Sinusoidal Functions: Niagara Falls Skywheel....

In summary, the homework statement is:-The highest point the rider will reach is 2.5 m + 50.5 m = 53 m-The lowest point on the Ferries Wheel is 2.5 m and the amplitude is 25.25 m-The Ferries Wheel completes a full cycle every 120 seconds and starts at the lowest point
  • #1
Evangeline101
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5

Homework Statement


upload_2016-8-5_20-58-5.png


Homework Equations


upload_2016-8-5_20-59-35.png


The Attempt at a Solution



a) Here is a sketch of the graph. The lowest point on the Ferries Wheel is 2.5 m and the highest point is 2.5 m + 50.5 m = 53 m. It completes a full cycle every 120 seconds and starts at the lowest point.

upload_2016-8-5_23-13-44.png


b) The highest point the rider will reach is 2.5 m + 50.5 m = 53 m

c) Okay so here is where my problem starts:

amplitude: 25.25 m a = 25.25

Period = 120 seconds k = 360/120 = 3

Vertical shift =
53 + 2.5/2= 27.75 c = 27.75

Phase shift = This is the part of the equation I am having trouble with. Can someone explain how to identify the phase shift for this equation, if there even is one?

Okay, so far the equation looks like this:

y= 25.25 sin (3(x - ____ )) + 27.75

( I am assuming the equation is sine, I am not sure if this is correct, it would be greatly appreciated if someone could explain how to identify a sine or cosine graph and let me know if this one is sine or cosine)

d) e) I haven't worked on these two since I have yet to complete the previous part.

Also, I have searched the internet for other explanations on this question, and they all explain using radians and pi ( to find k), I have not learned this method in my course, and think I should stick to the way I have been taught while answering this question. Thanks :)
 
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  • #2
MAJOR EDIT: Forget everything I previously wrote. I have to accommodate your way of doing things.
You have y = 25.25sin(3(x-__)) + 27.75

That looks great? x is time. So at time x=0, I assume the chair is at its lowest height. So somehow, y has to be equal to 2.5 m at that time. If you have a 0 offset, and x=0, then sin(0) will equal 0, which would make y=27.75. That isn't right so it tells us that we need some kind of an offset. So the portion of the equation 25.25sin(3(x-__)) is going to have to be some negative value in order for y to be equal to 2.5 m. So it appears that if sin(3(x-__)) is equal to -1, that would make y = 2.5. So what value results in sin equal to -1. And since x = 0, we are left with trying to find what offset d results in sin(3d)=-1?
 
Last edited:
  • #3
Evangeline101 said:
Here is a sketch of the graph
Not bad, but it would be even better if you made the gradient correct at the endpoints. E.g., suppose you were to continue it through another cycle by making a copy for the 120 to 240 second period. As you have sketched it, youwould have a sharp V at the join.
Evangeline101 said:
53 + 2.5/2
It would be easier to follow you working if you were to use parentheses correctly. What you have written is ##53+\frac{2.5}2##.
Evangeline101 said:
k = 360/120
i see you are taking the phase angle to be in degrees. Perhaps you have been taught that way, but it is unusual, and not a good idea. Generally one assumes that the argument to a trig function is in radians.
Evangeline101 said:
Phase shift = This is the part of the equation I am having trouble with.
The question as stated does not specify the height at time 0, so you can take it to be where you like. But for consistency, you should probably try to match your sketch.
With hindsight, it might have been simpler to choose a phase shift of zero then draw the graph accordingly.
It can be a bit tricky finding the shift that matches the graph. One way would be to sketch separately a graph of sin(x) then see how far you have to shift it and in what direction to match. If you add a small positive phase shift into your equation, which way will the graph shift?
 
  • #4
TomHart said:
So what value results in sin equal to -1. And since x = 0, we are left with trying to find what offset d results in sin(3d)=-1?

Ok so with some trial and error, I found that : sin(3(90) = -1

This is my new equation with d = 90

-25.25 sin(3(x-90) + 27.75

Is this an improvement to my last attempt at an equation?
 
  • #5
Evangeline101 said:
Ok so with some trial and error, I found that : sin(3(90) = -1

This is my new equation with d = 90

-25.25 sin(3(x-90) + 27.75

Is this an improvement to my last attempt at an equation?
Please make sure the parentheses match. I assume you mean sin(3(x-90)).
Does it give the right answer for x=0, 60, 120?

