sin and cosine graphing word problem

[theone4u96]

The problem is as follows:
A buoy in the harbor of San Juan, Puerto Rico, bobs up and down. The distance between the highest and lowest point is 3 feet. It moves from its highest point down to its lowest point and back to its highest point every 8 seconds.
(a) Find the equation of the motion for the buoy assuming that it is at its equilibrium point at t=0 and the buoy is on its way down at that time.
(b) Determine the height of the buoy at 3 seconds.
(c) Determine the height of the buoy at 12 seconds.

This is what I have so far for problem A.
period = 8seconds/6 ft (the buoy goes 3 ft down and 3 ft back up)
so then 2pi/k = 8/6 ; when i cross multiply i get k=3pi/2.
Would the amplitude be 3 since (3 - -3)/2 equals 3?
my guess is that the equation is something like y= -3sin ((3pi/2)pheta)

can you please tell me if this is right or at least push me in the direction that is right?

 

[theone4u96]

Thank You! i'm like not smart enough to know how to message you a thank you yet so yea. well thnx galactus.

 

[galactus]

No, no. I made a mistake. That is why I deleted it. The problem says assume the buoy is at equilibrium and is on its way down. I had it on the way up.

I also had 3 feet up and 3 down. That is what I get for not reading it carefully enough.

 

[theone4u96]

yea. I figured you just overlooked that part of it. and i didnt want to be rude and correct you when you sooo kindly just helped me out. well thnx again.

 

[galactus]

OK. Since it is 3 feet between the highest and lowest point, then the amplitude is 3/2.

The period is \(\displaystyle \frac{2\pi}{|b|}\)

\(\displaystyle \frac{2\pi}{8}=\frac{\pi}{4}\)

But, the problem also says it starts at equilbibrium and is on the way down.

\(\displaystyle y=-\frac{3}{2}sin(\frac{{\pi}x}{4})\)

or

\(\displaystyle y=\frac{3}{2}cos(\frac{{\pi}x}{4}+\frac{\pi}{2})\)

 

[theone4u96]

right. my work may have confused you. sry. im the one who messed up.

 

[theone4u96]

um. i was talking to someone else who has the class with me. i'm not trying to correct you here. I just need more help. He claims that we are supposed to make use of the fact that a period is the reciprocal of the frequency. ( 1 pd = 1/frequency). so if you refer back to my original posting you'll see that i arrived at the answer of k equaling 3pi/2. (not the period, k). can you please explain to me in more detail how you arrived at your answer for the period of the equation. Since our formula for the equation is (A)sin(k)(pheta). we don't actually put the period into the equation. a period equals 2pi/k. Once again, correct me if I'm wrong.

 

[galactus]

Here is a graph of your \(\displaystyle -3sin(\frac{3{\pi}x}{2})\) in blue, and \(\displaystyle -\frac{3}{2}sin(\frac{{\pi}x}{4})\) in red.

Note, the model you came up with has a period of about 1.33 seconds.

\(\displaystyle \frac{-3}{2}sin(\frac{{\pi}x}{4})\) starts at equilibrium and is starting down, and has a period of 8 seconds.

The period is given by \(\displaystyle \frac{2\pi}{|b|}\). Since it completes one period in 8 seconds, we have \(\displaystyle \frac{2\pi}{8}=\frac{\pi}{4}\).

The reason it is 3/2, is because it is 3/2 from the x-axis to the top or bottom of the wave. Thus, 3/2+3/2=3. The problem states that it is 3 unitsTOTAL from the highest to lowest points.