Amplitude of a wave hitting a boat

[Redbelly98]

No.

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

 

[itryphysics]

avg = 3.3 m

 

[Redbelly98]

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)?

 

[itryphysics]

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

so the climb will be 7.5 m?

 

[Redbelly98]

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

 

[itryphysics]

How do I know what the amplitudes are?

 

[Redbelly98]

You don't. That's why I called it "A". It is unknown until you find it and solve the problem.

 

[itryphysics]

does the solution involve the quadratic equation by any chance?

 

[Redbelly98]

No.

Refering back to my post #33, let's try to imagine/picture this another way.

Suppose there is a vertical spring or elastic cord stretched out, from the deck of the shorter boat up to the deck of the taller boat. Normally, the spring is 3.3 m long (the distance between the 2 decks, with no waves present).

Now, take the top end of that spring and move it by a distance A upward. At the same time, move the bottom end of the spring downward by the same distance A.

By how much has the spring's length been increased?

 

[itryphysics]

increases by 2A

 

[Redbelly98]

Yes, good. The spring's length has increased by 2A, from 3.3m to 3.3m+2A.

If you equate that last expression with the maximum climb given in the problem statement, you can find the amplitude A.

 

[itryphysics]

3.3 m + 2A = 4.5 m

A= 0.6 m

Thank you for all your help!

 

[Redbelly98]

Looks good. You're welcome.