Does Wave Intensity Change with Distance in Circular Waves?

[desmond iking]

Homework Statement

when a stone is thrown to position O (in a pond), a wave is generated and the wave eventually spread out form the center, my question is will the amplitude of particle of A, B and C change?
the wave intensity is I=
0.5m(w^2)(a^2)s^-1/ 2 pi r

so i have i got I is directly propotional to (a^2)/ r

as the r change , a change as well , but I change as well .

can i say that the a and r are not changed in the same ratio from position A to C ?

so this cause the intensity of A , B and C to change?

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

Water waves are actually quite complicated:
http://web.mit.edu/flowlab/NewmanBook/Tulin.pdf

More detail on classical waves:
http://galileo.phys.virginia.edu/classes/252/Classical_Waves/Classical_Waves.html

But modeled as just transverse mechanical waves, the amplitude decreases with radius because the energy is spread out over a larger volume as the wave travels ... and, as you have learned before, the intensity is proportional to the square of the amplitude.

To understand the equation you have, you need to make clear what the variables mean.
If it is like your other question, then isn't "a" a constant?

 

[desmond iking]

Water waves are actually quite complicated:
http://web.mit.edu/flowlab/NewmanBook/Tulin.pdf

More detail on classical waves:
http://galileo.phys.virginia.edu/classes/252/Classical_Waves/Classical_Waves.html

But modeled as just transverse mechanical waves, the amplitude decreases with radius because the energy is spread out over a larger volume as the wave travels ... and, as you have learned before, the intensity is proportional to the square of the amplitude.

To understand the equation you have, you need to make clear what the variables mean.
If it is like your other question, then isn't "a" a constant?

this is beyond too complicated for me at this level. can you briefly explain what is going on for wave.

 

[Simon Bridge]

The energy of the wave spreads out as the wave travels outwards - therefore the energy density decreases ... how is the amplitude related to the energy-density?... how is the intensity related to the amplitude?

 

[desmond iking]

The energy of the wave spreads out as the wave travels outwards - therefore the energy density decreases ... how is the amplitude related to the energy-density?... how is the intensity related to the amplitude?

energy density decreases, amplitude decreases.

 

[Simon Bridge]

how is intensity related to amplitude?

 

[desmond iking]

how is intensity related to amplitude?

I is direclty proportional to A^2

for me , intensity should be direclty proportional to A^2 / x for circular wave.

am i correct?
x = radius of circular wave

 

[Simon Bridge]

for me , intensity should be direclty proportional to A^2 / x for circular wave.

No! Intensity is proportional to amplitude-squared for ANY wave.

 

[desmond iking]

No! Intensity is proportional to amplitude-squared for ANY wave.

but intensity= power/ area for spherical wave...
so i would have $I \propto A^2/x^2$ ... (area= 4 pi x^2)

 

[desmond iking]

No! Intensity is proportional to amplitude-squared for ANY wave.

here's another example, the problem is solve using $I \propto1/x^2$ .. and the intensity isn't solved using

$I \propto A^2$

 

[Simon Bridge]

$$I=\frac{\bar P}{\text{Area}}\propto \text{amplitude}^2$$
For a point source $$A\propto\frac{1}{r}$$... therefore $$I\propto\frac{1}{r^2}$$

 

[desmond iking]

$$I=\frac{\bar P}{\text{Area}}\propto \text{amplitude}^2$$
For a point source $$A\propto\frac{1}{r}$$... therefore $$I\propto\frac{1}{r^2}$$

i think i can understand what do you mean now. by saying that I is direclty proportinal to A^2 , do u mean the wave intensity is solely depend on the power , but the area has the impact on it?

 

[Simon Bridge]

The definition of intensity is the average power per unit area (or length for 2D waves). It is the rate that energy arrives at or crosses through a surface. That is physically what the word means.

When the wave is described by sine function, then it turns out that the average power is proportional to the amplitude-squared of the sine wave. This is a consequence of the definition when it is applied to a sine wave.
This is what I'm getting you to explore in the other thread.

 

[desmond iking]

The definition of intensity is the average power per unit area (or length for 2D waves). It is the rate that energy arrives at or crosses through a surface. That is physically what the word means.

When the wave is described by sine function, then it turns out that the average power is proportional to the amplitude-squared of the sine wave. This is a consequence of the definition when it is applied to a sine wave.
This is what I'm getting you to explore in the other thread.

well , you said that When the wave is described by sine function, then it turns out that the average power is proportional to the amplitude-squared of the sine wave. , this is only the average power, am i right? the average power not yet divided by the length (2D) or surface area (3D) am i right?


so why you abandoned r (length) or area (r^2) ? which means I is only direclty proportional to A^2 only?

but not I is directly proportional to 1/r or 1/r^2