Ferris Wheel - Trigonometric Function

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Homework Help Overview

The discussion revolves around a problem involving the height of a rider on a pendulum ride, modeled using a trigonometric function. The original poster presents data points for height over time and seeks to determine parameters such as amplitude, period, vertical translation, and phase shift.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the amplitude and vertical translation based on maximum and minimum heights. They also explore the period and phase shift using a sine function model. Some participants question the correctness of substituting 2π with 360 in the context of period calculations, while others clarify the relationship between radians and degrees.

Discussion Status

Participants are actively engaging with the original poster's calculations, providing feedback on potential errors and clarifying concepts related to the period of trigonometric functions. There is an ongoing exploration of the implications of using different modes (degree vs. radian) in calculations, but no consensus has been reached regarding the original poster's approach.

Contextual Notes

The original poster's calculations are based on incomplete data for the pendulum's motion, which may affect the accuracy of their derived parameters. There is also a mention of prior successful calculations using a different method, which adds to the complexity of the discussion.

Sabellic
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Pendulum Ride - Trigonometric Function

Homework Statement


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

Homework Equations


y=a sin [b(x-c)] + d


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
28=360/b
b=360/28
b=12.86

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)

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

36=25sin[12.86(3-c)]+30
6=25sin[12.86(3-c)]
6/25=sin[12.86(3-c)]
13.89=12.86(3-c)
1.08=3-c
c=3-1.08
c=1.92

And so, my final equation reads as:

y=25sin[12.86(x-1.92)]+30

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:
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Therefore Period= 2pie/b

which becomes:
28 seconds = 2pie/b
28=360/b
b=360/28
b=12.86
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.
 
Mark44 said:
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.

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:
 
Sabellic said:
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:
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.
 
Mark44 said:
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.
Typo alert: 2 pi radians is the same angle measure as 360 degrees.
 
HallsofIvy said:
Typo alert: 2 pi radians is the same angle measure as 360 degrees.
Thanks, Halls. I actually had a factor of 2 in there momentarily and took it out.
 

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