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When you double your distance from a sound source

  1. Jun 7, 2003 #1


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    This cant be right.

    I missed this test question here.

    You shout at a cliff and hear the echo in 4s. the temp is 0 degrees C. How far away is the cliff?

    I think the answers were a)662 b) 680 c) 1320 d)1760

    I chose c because

    t=d/s or d=t*s

    now, i substitued d = 4s * 331m/s = 1324

    why did i miss this? I am pretty sure the formula and math is

    also one more question

    when you double your distance from a sound source radiating in all directoins. what happens to the intensity of the sound? is reduces to 1/2 its orginal value.

    Thats what i said unless its 1/4. what is the right answer.

    Dx :wink:
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jun 7, 2003 #2
    You computed:
    d = 4s * 331m/s = 1324

    That was exactly correct -- for the distance from you to the cliff and back. D is twice the distance from you to the cliff. So, the correct answer is a)662 m.

    Sound intensity varies inversely with the square of the distance, so the correct answer to the second question was 1/4.
  4. Jun 7, 2003 #3


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    thanks gnome!
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