In my post #3 I suggested you should be using radians, not degrees. Have you been taught radians?
 
  • #6
So let's check out your equation. If I plug in some values for x (time), what do we get?
For x = 0 seconds: -25.25sin(3(0-90)) + 27.75 = 2.5 m (lowest point)
x = 30 seconds: -25.25sin(3(30-90)) + 27.75 = 27.75 m (mid-point)
x = 60 seconds: -25.25sin(3(60-90)) + 27.75 = 53.0 m (highest point)
x = 90 seconds: -25.25sin(3(90-90)) + 27.75 = 27.75 m (mid-point)
x = 120 seconds: -25.25sin(3(120-90)) + 27.75 = 2.5 m (back to lowest point)

So we see that it does indeed take 2 minutes to make one complete revolution. Looks like you got it right! :)

When I first responded to your post, I started working with radians then I changed back to degrees. Radians is really the most logical units for these kinds of problems. The conversion from degrees to radians is 360° = 2π radians. But if you haven't covered that yet, then it probably makes sense to continue with what they are teaching you - degrees.

Also, like Haruspex pointed out, be careful that your parentheses match.
You wrote: sin(3(x-90). You are missing a closing parentheses. It should be sin(3(x-90)).
 
  • #7
haruspex said:
Please make sure the parentheses match. I assume you mean sin(3(x-90)).

Okay :thumbup:

haruspex said:
In my post #3 I suggested you should be using radians, not degrees. Have you been taught radians?

No unfortunately I have not been taught this method, although I researched briefly about using radians on my own, but I should probably stick to degrees since that is what the course material has taught me :/

Here is a section from my lesson:
upload_2016-8-10_23-54-56.png
 

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  • #8
Evangeline101 said:
Okay :thumbup:
No unfortunately I have not been taught this method, although I researched briefly about using radians on my own, but I should probably stick to degrees since that is what the course material has taught me :/

Here is a section from my lesson:
View attachment 104516
Ok, thanks for clarifying.
 
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  • #9
TomHart said:
So let's check out your equation. If I plug in some values for x (time), what do we get?
For x = 0 seconds: -25.25sin(3(0-90)) + 27.75 = 2.5 m (lowest point)
x = 30 seconds: -25.25sin(3(30-90)) + 27.75 = 27.75 m (mid-point)
x = 60 seconds: -25.25sin(3(60-90)) + 27.75 = 53.0 m (highest point)
x = 90 seconds: -25.25sin(3(90-90)) + 27.75 = 27.75 m (mid-point)
x = 120 seconds: -25.25sin(3(120-90)) + 27.75 = 2.5 m (back to lowest point)

So we see that it does indeed take 2 minutes to make one complete revolution. Looks like you got it right! :)

Is my graph consistent or accurate with the equation? should I redo it?

Ok since I have an equation now, I went on to attempt d) and e):

d) y= -25.25 sin(3(x-90)) + 27.75

y= -25.25 sin(3(10-90)) + 27.75

y = 5.88 m

The height of the rider after 10 seconds is 5.9 m.

e) Total guess on this part:

If the SkyWheel rotates 1.5 revolutions every 1 minute, it takes less time to complete one cycle and a half, so the graph compresses horizontally. The period is 60 (1 minute = 60 seconds). The equation is y= -25.25 sin(6(x-90)) + 27.75
 
  • #10
Evangeline101 said:
Is my graph consistent or accurate with the equation?
What do you think? Does it match the graph at all the points Tom listed?
Evangeline101 said:
The height of the rider after 10 seconds is 5.9 m.
Yes.
Evangeline101 said:
The period is 60
No, it does 1.5 revolutions in one minute, not one revolution.
 
  • #11
haruspex said:
What do you think? Does it match the graph at all the points Tom listed?

Okay, I re-did the graph to accurately match the points.

haruspex said:
No, it does 1.5 revolutions in one minute, not one revolution.

So the period for one revolution would be less than 60 seconds, could it possibly be 40?

Evangeline101 said:
If the SkyWheel rotates 1.5 revolutions every 1 minute, it takes less time to complete one cycle and a half, so the graph compresses horizontally.

Does this part look good?

Also, could y= 25.25 sin (3(x-30)) + 27.75 also be an accurate equation to model the height of the rider over time? it appears to match my sketch better.
 
