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Homework Help: Speed of sound through an aluminum rod

  1. Apr 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A man strikes a long aluminum rod at one end. Another man, at the other end with his ear close to the rod, hears the sound of the blow twice (once through the air and once through the rod), with a 332ms interval between them. How long is rod?

    2. Relevant equations
    unsure about relevant equations. But because I think this might be
    open end air column or a closed end

    speed= frequency*wavelength
    (for open end) wavelength= (2/n)*Length
    (for closed end) wavelength= (4/n)*Length

    3. The attempt at a solution

    At first I attempted this by doing v=x/t
    I assumed velocity of sound to be 340m/s, and the time .332s, so
    340m/s *.332s= x = 112.8 but I got that wrong.

    Then I thought maybe I should treat this as an air column, but I can't figure out if this is a open end or closed end air column. I'm reading a section on these concepts but I'm just confusing myself even more. The aluminum rod is struck and the man on other end hears the sound once in the air
    but just once through the rod. What does this mean??:confused:
  2. jcsd
  3. Apr 15, 2007 #2
    I think this might be an open end air column. And that it is the first harmonic
    So I tried
    speed/frequency= wavelength
    frequency= 1/.332=3.012

    (340m/s)/3.012Hz= 112.8 m
    112.8/2= L
    56.4m= L

    The question does not give the speed of sound, so I assumed it was 340m/s...but maybe I shouldn't. Is there a specific speed of sound for an aluminum rod?
  4. Apr 15, 2007 #3
    Hey FL, hello again. To your question 340 is good for air only, and that itself varies with temp, density, etc.
    So answer, for wrt to Al, yes there is and can be found by googling speed of sound, so this is more like a lightent/thunder problem--see a flash, hear a boom, how far away is lightening and nothing to do with resonance phenom.
  5. Apr 15, 2007 #4


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    I have no idea why you are worrying about "open" or "closed" end or "wavelength". You are not asked for a wave length and it is irrelevant. As denverdoc said, look up the speed of sound in aluminum. The speed of sound in air is 340 m/s so if the rod has length x, in meters, the man will hear the sound through the air in x/240 seconds. He will hear the sound through the rod in x/v seconds where v is the speed of sound in aluminum. The difference between those times is 332 ms= 0.322 s. Set x/340- x/v= .322 and solve for x.
  6. Apr 15, 2007 #5
    HallsofIvy, I thought maybe the problem involved something more, that's why I brought up open and closed air columns.

    I looked online and found the speed of sound for aluminum rod is 5100m/s

    so I just tried

    x/340 - x/5100=.332
    x(1/340 -1/5100)=.332
    x= .332/[(1/340)-(1/5100)= 117.3 m
    but that was incorrect.

    denverdoc, I'll look for those examples online.
  7. Apr 15, 2007 #6
    The thunder, lightning example kind of confused me. The example I'm reading talks about a time delay and echolocation. If I am interpreting the info correctly, then the sound of the bar striking the ground in the aluminum bar is .332s after the sound of the bar strking the ground in the air. So if I found the correct speed of sound for aluminum, the length of bar is velocity*time=5100m/s *.332 s = 1693.2 m. But that number is so huge, it just seems like that can't be the correct answer
  8. Apr 15, 2007 #7
    Forgive me if I redo your work, most of which looks sound, just easier for me as I'm lousy at proofreading math work.

    ping 1 time=x/5100
    ping 2 time=x/340

    ping 2 time-ping 1 time=.332 s (are you sure thats 332ms?)

    x/340-x/5100=0.332 hence x(1/340-1/5100)=.332.....times (364.3)=120.9m

    so method looks good, but a few percent diff in answer.
  9. Apr 15, 2007 #8
    yes I checked. 1 s = 1000ms. It is .332 s.
    Well, I guess I calculated incorrectly because I ended up getting 120.9m instead of my previous answer. Now I looked at the speed of sound again, and some sites say 343m/s some say 340 m/s. Using 340m/s I get 120.9 m
    but using 343m/s I get 122.08 m. I guess it doesn't really matter.
  10. Apr 15, 2007 #9
    Ok, I used 340m/s and I got it right. Thanks denvnerdoc and HallsofIvy.
  11. Apr 15, 2007 #10
    My bad. I meant checking that it was 332 ms, not the conversion factor. :tongue: Good work. Cuz your answer was slightly off, you went thru all kinds of headaches. I would consider buying for like 20 bucks a Schaums outline which has hundreds of problems with answers, so when the computer tells you you are wrong, you can be confident that your methods are sound and recheck math, before questioning your logic or looking for deeper problems. This is not to be used as cookbook--find problem like that presented and borrow method, much better to understand principles and apply those correctly. But for practice and checking logic, a good resource IMO. At least with problems in the back of the book, if you're small amt off, its a clue you're close. With on-line resources, no clues, just binary output:devil:

    (This only reminds me of going "batty" myself as in an echolocation problem posted recently involving doppler shifts--a small math error led to another 10 posts and all kinds of "novel" interpretations of physical situation including a couple on my part. But thats the temptation, when told the answer is wrong, we sometimes too quickly assume reasoning is at fault.)
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