Intensity of Waves: Proportional to y^2/x^4?

[desmond iking]

Homework Statement

since we know that the intensity of wave is I=P/A P =power, A=area

so I = 0.5m(w^2)(y^2)/4pi (x^2 )


so i got I is directly propotional to (y^2)/ (x^2 )

but there's variable x in the eqaution of y...

so can i say that I is directly propotional to y^2 / x^4 as in the third photo?

p/s : in here y also represnt A .

hopefully you guys can understand .

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

The problem statement (attachment 1) suggests maybe a longitudinal wave (oscillations in the x direction) propagating in the +x direction with equation:

##y=\frac{a}{x}\sin(\omega t -kx)## ... so the amplitude falls off with distance.

You are asked to determine how the intensity of the wave falls off with distance.

You have a model answer (attachment 2) which says $$I\propto\frac{1}{x^2}$$

i.e. your own $$I\propto \frac{y^2}{x^4}$$ is incorrect.

It is a serious mistake to use y to also be A ... you have to be consistent with your notation.
A, in your post, is area. The model answer uses to to mean amplitude, as in ##y=A\sin(\omega t - kx)##

The question is checking that you know that intensity is proportional to the square amplitude of the wave.

 

[desmond iking]

sorry. now my question is since we know that I is directrly proportional to 1/(x^2) and I also directly proportional to A^2 ... here my A=AMPLITUDE. so i get I is directly proportional to (A^2)/(x^2). since A= (a/x)... after substituting into the equation , i get I is directly proportional to (a^2 )/ (x^4)... why is it wrong?

 

[Simon Bridge]

since we know that I is directrly proportional to 1/(x^2)

... how do you know that?

 

[desmond iking]

... how do you know that?

here.. it's I is directly proportinoal to 1/ r^2 here.

i know it's not logical to say that I is directly proportional to (a^2 )/ (x^4)... but using mathematical treatment , I = 0.5m(w^2)(A^2)s^-1 /4pi (x^2 )... which means the power will be the same at that particular point only the surface area which is perpendicular to the direction of vibration changed, so the intensity change. but after thinking logically, the amplitude at that point should change am i right?what's wrong with the mathematical treatment?

or the amplitude of particle at certain distance will change for longitidunal waves only? not for transverse waves?

 

[Simon Bridge]

here.. it's I is directly proportinoal to 1/ r^2 here.

... you mean in the attachment <looks> I see.
The calculation in the attachment is not general.

The wave in that example is spherical. In the problem in post #1 above, the wave is not spherical.

 

[desmond iking]

... you mean in the attachment <looks> I see.
The calculation in the attachment is not general.

The wave in that example is spherical. In the problem in post #1 above, the wave is not spherical.

well, you said the wave is not spherical, so it is circular? , so I = P/2∏r , so i would get I is directly propotional to (a^2 )/ (x^3 )... since i is directly propotional to a^2 / r

amplitude is (a/x) as in the equation of y.

 

[Simon Bridge]

It is not me who says that the wave is nor spherical - it is the problem statement that says that.
Do not assume anything not in the problem statement.

You are also using faulty reasoning.
The ##I\propto A^2## is not something extra to the power relationship.
It is a consequence of it!

It is very puzzling that you have the correct answer with the correct working supplied and you still insist on incorrect answers and faulty reasoning.

 

[desmond iking]

It is not me who says that the wave is nor spherical - it is the problem statement that says that.
Do not assume anything not in the problem statement.

You are also using faulty reasoning.
The ##I\propto A^2## is not something extra to the power relationship.
It is a consequence of it!

It is very puzzling that you have the correct answer with the correct working supplied and you still insist on incorrect answers and faulty reasoning.

i just want to know why would I make such mistake so that i wouldn't make it in th efuture.

 

[Simon Bridge]

I don't know why you would make that mistake. The reasoning is in the model answer, and the theory is in your notes. The intensity is always proportional to the square amplitude.
That's all there is to it.

 

[desmond iking]

It is not me who says that the wave is nor spherical - it is the problem statement that says that.
Do not assume anything not in the problem statement.

You are also using faulty reasoning.
The ##I\propto A^2## is not something extra to the power relationship.
It is a consequence of it!

It is very puzzling that you have the correct answer with the correct working supplied and you still insist on incorrect answers and faulty reasoning.

sorry for interrupting again.
I = P/2∏r (for circular waves) , so i would get I is directly propotional to (a^2 )/ (r^3 )... since I is directly propotional to a^2 / r but the amplitude = (a/x )

 

[Simon Bridge]

sorry for interrupting again.
I = P/2∏r (for circular waves) , so i would get I is directly propotional to (a^2 )/ (r^3 )... since I is directly propotional to a^2 / r but the amplitude = (a/x )

No. That is not logical:

You have
(1) ##I \propto A^2##
(2) ##A=a/x## : "a" is a constant.

(2) → (1): ##I \propto 1/x^2##

This is just normal algebra.

