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Am I wrong, or is my professor? Intensity Question

  1. Feb 19, 2016 #1
    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 = 2π2ρƒ2vx2max 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?
     
  2. jcsd
  3. Feb 19, 2016 #2

    NascentOxygen

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    What would be the units of intensity?
     
  4. Feb 19, 2016 #3
    Intensity = Poweravg/Area so W/m2 or J/sm2
     
  5. Feb 19, 2016 #4

    NascentOxygen

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    So in the absence of any converging lens effect, how could W/m2 be increased without adding watts?
     
  6. Feb 19, 2016 #5
    Well.. Power = Force x velocity. So you have a constant force but increase the velocity so you have higher watts.
     
  7. Feb 19, 2016 #6
    and I'm a little unclear on what a converging lens effect does/is in this case..
     
  8. Feb 19, 2016 #7

    NascentOxygen

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    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.
     
  9. Feb 19, 2016 #8
    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 thats a different story.
     
  10. Feb 19, 2016 #9

    NascentOxygen

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    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.
     
  11. Feb 19, 2016 #10
    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.
     
  12. Feb 19, 2016 #11

    A.T.

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    You have to.

    Because there is no "Conservation of Length", but there is a "Conservation of Energy".
     
  13. Feb 19, 2016 #12
    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.
     
  14. Feb 19, 2016 #13
    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..
     
  15. Feb 19, 2016 #14

    A.T.

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    Yes, transferred, not created. So the input rate must equal the output rate.
     
  16. Feb 19, 2016 #15

    sophiecentaur

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    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.
     
  17. Feb 19, 2016 #16

    A.T.

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    And that the intensity will definitely not increase.
     
  18. Feb 19, 2016 #17
    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.
     
  19. Feb 19, 2016 #18

    sophiecentaur

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    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.)
     
  20. Feb 19, 2016 #19

    russ_watters

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    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.
     
    Last edited: Feb 19, 2016
  21. Feb 19, 2016 #20

    sophiecentaur

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    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.
     
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