Intensity of circular waves

In summary, when a stone is thrown into a pond, a wave is generated and spreads out from the center. The amplitude of the particles A, B, and C will change as the wave travels and the intensity of the wave is directly proportional to the square of the amplitude. The intensity also decreases as the wave travels outwards due to the energy spreading out over a larger area. The relationship between intensity and amplitude is the same for all types of waves, regardless of their shape or size.
  • #1
desmond iking
284
2

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

 

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  • #2
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?
 
  • #3
Simon Bridge said:
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.
 
  • #4
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
Simon Bridge said:
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.
 
  • #6
how is intensity related to amplitude?
 
  • #7
Simon Bridge said:
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
 
  • #8
for me , intensity should be direclty proportional to A^2 / x for circular wave.
No! Intensity is proportional to amplitude-squared for ANY wave.
 
  • #9
Simon Bridge said:
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)
 
  • #10
Simon Bridge said:
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##

E19rSNS.jpg

oHGQRAt.jpg
 
  • #11
$$I=\frac{\bar P}{\text{Area}}\propto \text{amplitude}^2$$
For a point source $$A\propto\frac{1}{r}$$... therefore $$I\propto\frac{1}{r^2}$$
 
  • #12
Simon Bridge said:
$$I=\frac{\bar P}{\text{Area}}\propto \text{amplitude}^2$$
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?
 
  • #13
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
Simon Bridge said:
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
 
Last edited:

What is the intensity of circular waves?

The intensity of circular waves is a measure of the amount of energy transmitted per unit time through a given area. It is proportional to the square of the amplitude of the wave.

How is the intensity of circular waves related to the amplitude?

The intensity is directly proportional to the square of the amplitude. This means that as the amplitude increases, the intensity also increases.

What factors affect the intensity of circular waves?

The intensity of circular waves can be affected by several factors, including the amplitude of the wave, the distance from the source, and any obstacles or barriers in the path of the wave.

Can the intensity of circular waves be measured?

Yes, the intensity of circular waves can be measured using specialized equipment such as a wave intensity probe or a sound level meter. It can also be calculated using mathematical formulas.

How does the intensity of circular waves change with distance from the source?

The intensity of circular waves decreases with distance from the source, following the inverse square law. This means that the intensity is inversely proportional to the square of the distance from the source.

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