Calculating the amplitude of waves in water

[Josh0768]

Suppose I have a perfectly circular pool which is four meters in radius, two meters in depth, and filled with water. Say I drop a steel ball with a radius of five centimeters into the middle of the pool from a height of five meters above the water's surface. After three seconds, what will be the approximate height (relative to the surface of the water before it was perturbed) be of a point on the water's surface that is two meters horizontal from the center of the pool? Ignore any contribution to the volume of water made by the steel ball.

*Please note: original specifications were a pool depth of three meters, radius of five meters, and time elapsed of six seconds. These parameters have been changed to those you see now.*

 

[256bits]

@Josh0768
Interesting problem.
Not sure how to actually solve it in the real world, so I think we would have to go for some ideal situation.
What assumptions of waves produced would you think would have to apply for this scenario?
What can you say about the ball when it is 5m high?
What can you say about the water waves when the ball has entered into the pool?
Anything come to mind?

 

[Josh0768]

What assumptions of waves produced would you think would have to apply for this scenario?

I'll be honest, I have no idea. I haven't really studied waves since high school, and that was only a basic, 2D overview of EM waves. But still, I'll take a guess: maybe we assume the surface waves are something along the lines of frictionless, or they propagate without being dragged by the air beneath it and water below it? That's about as much as I can think of.

What can you say about the ball when it is 5m high?

Describe its potential energy maybe? In which case, mgh = density*volume*9.8*5 = 4.215 J

What can you say about the water waves when the ball has entered into the pool?

Calculating some sort of energy transfer is probably in order. It would be a matter of converting the kinetic energy of the ball into kinetic energy in the water - but not all of the ball's energy, since the drag through water will only decelerate it so much. The energy the ball loses then gets dispersed throughout the water, which oscillates in some fashion, and eventually all the energy is lost to friction and the pool returns to equilibrium. Unfortunately, I do not know how to quantify any of these processes lol. I hate to leave all the hard work up to someone else, but after watching ripples in my pool I wanted to know how these waves really worked. If the least anyone could do is tell me where to read up on the math and physics I need to learn in order to figure out this problem myself, I'd be grateful.

 

[256bits]

That is a pretty good start.
That is what I was thinking also.

Calculating some sort of energy transfer is probably in order

 

[anorlunda]

It is not an easy problem. Obviously, the waves will be larger in the middle, and gradually decrease the further you get away from the center. How fast they decay depends on properties of water like viscosity and temperature and initial state of motion.
I would say this is a case where calculation is much harder than experiment. Even if you did calculate, how would you have confidence in your answer without experiment. Are you sure that you want to calculate?

 

[Vanadium 50]

I don't think this is going to be practical. The question is essentially "the water in the pool has N modes of oscillation. A disturbance transfers energy to each of N modes, subject to a time-varying boundary condition (the entry of the ball). What is the summed amplitudes of these N modes over time"?

N is very large. The first step in solving this would be to count N. I don't think I even have a good way to start on that.

 

[Vanadium 50]

Looking at pictures of this phenomena, e.g.

I am more convinced that this will not be simple. These don't even look like sines and cosines. They might be Bessel functions.

 

[sophiecentaur]

It may be easier to start with a simpler 2D model with a narrow trough with parallel sides.
There are (quite) a few different factors involved, like the volume of the ball and its entry speed. As mentioned above, the Kinetic Energy of the ball will be shared between the wave generated and the stirring of the water as the ball passes under the surface. If you use a disc of the same mass, more energy will be matched into the wave because more surface water will be displaced on contact.
I remember spending hours at the sea side, as a boy, trying to make "dead man's drops" by spinning flat stones on a horizontal axis (the opposite of skimming stones) and getting a satisfying 'thunk' sound as they entered the water. Very little wave formation that way. At least a ball is a simple model to start with but some Hard Sums involved, even with that.

 

[shivaprasadvh]

Find the solution here.
This can give the amplitude of the wave when steel ball just touches the water surface. One last step is remaining in this. Just check if it can be good to go.

 

[Vanadium 50]

Apart from the impossible-to-read light-light-light blue on white color scheme that you chose, your units aren't even correct. Height is not measured in the square root of joules.

 

[shivaprasadvh]

Apart from the impossible-to-read light-light-light blue on white color scheme that you chose, your units aren't even correct. Height is not measured in the square root of joules.

Okie. My equation never says height is measured in the square root of joules. But I have told amplitude is proportional to square root of joules. Hope this is clear. If you have to say anything on solution kindly comment. Will be happy to learn.

Also I have mentioned, I have not arrived at solution but one or two steps to this relation can lead to solution.
Is distance proportional to speed? Does it mean you measure speed in meters or distance in m/s.

 

[shivaprasadvh]

View attachment 266577

Find the solution here.
This can give the amplitude of the wave when steel ball just touches the water surface. One last step is remaining in this. Just check if it can be good to go.

If anything not okie in this derivation / part solution pls suggest me.

 

[etotheipi]

I don't think it will be so simple. To me at least it would seem that the amplitude will depend on lots of different factors, for instance viscosity and surface tension - which are not part of your working. Also, the actual period of contact looks very complicated. And the resulting pattern of the water, as @Vanadium 50 alluded to in #7, doesn't look sinusoidal at all!

 

[Vanadium 50]

My equation never says height is measured in the square root of joules. But I have told amplitude is proportional to square root of joules.

We were asked about height.

After three seconds, what will be the approximate height

You aren't answering the question.

 

[shivaprasadvh]

We were asked about height.
You aren't answering the question.

Yeah true not a solution. But this can be extended to the solution. Want to know if the assumption i made and the thought process is right? If so can extend the solution.

 

[sophiecentaur]

Find the solution here.

If you want to be sure that people can actually read what you have written then you could very easily take that hardly-readable image and give it some help. Put the jpeg into your 'photo application, increase the contrast and take the gain so the the white is actually white and the writing is very dark grey. That's what the most basic photocopier does automatically if you tell it to do a black and white image; you are computer savvy so you could do it too and the result would be that more people would be bothered to read your posting.

 

[shivaprasadvh]

If you want to be sure that people can actually read what you have written then you could very easily take that hardly-readable image and give it some help. Put the jpeg into your 'photo application, increase the contrast and take the gain so the the white is actually white and the writing is very dark grey. That's what the most basic photocopier does automatically if you tell it to do a black and white image; you are computer savvy so you could do it too and the result would be that more people would be bothered to read your posting.

Sure, I will do this next time. Thanks...