a solution

    To find "a" (amplitude)= (max - min) / 2 = (55 - 5)/2 = 25
    To find "d" (axis of symmetry)= (max + min) / 2 = (55 + 5)/2 = 30

    To find "b" find the Period

    Period = 2p/absolute value of b

    However, because this ride works as a pendulum, one cycle will be completed when there are 2 highs and 2 lows. So the bob starts at a height of 55 metres; it will then reach a low of 5 metres; it will (hypothetically) reach a height until it runs out of speed; it will then (hypothetically) return to the minimum height of 5 metres; and then, finally it will return to its start position. That is the completion of 1 cycle in a pendulum.

    Because the graph is incomplete, we have just one maximum and one minimum. The maximum starts at 55 metres, and then there is a minimum at 5 metres. Therefore, we have only completed 1/4 of the cycle at 7 seconds. Roughly, a complete cycle will take 28 seconds.

    Therefore Period= 2pie/b

    which becomes:
    28 seconds = 2pie/b

    So the "b" value is 12.86.

    To find the value of "c", I will plug in a co-ordinate value into the equation. Let us take the co-ordinate (3, 36)

    y=asin[b(x-c)]+d; becomes:


    And so, my final equation reads as:


    However, when I enter this value into my graphing software, it looks nothing like the graph I did on paper! Do you guys see any errors in what I did?

    Thank you so much in advance.
    Last edited: Oct 25, 2008
  2. jcsd
  3. Oct 25, 2008 #2


    Staff: Mentor

    You have replaced 2*pi with 360. While it's true that 2*pi radians is the same angle as 360 degrees, it's not true that 2*pi = 360.
  4. Oct 25, 2008 #3
    I don't understand. My teacher said that The period of the graphs on transformed sine and cosine functions can be found by the following formula: 2pi/|b|.

    As well, whenever I did my previous questions using 360/|b| I got the correct answer.:confused:
  5. Oct 27, 2008 #4


    Staff: Mentor

    If you're doing calculations with a calculator in degree mode, that will work. It won't work if the calculator is in radian mode.

    Just think about it: pi is about 6.28, which is nowhere near 360, But pi radians is the same angle measure as 360 degrees.
  6. Oct 27, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Typo alert: 2 pi radians is the same angle measure as 360 degrees.
  7. Oct 27, 2008 #6


    Staff: Mentor

    Thanks, Halls. I actually had a factor of 2 in there momentarily and took it out.
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