# Homework Help: Intensity of waves

1. Jul 12, 2014

### desmond iking

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

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 .

2. Relevant equations

3. The attempt at a solution

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2. Jul 12, 2014

### 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.

3. Jul 13, 2014

### 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?

4. Jul 13, 2014

### Simon Bridge

... how do you know that?

5. Jul 13, 2014

### desmond iking

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?

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6. Jul 13, 2014

### Simon Bridge

... 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.

7. Jul 13, 2014

### desmond iking

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.

8. Jul 13, 2014

### 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.

9. Jul 13, 2014

### desmond iking

i just wanna know why would I make such mistake so that i wouldnt make it in th efuture.

10. Jul 13, 2014

### 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.

11. Jul 13, 2014

### desmond iking

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 )

12. Jul 13, 2014

### Simon Bridge

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.

13. Jul 13, 2014

### desmond iking

sorry, i really dont 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

14. Jul 13, 2014

### desmond iking

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

since intensity = power / 2 pi r ....

Last edited: Jul 13, 2014
15. Jul 13, 2014

### 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 travelling plane-wave with equation:

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

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

16. Jul 14, 2014

### desmond iking

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

17. Jul 14, 2014

### Simon Bridge

... 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: $y(x,t)=\frac{b}{x}\sin(\omega t - kx)$

18. Jul 14, 2014

### desmond iking

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

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19. Jul 14, 2014

### Simon Bridge

why do you have "second" in your equation?
There should be no units in an equation.
... I never said that the source was a vibrating mass.

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.

20. Jul 14, 2014

### 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)

21. Jul 14, 2014

### 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.

22. Jul 17, 2014

### 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.

23. Nov 16, 2014

### 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

24. Nov 16, 2014

### Simon Bridge

What do you need help with?

25. Nov 16, 2014

### Akashsinha

Why not multiplying with frequencies as intensity is proportional to frequency

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