Amplitude of a wave hitting a boat

In summary: The distance from the center of each boat is 2.1m-4.5m = 2.4 m How can that be twice the distance between them?In summary, the amplitude of the waves is twice the distance between the boats.
  • #1
itryphysics
114
0

Homework Statement


You are finding it a challenge to climb from one boat up onto a higher boat in heavy waves.
If the climb varies from 2.1 m to 4.5m , what is the amplitude of the wave? Assume the centers of the two boats are a half wavelength apart.



Homework Equations



just used simple arithmatic

The Attempt at a Solution


so since the starting point will be 2.1 m, i assumed that to get the amplitude i would only have to subtract that from the height of the higher boat. so 4.5 - 2.1 = 2.4
This answer is incorrect. Please explain why.
 
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  • #2


HINT: The boats are half a wavelength apart.
 
  • #3


so 2.4 divided by 2? :S
 
  • #4


itryphysics said:
so 2.4 divided by 2? :S
Not quite. The question states the difference in the height between the two boats varies from 2.1m to 4.5m. This is not the same as the positions of the boats. In other words, the maximum vertical distance between the two boats is 4.5m and the minimum is 2.1m.

Do you follow?
 
  • #5


hmmm... so if 2.1 is the minimum does that mean from the bottom of one boat to the bottom of the other? and then maximum is from the bottom of the first boat to the top of the second boat?

actually I am not quite sure I follow
 
  • #6


itryphysics said:
hmmm... so if 2.1 is the minimum does that mean from the bottom of one boat to the bottom of the other? and then maximum is from the bottom of the first boat to the top of the second boat?

actually I am not quite sure I follow
No. As the question says, these distances are measured to the centre of each boat. The vertical distance between the boats depends on their position of the wave.

To illustrate this, sketch a sine wave and draw two dots on the wave separated by half a wavelength. Now shift both points the same distance to the right along the wave and draw two new points. Now measure the vertical distance between the first two dots and then do the same for the second two dots. Compare the vertical distances between the two sets of dots.

Do you see?
 
  • #7


The vertical distance between the two points in each set remains the same. right?
 
  • #8


itryphysics said:
The vertical distance between the two points in each set remains the same. right?
No, not in general.
 
  • #9


well the points swap positions. I am not sure what you are getting at. can you please give another example or perhaps elaborate more on this one
 
  • #10


itryphysics said:
well the points swap positions. I am not sure what you are getting at. can you please give another example or perhaps elaborate more on this one
Okay, more specific.

  1. Plot y=sin(x)
  2. Plot the points (x,y)=(0,0) and (x,y)=(pi,0) - this is group A
  3. Plot the points (x,y)=(pi/2,1) and (x,y)=(3pi/2, -1) - this is group B
  4. What is the vertical distance between the points in group A?
  5. What is the vertical distance between the points in group B?

Do you see now?
 
  • #11


vertical distance b/w group A is zero and b/w group B is 2
 
  • #12


itryphysics said:
vertical distance b/w group A is zero and b/w group B is 2
Correct. So even though the points in both groups are separated by half a wavelength (pi) their vertical distances are not necessarily equal - they depend on the position of the two points, not just their relative position.

Does that make sense now?
 
  • #13


yes I understand now. How do I go about figuring out the position of the two points?
 
  • #14


itryphysics said:
yes I understand now. How do I go about figuring out the position of the two points?
Again think of the sine curve. At which two points on the curve is the vertical distance between these point maximum and what value does it correspond to?
 
  • #15


max distance would be (pi/2, 1) (3pi/2,-1)
so a vertical distance of 2
 
  • #16


itryphysics said:
max distance would be (pi/2, 1) (3pi/2,-1)
so a vertical distance of 2
Correct. So how does this value of 2 correspond to the amplitude?
 
  • #17


oh so the amplitude will be twice the distance between them?
 
  • #18


itryphysics said:
oh so the amplitude will be twice the distance between them?
Not quite. What is the definition of amplitude?
 
  • #19


amplitude: The maximum height of a crest or depth of a trough, relative to the normal level.
 
  • #20


itryphysics said:
amplitude: The maximum height of a crest or depth of a trough, relative to the normal level.
And where is the normal level in this case?
 
  • #21


zero?
 
  • #22


itryphysics said:
zero?
If that were the case, then the minimum separation between the boats would be zero, as for the sine exercise we did earlier.
 
  • #23


Im a bit lost. so we established that the maximum vertical distance between the two boats is 2. The distance from the center of each boat is 2.1m-4.5m = 2.4 m How can that be?
 
  • #24


itryphysics said:
Im a bit lost. so we established that the maximum vertical distance between the two boats is 2.
No we didn't, that was just in the exercise that I gave you to illustrate that the vertical distance depends on the absolute position of each boat, not the relative position.

Edit: Hang on a minute, I made need to re-think our approach.
 
  • #25


so is the normal level 2.1 then?
 
  • #26


Hootenanny, were you able to sketch a different route?
 
  • #27


I'm thinking of a different approach to this problem.

First of all, I interpret "the climb" as the height difference between the decks of the two boats. The person has to climb from the deck of the lower boat up to the deck of the higher boat.

Next, think: what would the climb be if there were no waves present? Hint ... it is not zero. Use the information about what the climb is in the presence of the wave.
 
  • #28


hmm how can i determine the deck position of the 2nd boat since no dimensions are given? would the climb without the waves be (2.1+x) ? :S
 
  • #29


hmm how can i determine the deck position of the 2nd boat since no dimensions are given?

We don't need the actual position of either deck, we only need the difference between the two decks, which is the same as the climb.

would the climb without the waves be (2.1+x)

I don't know what "x" is in your expression.

The amount of climb in the absence of a wave would be the same as the average climb when the waves are present, and that is _____ meters?
 
  • #30


2.4 m
 
  • #31


No.

The climb is changing, but stays in the range between 2.1m and 4.5m. What would be the average?
 
  • #32


avg = 3.3 m
 
  • #33


Okay, good. With no wave present, the height difference (or the climb, as they put it) will be 3.3 m.

Now let's add the waves. The boats are 1/2 wavelength apart, so when the taller boat height is higher by A (the amplitude), the shorter boat is lower by A.

So take that 3.3m climb, and:
1. raise the tall boat by a distance A
2. lower the shorter boat by a distance A

What will the climb be then (in terms of A and the 3.3m)?
 
  • #34


tall boat will be 8.7 m
shorter boat will be 1.2 m

so the climb will be 7.5 m?
 
  • #35


Well, that would be impossible given that the climb is never more than 4.5 m maximum.
 

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