  • #12
As I was writing this, I just saw your latest response. Yes, 40 seconds is right for the period. So if I plug in 0 seconds into your new equation, it looks like I get the minimum height of 2.5 m. Now if you plug 40 seconds into your new equation, does that also result in the minimum height?

-------------------------------------
Evangeline, I think you ought to just work the problem over from the start with the new period. It would be good practice. Just repeat the same process. And I haven't checked for sure, but you may have to come up with a new phase shift. And I think that after you come up with the new equation, you should make a new graph. Just plug in some times, such as x=5, 10, 15, 20 seconds . . . or something like that - and plot those individual data points on the graph. Then you can fill in the shape of the curve from that. And as Haruspex pointed out, the slopes at the minimum points on your original graph are not quite right. For sines and cosines, the slope at the maxes and mins is ALWAYS 0 (zero).

Had Haruspex not mentioned that it was 1.5 revolutions in one minute, I probably would have misread the problem to be 1.5 minutes per revolution.

You may already know this, but in case you don't, here is something that is useful in various situations:
If you have something like 1.5 revolutions per minute, which is a frequency, the period is simply the reciprocal of that.
So to make it easy for me, I would write it as 1.5 revolution/1 minute. So I just swap the top and bottom to get 1 minute/1.5 revolution.
Then divide 1 by 1.5 to get 0.667 minutes/revolution, which is the period. So for your situation, you would want to convert 0.667 minutes (or 2/3 of a minute) to seconds.
So 0.667 minutes/revolution = ? seconds/revolution

Another example: 50 miles per hour. For me, I would write that 50 miles/1 hour. Now swap top and bottom to give 1 hour/50 miles. Then divide 1 by 50 = 0.02.
So 50 mi/hr is equivalent to 0.02 hr/mi. It probably is more intuitive to convert hours to minutes since the hours is such a small number.
So how many minutes is 0.02 hours? 0.02 hours = (0.02 hr)(60 min/1 hr) = 1.2 minutes.
So 50 mi/hr is equivalent to 1.2 min/mile.

Another example: Donuts cost $5 for a dozen. This one is a little trickier, but I am going to write dollars and donuts as the units.
5 dollar/12 donuts is equivalent to 12 donuts/5 dollar. 12/5 = 2.4, so $5 for a dozen donuts is equivalent to 2.4 donuts/dollar.
In other words, you can buy 2.4 donuts for one dollar - that is, if they would let you.

Sorry for going on for so long.
 
  • #13
TomHart said:
As I was writing this, I just saw your latest response. Yes, 40 seconds is right for the period. So if I plug in 0 seconds into your new equation, it looks like I get the minimum height of 2.5 m. Now if you plug 40 seconds into your new equation, does that also result in the minimum height?

-------------------------------------
Evangeline, I think you ought to just work the problem over from the start with the new period. It would be good practice. Just repeat the same process. And I haven't checked for sure, but you may have to come up with a new phase shift. And I think that after you come up with the new equation, you should make a new graph. Just plug in some times, such as x=5, 10, 15, 20 seconds . . . or something like that - and plot those individual data points on the graph. Then you can fill in the shape of the curve from that. And as Haruspex pointed out, the slopes at the minimum points on your original graph are not quite right. For sines and cosines, the slope at the maxes and mins is ALWAYS 0 (zero).

Had Haruspex not mentioned that it was 1.5 revolutions in one minute, I probably would have misread the problem to be 1.5 minutes per revolution.

Okay let me clarify my answers to the question from part a) to e)

a) Here is a sketch of the graph. The lowest point on the Ferries wheel is 2.5 m and the highest point is 2.5 m + 50.5 m = 53 m. It completes a full cycle every 120 seconds and starts at the lowest point.
(I re-did the graph)
upload_2016-8-12_0-17-17.png


b) The highest point the rider will reach is 2.5 m + 50.5 m = 53 m.

c) Period = 120 seconds, amplitude is 25.25 m, vertical shift is 27.75 m, phase shift is 30 seconds to the right.

y= 25.25 sin(3(x-30)) + 27.75

d) y = 25.25 sin(3(x-30)) + 27.75
y= 25.25 sin(3(10-30)) + 27.75
y = 25.25 sin(-60) + 27.75
y = 5.88 m

The height of the rider after 10 seconds is 5.9 m.

e) If the wheel rotates and completes 1 cycle and a half every minute, it takes less time to complete one cycle, so the graph compresses horizontally. The period decreases and is 40 (60/1.5 = 40 seconds). The (changed) equation is
y= 25.25 sin (9(x-30)) + 27.75 These are my answers to the question in full, can you please look over it and explain where and why I should start over?
 
