+1995 If someone is willing to help me I would greatly appreciate it .
Question 1 A researcher observes and records the height of a weight moving up and down on the end of a spring. At the beginning of the observation the weight was at its highest point. From its resting position, it takes 20 seconds for the weight to reach its highest position, fall to its lowest position, and return to its resting position. The difference between the lowest and the highest points is 6 in. Assume the resting position is at y = 0.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.
Question 2
At an ocean depth of 8 meters, a buoy bobs up and then down 5 meters from the ocean's depth. Sixteen seconds pass from the time the buoy is at its highest point to when it is at its lowest point. Assume at x = 0 the buoy is at normal ocean depth.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.
+37146 Period = 20 seconds (given) Frequency = 1/period = 1/20
Starts at its highest point = cosine function
resting point = x axis (given)
Amplitude = 6/2 = 3 (1/2 of the height of the 'wave')
3 cos x has period of 2pi to get period of 20 : 2pi/? = 20 then ? = 2pi/20 =.1 pi
3 cos (1/10 * pi * x)
Sorry....I see Chris has pointed out that you may need to use the sine function....we will need to shift the curve 1/4 period to the left
20 sec * 1/4 = 5
3 sin (1/10 pi (x+5))
+129852 Thanks, EP
Note Jenny : If you have to use the sine function, here's a possibility :
https://web2.0calc.com/api/ssl-img-proxy?src=%2Fimg%2Fupload%2Fc3c1e3de963ed0ca9b1000f1%2Fgraph.png
+37146 Here is a graph (with sin function) : object starts at its highest position at t= 0 takes 20 sec for full cycle
+129852 For the second one, the first sentence is a little confusing.....I'm assuming that the amplitude is 5 meters...and the midline is 8 metrers below the water's surface....thus the highest point of the buoy is 3 meters below the surface of the water and the low point is 13 meters below the water's surface
Here's the graph : https://www.desmos.com/calculator/6umbpfydcn
+37146 Number 2 (using sine function)
I think the amplitude is up 5 and down 5 from the midline Amplitude = 5 (Chris has a different interpretation)
5 sin ?x period is 32 seconds 2pi/32 = 1/16 pi
5 sin( 1/16 pi x) and it is shifted UP 8 meters
5 sin (1/16 pi x) + 8 SHould fit the bill !
+1995 I cannot see Chris's graph , is it the same answer but different interperation as you had said ?
+1995 Hmm im not sure myself . Both seem as if they could possibly work but there must be a word that differentes it to being the answer. I will try reading it over using both of yours explanations if you guys end up finding which one is correct could you guys let me know?
+37146 Yah, the wording of the question is a bit unclear...and we both interpreted/read it differently....sorry for the confusion! 
+37146 I read : Ocean depth 8 Up and down 5 from ocean depth (8) which would mean it is bobbing up to 13 and down to 3....
+1995 would 8 meters be the midline ? since from 8 meters (0,8) the buoy bobs up and down 5 from the midline?
+1995 oh okay so the difference between your guys is whether the midline is posititve 8 or negatitive ? Hmmm okay .
What does the x=0 , could that possibly give a hint ?
+129852 I'd go with EP's.....I think the buoy is floating on the water's surface......and the depth of the water = 8ft
So....the buoy would rise to a height of 13 ft above the ocean bottom and fall to a point 3 ft from the ocean's bottom
This makes more sense than mine.....
+1995 That is what I am thinking it makes a bit more sense also just because it is a "buoy" , lets say if the ocean was 8 meters like how it says ( 8 meters depth" it wouldnt be able to go down any further since that is the "depth"( how far down it does) if that makese sense.
+1995 I think we did haha, thank you both for your help! 
i apprectiate it a lot
