Intensity of circular waves

1. Jul 13, 2014

desmond iking

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. The attempt at a solution

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2. Jul 13, 2014

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 modelled 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?

3. Jul 13, 2014

desmond iking

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

4. Jul 13, 2014

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?

5. Jul 13, 2014

desmond iking

energy density decreases, amplitude decreases.

6. Jul 13, 2014

Simon Bridge

how is intensity related to amplitude?

7. Jul 13, 2014

desmond iking

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

8. Jul 13, 2014

Simon Bridge

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

9. Jul 13, 2014

desmond iking

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

10. Jul 13, 2014

desmond iking

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

$I \propto A^2$

11. Jul 13, 2014

Simon Bridge

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

12. Jul 14, 2014

desmond iking

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?

13. Jul 14, 2014

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.

14. Jul 14, 2014

desmond iking

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

Last edited: Jul 14, 2014