  • #14
Evangeline, sorry for the confusion. I started to write my last response, but before I finished, I saw your response come up. I separated what I wrote before and after I saw your response with a dashed line. My response above the dashed line is what I wrote AFTER I saw your response. So you could disregard a lot of what I wrote below the dashed line.

In part e), the multiplication factor '9' in your equation looks right - in other words, it completes one cycle in 40 seconds - but the phase shift doesn't seem to work. It looks like at time x=0, the rider is at the maximum height, at x=20, rider is at minimum, and at x=40, rider is back to maximum.

I think that once you fix that, it looks to be correct. Good job!
 
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  • #15
TomHart said:
In part e), the multiplication factor '9' in your equation looks right - in other words, it completes one cycle in 40 seconds - but the phase shift doesn't seem to work. It looks like at time x=0, the rider is at the maximum height, at x=20, rider is at minimum, and at x=40, rider is back to maximum.

Ok so for part e)

Here is the changed equation: (i fixed the phase shift)

y = 25.25 sin(9(x-10)) +27.75

At x=0 , y= 25.25 sin(9(0-10)) + 27.75 = 2.5 m
At x= 20, y= 25.25 sin(9(20-10)) + 27.75 = 53 m
At x = 40, y= 25.25 sin(9(40-10)) + 27.75 = 2.5 m

At x= 0, the rider is at the lowest point, at x=20, the rider is at the highest point, and at x = 40, the rider is back to the lowest point ( one cycle has occurred)

Does this look right?
 
  • #16
Looks right. Great job!
 
  • #17
TomHart said:
Looks right. Great job!

Thanks (TomHart/haruspex) for being awesome and helping me figure out this question! I really appreciate it :biggrin:
 
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  • #18
I believe it's y=25.25sin3(x-30)+27.75. If you look at the midline at 27.75 and go from that value on the y axes to the curve that you plotted you can see that it's 30 units shifted to the right.
 
  • #19
It should be y=25.25sin(3(x-30))+27.75
 
  • #20
Denver said:
It should be y=25.25sin(3(x-30))+27.75
Is that for part c or part e? What you propose matches the answer for part c in post #13.
 
  • #21
That would be for part c .
 
  • #22
Denver said:
That would be for part c .
So in post #18, are you merely confirming the answer in post #13 or do you think you have found an error in the answer given there?
 
  • #23
I'm just confirming that #13 is correct.
 
  • #24
Denver said:
I'm just confirming that #13 is correct.
Ok. But the thread is five years old.
 
  • #25
yea so?
 
  • #26
Denver said:
yea so?
Just trying to understand why you made the post.
 

1. What is a sinusoidal function?

A sinusoidal function is a mathematical function that describes a periodic oscillation, or movement back and forth, resembling a sine wave. It is commonly used to model natural phenomena such as sound waves, light waves, and motion of objects.

2. How is a sinusoidal function related to the Niagara Falls Skywheel?

The Niagara Falls Skywheel is a large ferris wheel located in Niagara Falls, Canada. Its circular motion can be described by a sinusoidal function, as it moves up and down in a periodic motion.

3. What is the period of the Niagara Falls Skywheel's sinusoidal function?

The period of the Niagara Falls Skywheel's sinusoidal function is the time it takes for one complete revolution of the wheel, or the time it takes for it to return to its starting point. This can vary depending on the speed of the wheel, but on average it takes about 10-12 minutes for the Skywheel to complete one full rotation.

4. How is the amplitude of the sinusoidal function related to the height of the Niagara Falls Skywheel?

The amplitude of a sinusoidal function is the maximum distance from the center of the function to its peak or trough. In the case of the Niagara Falls Skywheel, the amplitude would represent the maximum height of the wheel. This can vary, but on average the Skywheel has an amplitude of 175 feet.

5. Can the Niagara Falls Skywheel's sinusoidal function be used to predict its future movements?

Yes, the sinusoidal function can be used to make predictions about the future movements of the Niagara Falls Skywheel. By analyzing the pattern of the function, we can determine when the wheel will reach its peak or trough, and at what time it will return to its starting point. This can be helpful for planning rides and managing wait times for visitors.

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