This graph looks like a trigonometric function. Which trigonometric function would be best to use for this model: cosine, sine, or tangent? Why?

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a. It’s time to start making your model fit the path of the mass. Start with amplitude. What is the amplitude of the weight’s motion? How can you modify the base trigonometric function you identified in part b to fit this amplitude? Do this now.

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b. What is the period of the weight’s motion? How can you modify the trigonometric function you identified in part c to fit this period? Do this now.

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c. Check how your model looks compared with the mass by graphing your model with the Edmentum Graphing Tool. Looking at the graph, what is different between your model and the graph of the mass? What needs to change to make your model more accurate? Make this change to your model, and type in the new equation for your model.

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[tex]\displaystyle \boxed{y = \frac{6}{25}cos\:(1\frac{1}{2}\pi{x} - \frac{\pi}{2})} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{1}{3}} \hookrightarrow \frac{\frac{\pi}{2}}{1\frac{1}{2}\pi} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{3}} \hookrightarrow \frac{2}{1\frac{1}{2}\pi}\pi \\ Amplitude \hookrightarrow \frac{6}{25}[/tex]

OR

[tex]\displaystyle \boxed{y = \frac{6}{25}sin\:1\frac{1}{2}\pi{x}} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{3}} \hookrightarrow \frac{2}{1\frac{1}{2}\pi}\pi \\ Amplitude \hookrightarrow \frac{6}{25}[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the sine graph, if you plan on writing your equation as a function of cosine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = \frac{6}{25}cos\:1\frac{1}{2}\pi{x},[/tex] in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the sine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the cosine graph [photograph on the right] is approximately shifted [tex]\displaystyle \frac{1}{3}\:unit[/tex]to the left, which means that in order to match the sine graph [photograph on the left], we need to shift the graph FORWARD [tex]\displaystyle \frac{1}{3}\:unit,[/tex]which means the C-term will be positive, and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{\frac{1}{3}} = \frac{\frac{\pi}{2}}{1\frac{1}{2}\pi}.[/tex]So, the cosine graph of the sine graph, accourding to the horisontal shift, is [tex]\displaystyle y = \frac{6}{25}cos\:(1\frac{1}{2}\pi{x} - \frac{\pi}{2}).[/tex]Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [-2\frac{2}{3}, 0],[/tex]from there to [tex]\displaystyle [-1\frac{1}{3}, 0],[/tex]they are obviously [tex]\displaystyle 1\frac{1}{3}\:unit[/tex]apart, telling you that the period of the graph is [tex]\displaystyle 1\frac{1}{3}.[/tex]Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = 0,[/tex]in which each crest is extended six-twenty-fifths unit beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**Now, it really does not matter which function you use because the result will always be the same, but IF the system only wants one function as an answer, then it would be the sine function sinse it intersects the origin. Other than that, it does not matter which one is chosen.

I am delighted to assist you at any time.

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1
2
3
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21
215