Tuesday, May 29, 2007

Periodic Functions

Whoa!! I almost forgot I was the scribe for today. Anyways...here it is..In class , we worked on a periodic function problem as groups of 3 or 4. It took us a while to figure out this problem so we had spies from each group to see what the other groups had done. Here is a recap on the problem.

A tsunami(commonly called a "tidal wave" because its effect is rapid change in tide) is a fast-moving ocean wave caused by an underwater earthquake. The water first goes down form its normal level, then rises an equal distance above its normal level, and finally return ti its normal level. The period is about 15 minutes. Suppose that a tsunami with an amplitude of 10 metres approaches the pier at Honolulu, where the normal depth of the water is 9 metres.

For this problem we first need to construct a graph and fill in "DABC".

This is how we start our graph. It stated that the normal depth of the water if 9 metres therefore it is the average value of function(sinusoidal axis). It is where our graph will start. We then add 10 metres to the sinusoidal axis since the amplitude is 10. It tell us how far to go from the sinusoidal axis. This will give us the maximum(19 metres). We then find the minimum by subtracting 10 from 9(-1).



We can now construct the graph.

The graph first goes down since it is stated in the information that the water first goes down from its normal depth. This wave has an amplitude of 10(as stated earlier) therefore the wave goes down its minimum depth which is -1. It also says that it then goes back to its normal depth at the same equal time. Therefore we divide the whole period(which is 15 ) into 4 parts because it first goes down then up, then up again and finally down.



We can now find "DABC"

A= -10 ( amplitude. it is negative because the first movement of the wave is down)

B=2(pi)/15 ( determines the period , we place 15 because a period is every 15 mins)

C=0 (since the wave started at 0 minutes and at the sinusoidal axis, there is no horizontal shift)

D= 9 ( this tells us that the whole graph will move 9 metres up, it is the sinusoidal axis)

Now we can put these informations into an equation.


The amplitude
2 Pi(3.14) / 15 = o.42
The vertical shift\

A) Assuming that the depth of the wave varies sinusoidally with time as the tsunami passes, predict the depth of the water at the following times after the tsunami first reaches the pier.

i) 2 minutes ii) 4 minutes iii) 12 minutes

For this problem, we could just insert these times into the equation in place of x to determine the depth of the water. For example, this is the equation for 2 minutes:




We do the same for the rest, place the minutes in place of x.

B) According to your model, what will be the minimum depth of the water? How do you interpret this answer in terms of what will happen in the real world?


We can see the lowest depth of the water by looking at our graph. It is -1 since it is the lowest the wave can go. In the real world this is kind of confusing. How can the lowest depth be -1 , this would mean it digs under see bottom. Therefore we state it as , for lenght of time the wave is at its lowest, we can see the bottom.

C) The wavelenght of the water is the distance a crest of the wave travels in one period. It is also equal to the distance between two adjacent crests. If a tsunami travels at 800 kilometers per hour, what is the wavelenght?

For this question. it is stated that the the wave is traveling 800 km/h. In our graph, one period of the wave is 15 minutes. We found the answer for this question by cross multiply.




THE NEXT SCRIBE IS CHRIS A.!!!!

5 comments:

  1. THANK YOU!!!
    THIS IS TOTALLY WHAT I WAS LOOKING FOR!

    ReplyDelete
  2. I LOVE YOU! THIS IS AWESOME! I CAN JUST MARRY YOU!!!

    ReplyDelete
  3. THANK YOU SO MUCH!!!!

    ReplyDelete
  4. This was very helpful but I don't understand why 2pi/15 isn't in the equation if it's the period

    ReplyDelete