 

[desmond iking]

I don't know why you would make that mistake. The reasoning is in the model answer, and the theory is in your notes. The intensity is always proportional to the square amplitude.
That's all there is to it.

sorry, i really don't understand the situation.

i know that intensity of waves = P/2∏r (for circular waves) , where power = 0.5m(w^2)(A^2)s^-1 , after substituiting the P inside , i have I = 0.5m(w^2)(A^2)s^-1 /2 pi r.

the amplitude is a or a/x ?

so i can say that I is directly proportional to (A^2)/r .

but in the question, the solution gives I is directly proportional to (A^2) but not I is directly proportional to (A^2)/r .

Taking I is directly proportional to (A^2)/r ,

provided a (amplitude ) is constant, so i have I is directly proportional to 1/x as in the working

 

[desmond iking]

No. That is not logical:

You have
(1) ##I \propto A^2##
(2) ##A=a/x## : "a" is a constant.

(2) → (1): ##I \propto 1/x^2##

This is just normal algebra.

well why ##I \propto A^2## not ##I \propto A^2/x## ?

since intensity = power / 2 pi r ...

 

[Simon Bridge]

It looks to me like you are getting mixed up between "a" and "A".
##I\propto A^2## because of the power relation.Start from the beginning:
If you have a traveling plane-wave with equation:

##y(x,t)=A\sin(\omega t -kx)##

How are you finding the average power carried by that wave?

 

[desmond iking]

It looks to me like you are getting mixed up between "a" and "A".
##I\propto A^2## because of the power relation.


Start from the beginning:
If you have a traveling plane-wave with equation:

##y(x,t)=A\sin(\omega t -kx)##

How are you finding the average power carried by that wave?

P=(0.5 mw^2 A^2) /s

 

[Simon Bridge]

P=(0.5 mw^2 A^2) /s

... I'm guessing that "w" stands for ##\omega## here.
What does the "m" and the "s" stand for?

But notice that this says that ##P\propto A^2##?
The intensity would be this average power, per unit area.
Therefore, ##I\propto A^2## as well.

Now try again for the wave: ##y(x,t)=\frac{b}{x}\sin(\omega t - kx)##

 

[desmond iking]

w = angular frequency, s stand for second ... why the radius of sphere is not taken into consideration? m=mass of vibrating source

 

[Simon Bridge]

w = angular frequency, s stand for second ...

why do you have "second" in your equation?
There should be no units in an equation.

m=mass of vibrating source

... I never said that the source was a vibrating mass.

why the radius of sphere is not taken into consideration?

Never said anything about a sphere either - since there is no sphere, there is no need to take it's radius into consideration.

If it is not in the problem statement, you need to say why you are introducing it.
Note: The effect of distances on the wave is already taken into account in the wave equation.

 

[desmond iking]

Or Maybe I can say in this way? Since the question doesn't state it's a circular wave or spherical wave, so we only consider I is directly propotional to A^2 only? ( part a ii)

 

[Simon Bridge]

Doesn't matter if the question say or not - I is proportional to A^2

I can also be proportional to other things ... like frequency.
But in your original question, you were asked only to consider how I changes with the x coordinate.
Since frequency stays the same, and amplitude changes with x, you only need to consider the amplitude.
You don't have to do anything else to take x into account because the amplitude has already done that.

The "proportional to" symbol hides a lot of extra influences.

 

[Simon Bridge]

How are you getting on with this?

In detail: ##I \propto \omega^2A^2/c## where c is the wave-speed in the medium.
This relation is true no matter what the medium is.

The amplitude A will have a dependence on space depending on the geometry of the wave.
i.e. for spherical waves, ##A\propto 1/r## and for cylindrical waves ##A\propto 1/\sqrt{r}##.

The constant of proportionality depends on the exact type of wave - i.e. water waves, sound waves, light waves etc.

i.e. mechanical waves in a medium: ##I=\kappa \omega^2A^2/c## where ##\kappa## is the coefficient of restitution for the medium.

 

[Akashsinha]

Many of questions about ratio of wave intensity of two waves like A1Sin(200wt) and A2sin(400wt) just do
A1^2/A2^2 and don't multiply it with their frequencies. Plz help

 

[Simon Bridge]

What do you need help with?

 

[Akashsinha]

Why not multiplying with frequencies as intensity is proportional to frequency

 

[Akashsinha]

What do you need help with?

Why not multiplying with frequencies as intensity is proportional to frequency

 

[Simon Bridge]

Why not multiplying with frequencies as intensity is proportional to frequency

Try this: $$I_1 \propto \omega^2 A_1^2\\
I_2 \propto \omega^2 A_2^2\\
\frac{I_1}{I_2}=\cdots$$

 

[Akashsinha]

Didn't got anything...

 

[Simon Bridge]

Hmmm - for some reason the LaTeX didn;t render in post #27 ... trying again:
$$I_1 \propto \omega^2 A_1^2\\
I_2 \propto \omega^2 A_2^2\\
\frac{I_1}{I_2}=\cdots$$
... what do you mean "didn't got anything"?
Please show your working.