Question

(30 Points) A buoy oscillates in simple harmonic motion according to the motion of waves at sea. An observer notes that the buoy moves a total of 6 feet from its lowest point to its highest point. The frequency of the buoy is ¼ (it completes ¼ of a cycle every second).

a. What is the amplitude of the function that models the buoy's motion?



b. What is the period of the function that models the buoy's motion?



c. Using cosine, write a possible equation to model the motion of the buoy

Explain your answers please

Answer 1

c.
`\displaystyle y = 3cos\:(\pi)/(2)x`

b.
`\displaystyle 4`

a.
`\displaystyle 3`

Step-by-step explanation:

`\displaystyle \boxed{y = 3sin\:((\pi)/(2)x + (\pi)/(2))} \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{-1} \hookrightarrow (-(\pi)/(2))/((\pi)/(2)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{4} \hookrightarrow (2)/((\pi)/(2))\pi \\ Amplitude \hookrightarrow 3`

OR

`\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{4} \hookrightarrow (2)/((\pi)/(2))\pi \\ Amplitude \hookrightarrow 3`

Keep in mind that although you are told write a cosine equation, if you plan on writing your equation as a function of sine, 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
`\displaystyle y = 3sin\:(\pi)/(2)x,` in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine 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 sine graph [photograph on the right] is shifted
`\displaystyle 1\:unit`to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD
`\displaystyle 1\:unit,`which means the C-term will be negative, and by perfourming your calculations, you will arrive at
`\displaystyle \boxed{-1} = (-(\pi)/(2))/((\pi)/(2)).`So, the sine graph of the cosine graph, accourding to the horisontal shift, is
`\displaystyle y = 3sin\:((\pi)/(2)x + (\pi)/(2)).`Now, with all that being said, in this case, sinse you ONLY have a wourd problem to wourk with, you MUST use the above formula for how to calculate the period. Onse you figure this out, the rest should be simple. 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
`\displaystyle y = 0,`in which each crest is extended three units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

b. To find the period [units of time], simply take the multiplicative inverce of the frequency, or use the formula below:

`\displaystyle F^(-1) = T`

a. To find the amplitude, simply split the height in half.

I am delighted to assist you at any time.

Answer 2

Answer: a) A = 3

b) P = 4

c)
`\bold{3\ cos(\pi)/(2)x}`

Step-by-step explanation:

`Amplitude (A) = (max-min)/(2)\\\\\\.\qquad \qquad \qquad =(6)/(2)\\\\\\.\qquad \qquad \qquad =\boxed{3}`

Period (P) is the reciprocal of the frequency (f)

f =
`(1)/(4)` → P =
`(4)/(1)\quad =\boxed{4}`

y = A cos (Bx - C) + D

• A = 3
• 
  `P=(2\pi)/(B)` →
  `4=(2\pi)/(B)` → 4B = 2π →
  `B=(2\pi)/(4)` →
  `B=(\pi)/(2)`
• C = none given
• D = none given

`\large{\boxed{y=3\ cos(\pi)/(2)x}}`