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Hello,Question.Two people stand beside a long fence. One places

  1. May 16, 2010 #1
    Hello,
    Question.
    Two people stand beside a long fence. One places her ear against the fence while the other person gives the fence a sharp knock. Two sounds seperated by 0.7s are heard by the listener. If the speed of sound is 335m/s in the air and 5300m/s in the metal fence, how far apart are the people?

    My attempt.
    335 * 0.7 = 234.5
    5300 * 0.7 = 3710

    3710 - 234.5 = 3475.5

    They are 3475.5m apart.

    is this right, please check
    Thanks
     
  2. jcsd
  3. May 16, 2010 #2
    Re: Waves

    The 1st sound reaches in a time T1 where T1=s/5300
    The 2nd sound reaches in a time T2 where T2=s/335
    s is the distance required.
    So T2 - T1 is 0.7s
    From this find s.
     
  4. May 16, 2010 #3
    Re: Waves

    Ummm.... How does that work.
    Is is like a equation.
    Sorry I'm confused.
    Thanks
     
  5. May 16, 2010 #4
    Re: Waves

    The time difference was 0.7 s according to the question.
    time is distance / speed
    time (T1) for the sound in air is distance / 335
    time (T2) for the sound in the metal was distance / 5300
    So T1 = s / 335 and T2 = s / 5300
    we know T1 - T2 = 0.7
    so 0.7 = s/335 - s/5300
    It's over to you to find the value of s, the answer to the question.
     
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