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Sounds propagation in different mediums

  1. Jan 30, 2005 #1
    I am utterly confused with this problem, I am not quite sure how to formulate an equation to solve for the answer

    :
    The speed of sound in steel is about 4.2 kilometers per second. A steel rail is struck with a hammer, and there is an observer some distance away with one ear to the rail. The observer measures 1.60 seconds between hearing the sound through the rail and hearing the sound through the air. If the air temperature is 20 degrees Celsius and there is no wind, how far away is the observer?
    Just looking for someone to show me where to begin
    any help appreciated?
     
  2. jcsd
  3. Jan 30, 2005 #2
    So far I know you use 343 m/s for speed of sound in air at 20 degrees celcius, and you use 4.2km/s (4200m/s) as speed of sound in the steel, for both you can use the equation v=d/t but both d and t are unknown, also you know t(air)-t(steel) = 1.6 seconds, from there i am not sure how to proceed, i have been trying to come up with a way to combine all of those equations but am comming up short...
     
  4. Jan 30, 2005 #3

    HallsofIvy

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    Yes, both d and t are unknown- that means you need two equations to solve for the two unknown. Those two equations are the motion through the air and through the rail. The distance sound has to travel in either case is d but there are two different times: call them tair and tsteel.

    The speed of sound through the air is 343 m/s so d/tair= 343 which is the same as tair= d/343.

    The speed of sound through steel is 4200 m/s so d/tsteel= 4200 which is the same as tsteel= d/4200.

    What you DO know is the difference in the times: t1- t2= 1.6 s.

    Okay subtract one equation from another:
    tair- tsteel= 1.6 s= d/343- d/4200. Now you have a very simple equation to solve for d.
     
  5. Jan 30, 2005 #4
    Thanks!

    Thank you, however right before I read this post i stumbled upon a way i think will work, it may be the same as what u said however, i just thought tat u know the dstances n ar and steel are equal, so d=vt; vt(air)=vt(steel) ; then vt(air)=v(t-1.6) and then solve for t of air, then plug that into the equation v=d/t to solve for d in air. does this work also?
     
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