Am I wrong, or is my professor? Intensity Question

[FerPhys]

Hello everyone, please help me understand this.
We were given a conceptual question that says "A sound wave goes from the air into the ocean. Which of the following applies (circle all that apply):
a)its intensity increases b) its wavelength decreases c) its frequency increases
d)its frequency remains the same e) its velocity decreases.

My answers were A & D.
I was correct on D but my professor said A was wrong. He didn't really give the class much of an explanation but when I asked him personally he said; you're saying intensity increases, where does the "new energy" come from? intensity=power/area
At the time I didn't really have an answer so I took him to be correct but after reviewing I still think I'm correct.
Since Intensity = 2π2ρƒ2v∆x2max and the density of water is greater than that of air, frequency remains constant, and velocity increases in water (I have no clue where Δx²max comes into play) I chose intensity to be increasing. Now, I think I found the flaw in his logic. He asked, "where is all that energy come from to make the intensity greater"? He believes intensity measures a quantity of how much energy is transfer but it is actually THE rate at which energy is transferred in an area! Therefore, from the original question wouldn't I be correct to say that the intensity in water increases? Meaning energy is transferred at a higher rate/area?
Thoughts?

 

[NascentOxygen]

What would be the units of intensity?

 

[FerPhys]

Intensity = Power_avg/Area so W/m² or J/sm²

 

[NascentOxygen]

Intensity = Power_avg/Area so W/m² or J/sm²

So in the absence of any converging lens effect, how could W/m² be increased without adding watts?

 

[FerPhys]

Well.. Power = Force x velocity. So you have a constant force but increase the velocity so you have higher watts.

 

[FerPhys]

and I'm a little unclear on what a converging lens effect does/is in this case..

 

[NascentOxygen]

and I'm a little unclear on what a converging lens effect does/is in this case..

Just as you can focus light to an intense spot with a glass lens, so you can also focus sound using a large lens cast in wax, for example. I was saying we can rule out using such means to increase the sound intensity in the scenario under discussion.

 

[FerPhys]

I see. So what do you think of my explanation so far? Since its a rate you don't have to ask, where did those joules come from? It's like me saying, v1= 10m/s then saying v2=20m/s , you don't ask, where did those meters come from. It's just a rate. similarly here, I believe its just a rate at which energy is transferred, not necessarily how much energy there is. Given initially values of you mechanical energy I'm sure you could find how long it would take for you (in area and seconds) to transfer all of your energy and therefore stop the wave. But that's a different story.

 

[NascentOxygen]

Well.. Power = Force x velocity. So you have a constant force but increase the velocity so you have higher watts.

This relation applies where that force is moving a body at velocity v. When sound travels from A to B through a medium, say, water, it doesn't transport the whole body of water from location A to location B; it just jostles nearby particles a little causing each to jostle the next and pass the wave along. It's the wave that travels from A to B, not any mass.

 

[FerPhys]

Right, exactly. The way sound waves move is do to a variation in pressure. The particles themselves oscillate but cause the one near them to begin to oscillate themselves. Hence, there is an acceleration caused by a group of particles on another group of particles and a mass of the particles, a force, along with a velocity of these particles, the speed of sound in that medium. So if you have the force which causes these particles to accelerate and the velocity at which they move you can find power, or energy transferred in seconds. Makes sense right? Energy has the be transferred in order for the wave to continue.

 

[A.T.]

Since its a rate you don't have to ask, where did those joules come from?

You have to.

It's like me saying, v1= 10m/s then saying v2=20m/s , you don't ask, where did those meters come from.

Because there is no "Conservation of Length", but there is a "Conservation of Energy".

 

[FerPhys]

right, but again, Intensity isn't telling you how much energy you have it simply tells you the rate at which energy is transferred through an area.
It doesn't tell me anything at all about the energy these particles have.

 

[FerPhys]

Which intuitively makes sense right? The rate at which energy is transferred in a wave should be greater in water since it's denser, meaning more friction etc..

 

[A.T.]

right, but again, Intensity isn't telling you how much energy you have it simply tells you the rate at which energy is transferred through an area.

Yes, transferred, not created. So the input rate must equal the output rate.

 

[sophiecentaur]

Without some impedance matching system, most of the sound energy, incident on the water surface will be reflected because of the wildly different impedances of air and water.
The only thing you can say about any interface situation is that frequency has to be unchanged.

 

[A.T.]

The only thing you can say about any interface situation is that frequency has to be unchanged.

And that the intensity will definitely not increase.

 

[mrspeedybob]

You have to be careful to answer the question that was asked.

The question implied by choice A is... Would the intensity of sound increase when the sound propagates into water from air.
The question you seem to be answering is... Would the rate at which sound energy is converted into heat increase when the the sound propagates into water from air.

 

[sophiecentaur]

This is the old problem of questions that are not well posed in the first place. (I don't mean the OP's question - I mean the one that was set in the first place.)

 

[russ_watters]

This is the old problem of questions that are not well posed in the first place. (I don't mean the OP's question - I mean the one that was set in the first place.)

I think these "not well posed" questions are usually written that way on purpose and done for good reason: correctly interpreting a problem is a critical skill in problem solving. In real life, you rarely get precisely/concisely written problems, with exactly the relevant input data to solve it. Problem interpretation is not only a critical engineering skill, it helps you with comprehension of any unclear information, from talking to a car salesman to interpreting political speech.

 

[sophiecentaur]

I think these "not well posed" questions are usually written that way on purpose and done for good reason: correctly interpreting a problem is a critical skill in problem solving. In real life, you rarely get precisely/concisely written problems, with exactly the relevant input data to solve it. Problem interpretation is not only a critical engineering skill, it helps you with comprehension of any unclear information, from talking to a car salesman to interpreting political speech.

