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Waves and sounds - freshwater and air.

  1. Nov 27, 2006 #1
    At a height of 15 meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the lake returns to the point of origin 0.130 s later. The air and water temperatures are 20°C. How deep is the lake?

    I know that the velocity of air at 20 degrees Celsius is 343 m/s.

    I know that the velocity of freshwater at 20 degrees Celsius is 1482 m/s.

    ... How in the heck do I figure this out??
  2. jcsd
  3. Nov 27, 2006 #2
    Ignore the air part of the question and concentrate on teh water part. Find the distance that the wave travels.
  4. Nov 27, 2006 #3
    Don't ignore the air part... find the time it would take the wave to travel through the air, multiply it by two to account for the trip down and back up, then subtract it from the total time.

    The depth of the lake will come out of how long it would take the wave to travel through water at its speed. Don't forget it makes two trips.
  5. Nov 27, 2006 #4
    This is what I've done so far ...

    velocity of air = [distance/time] ...

    343 = 15 / time

    time = .0437 seconds ... multiply by 2 (to take the account for the return trip) ... therefore, time = .0875 seconds.
  6. Nov 27, 2006 #5
    My next guess is to take the account for when the sound wave penetrates the fresh water.

    ... so, my guess is ...

    velocity of fresh water = lambda * frequency ...

    ... 1 / time = frequency ...

    ... using .0875 as a reference ... frequency = 23.5294 Hz.

    ... am I missing something ... subtracting from total time ???
  7. Nov 28, 2006 #6
    Why are you treating water differently than air? You get the distance in just the same way.
  8. Nov 28, 2006 #7
    Ah, my mistake. Misread there. I do not think you really need to find the frequency of the sound wave. Moreover you don't know whether the time you got was actually the period.
    Just work backwatds from what you got b4. Subtract the total time from the time in air which gives you the total time in water. You should be able to get it from here.
  9. Nov 28, 2006 #8
    you are given the two way travel-time.
    subtract from this the time spent in the air using: time=distance/speed.
    then use what's left over to find the distance travelled in the water.

    Don't forget you're dealing with 2 way travel time, meaning you have to be careful with what distances you use, draw a diagram if this isn't clear.
  10. Nov 28, 2006 #9
    I understand the drawing that I made - but the numbers don't seem right ... this is a tough question!
  11. Nov 28, 2006 #10
    Anyone else want to take a chance at this?
  12. Nov 28, 2006 #11
    Don't I have to multiply the time in the water by 2?
  13. Nov 28, 2006 #12

    Like I said... once you subtract the time it takes the wave to travel in the air, you have the time it takes to travel in the water. You know how fast it moves and how much time it takes... use kinematics to find the distance, just like you did when you found time with air, only your variable is now distance, not time.
  14. Nov 28, 2006 #13
    So, I multiplied the time in the water by 2.

    ... 0.130 * 2 = .260 seconds in the water.

    ... I subtracted the calculated time from the air ...

    ... .260 - .0875 = .1725 total seconds.

    ... taking the kinematic equation [displacement = (total time * velocity of water at 20 degrees Celsius)] ...

    ... displacement = (.1725 * 1482)

    ... displacement = 255.65 meters ...

    ... I still got this wrong ... what did I do wrong?
  15. Nov 29, 2006 #14
    The wave took, in total, 0.13 seconds to go down and up.

    It spent 0.0875 seconds in the air. This leaves 0.13 - 0.0875 = 0.0425 seconds that the wave spent in the water.

    The wave had to go all the way down, then all the way up. The same distance each way. This means that half the time it was moving down, the other half up.

    0.0425/2 = 0.02125 seconds moving down means that it took 0.02125 seconds for the wave to travel some distance at a speed of 1482 m/s.

    Now use the kinematics.
  16. Nov 29, 2006 #15
    Oh, duh!! Now, I see what you meant ... I finally drew a picture of it to understand it ... LOL

    ... from your calculated time of 0.02125 seconds and the velocity of water at 20 degrees Celsius ...

    ... displacement = velocity of water * total time

    ... displacement = 1482 * .02125

    ... displacement = 31.4925 meters

    ... Thanks, Ethan ... I just didn't see it until I drew it. I should take your advice to heart next time ... LOL
  17. Nov 29, 2006 #16
    yes that's right, i just checked it myself
  18. Nov 29, 2006 #17
    LOL - it's all good!
  19. Nov 29, 2006 #18
    More importantly, draw it next time! =)
  20. Nov 29, 2006 #19
    That's for sure ... [sigh of relief]
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