Amplitude of a circular ripple at distance r from source

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

Does anybody please know another way to solve this problem?

EDIT: Why do they assume that no energy is absorbed by the water?

Many thanks!

 

[nasu]

What is wrong with this one?

 

[ChiralSuperfields]

What is wrong with this one?

Thank you for your reply @nasu! I don't understand why they can assume that no energy is absorbed by the water. Do you please know why?

 

[nasu]

But the problem tells you to use energy.

 

[ChiralSuperfields]

But the problem tells you to use energy.

Thank you for your reply @nasu! I guess their statement could be interpreted as energy is not absorbed. I though the problem could be a bit clearer in that manner thought.

 

[nasu]

No, it means that the energy is spread over larger and larger perimeters (because is a 2D wave).
In 3D the energy of a point (or spherical) source is spread over larger and larger spherical surfaces and so the energy per unit area decereases as 1/r ²(inverse square law). As the energy is proprtional to amplitude squared, the energy per unit area (in 3D) decreases as 1/r.
For the 2D case you can make a similar deduction. Here the energy is spread over a circumference so the energy over unit length of circumference decreases as 1/r.

 

[malawi_glenn]

I don't understand why they can assume that no energy is absorbed by the water. Do you please know why?

Otherwise the problem is not solveable unless we know a whole lot about water.

 

[ChiralSuperfields]

No, it means that the energy is spread over larger and larger perimeters (because is a 2D wave).
In 3D the energy of a point (or spherical) source is spread over larger and larger spherical surfaces and so the energy per unit area decereases as 1/r ²(inverse square law). As the energy is proprtional to amplitude squared, the energy per unit area (in 3D) decreases as 1/r.
For the 2D case you can make a similar deduction. Here the energy is spread over a circumference so the energy over unit length of circumference decreases as 1/r.

Thank you for your reply @nasu! Sorry for the late reply, I completely forgot about this thread!

Many thanks!

 

[ChiralSuperfields]

Otherwise the problem is not solveable unless we know a whole lot about water.

Thank you for your reply @malawi_glenn ! Sorry for the late reply, I completely forgot about this thread!

Many thanks!

 

[haruspex]

Why do they assume that no energy is absorbed by the water?

Some is, of course, due to viscosity, but the viscosity of water is very low, allowing waves to propagate a long way. Compare with trying to make ripples across the top of a pot of treacle.

 

[ChiralSuperfields]

Some is, of course, due to viscosity, but the viscosity of water is very low, allowing waves to propagate a long way. Compare with trying to make ripples across the top of a pot of treacle.

Thank you for your reply @haruspex ! Sorry about the late reply, I forgot about this thread again.

I have never tried to seen treacle, but searching it up, it looks it has quite a high viscosity. I guess that means that the kinetic energy of the particles from a wave source is quite rapidly dissipated as friction between the layers of the liquid so the ripples won't travel far.

https://link.springer.com/referencework/10.1007/978-0-387-92897-5

Many thanks!