I agree with you , in principle, of course. However, writing a suitable question does involve a lot of extra skill plus a good knowledge of the students. Teachers can often verge on the smartarse side of things and that can give students real confidence problems. I would say that 'those types' of question are very good in situations where you are actually with a class and can feed them with suitable clues when necessary. When students are alone with such a problem they can spend hours and hours of frustration and get nowhere. Real life problems are different because no one expects all the necessary facts to be readily available. Even students who make it successfully through their courses are not instantly capable of solving ill conditioned problems and they need experience before they can cope. TGFPF, I say.

 

[fresh_42]

I think these "not well posed" questions are usually written that way on purpose and done for good reason: correctly interpreting a problem is a critical skill in problem solving. In real life, you rarely get precisely/concisely written problems, with exactly the relevant input data to solve it. Problem interpretation is not only a critical engineering skill, it helps you with comprehension of any unclear information, from talking to a car salesman to interpreting political speech.

I have never seen it that way and knowing how didactic departments work I have my doubts that it is on purpose. Nevertheless it is a challenging extra. Your view on this will certainly prevent me from some severe and angry tantrums next time I help someone with his schoolbooks. Thank you.

 

[russ_watters]

I have never seen it that way and knowing how didactic departments work I have my doubts that it is on purpose. Nevertheless it is a challenging extra. Your view on this will certainly prevent me from some severe and angry tantrums next time I help someone with his schoolbooks. Thank you.

Maybe I'll start a new thread on this. I recently took my PE exam and it had two halves:
Part 1 was school type, cookie cutter problems. Often difficult, but usually straightforward.

Part 2 is practical problems. They are usually easy to solve once you identify the question and relevant data, but they bombard you with extra and often misleading data and bury the actual question in a paragraph of irrelevant discussion of something else. The test has a pretty high failure rate because many people get buried by such questions and can't properly and quickly parse them.

 

[fresh_42]

The test has a pretty high failure rate because many people get buried by such questions and can't properly and quickly parse them.

You already gave away the correct strategy to solve those tests by using the word "parse" instead of "read".

 

[litup]

Well, suppose 1/10th of the air intensity is transferred to the water because 90% is reflected. That would be in RF terms a VSWR of about 10:1 or so. But the sound velocity in water is something like 10 times that of air so the wavelength would have to increase for the same frequency, right? If you have a device that converts green light at 550 Nm to IR at 1100 Nm the IR wave represents a lower energy. So why would the energy not be lower in the water? I can see the pressure wave influencing the water at the same rate so the frequency of the resultant water wave should be the same but the resultant wave would be stretched out in physical space since the speed of sound in water is so much greater than the speed of sound in air. Why wouldn't that equate to a lower energy wave?

 

[jbriggs444]

But the sound velocity in water is something like 10 times that of air so the wavelength would have to increase for the same frequency, right? If you have a device that converts green light at 550 Nm to IR at 1100 Nm the IR wave represents a lower energy. So why would the energy not be lower in the water? I can see the pressure wave influencing the water at the same rate so the frequency of the resultant water wave should be the same but the resultant wave would be stretched out in physical space since the speed of sound in water is so much greater than the speed of sound in air. Why wouldn't that equate to a lower energy wave?

Reasoning about the classical behavior of sound by analogy to the quantum behavior of light is unsound. In addition, reasoning that "because the wavelength increased, the energy decreased" is unsound even for light if you are not holding the medium constant.

 

[sophiecentaur]

If you have a device that converts green light at 550 Nm to IR at 1100 Nm the IR wave represents a lower energy

It is the frequency that tells you the energy of a photon and the frequency of green light is the same, wherever. As JBriggs says, there is no valid link between the behaviour of sound and light in this context but, whatever the situation, the frequency of any wave at a (stationary) interface is unchanged.
You won't get a passive device that 'converts green to IR'.

 

[FerPhys]

still haven't heard a compelling answer to my question. Anyone?

 

[fresh_42]

still haven't heard a compelling answer to my question. Anyone?

You have. Your professor was right.

 

[olgerm]

Anyone?

Maybe you understand this wrongly because you imagine intensity as energy density. Energy density in the wave decreases when soundwave goes to water and wave speed increases, because the volume water which includes the wave is bigger than volume of air which includes same wave, but energy of wave must be same in water and in air. But intensity do not change when wave goes to water.

$E_{wave\ in\ air}=\iiint(dV \cdot U_{wave\ in\ air}(x;y;z))$
$E_{wave\ in\ water}=\iiint(dV \cdot U_{wave\ in\ water}(x;y;z))$
Since energy of wave must stay same. $E_{wave\ in\ air}=E_{wave\ in\ water}$
so $\iiint(dV \cdot U_{wave\ in\ air}(x;y;z))=\iiint(dV \cdot U_{wave\ in\ water}(x;y;z))$
since shape of wave is same in both mediums we can write:
$V_{wave\ in\ air} \cdot U_{wave\ in\ air}=V_{wave\ in\ water} \cdot U_{wave\ in\ water}$
If it is a planewave then $\frac{V_{wave\ in\ water}}{V_{wave\ in\ air}}=\frac{v_{wave\ in\ water}}{v_{wave\ in\ air}}$
so $\frac{U_{wave\ in\ water}}{U_{wave\ in\ air}}=\frac{v_{wave\ in\ air}}{v_{wave\ in\ water}}$

U is energy density;
v is wave speed;
E is energy;
V is volume.

 

[sophiecentaur]

But intensity do not change when wave goes to water.

But what about the majority of the energy in the incident wave - that is reflected back into the air? The power transferred into the water is a very small fraction.