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Pulses in wire - waves

  1. Apr 11, 2006 #1
    Hi, I am having difficulty understanding this question:

    A wire 5.0 m long and having a mass of 130 g is stretched under a tension of 220 N. If two pulses, separated in time by 30.0 ms, are generated, one at each end of the wire, where will the pulses first meet?

    m (distance from the left end of the wire)

    From my deduction the each wave starts from the opposite side.
    I figured out the v=91.9866 b/c the square root of 220/(.13/5)=91.9866 m/s

    I then set up equations
    t=time x=distance

    I got x to equal 2.6379799 m
    However, I am not certain if this is the correct answer and I am down to my last submission. I was wondering if anyone could agree with what I got.
  2. jcsd
  3. Apr 12, 2006 #2


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    I get 1.12m for my answer. Here's my calcs;

    [tex]v = \sqrt{\frac{T}{\frac{m}{L}}} = \sqrt{\frac{220}{\frac{0.13}{5}}} = 91.9866 m/s[/tex]

    [tex]x = vt \Rightarrow t = \frac{x}{v}[/tex]
    [tex]5 - x = v(t + 0.03)[/tex]

    Subbing [itex]t = \frac{x}{v}[/itex] into [itex]5 - x = v(t + 0.03)[/itex] gives;

    [tex]5 - x = v\left( \frac{x}{v} + 0.03 \right) = x + 0.03v[/tex]

    [tex]2x = 5 - 0.03v \Rightarrow x = \frac{5 - 0.03\times 91.9866}{2}[/tex]

    [tex]\fbox{ x = 1.120201m }[/tex]

    I'm not sure that I'm right though :confused:

  4. Apr 12, 2006 #3


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    I disagree with your answer and agree with Hootenanny. Your mistake is the sign of the 3*10^-3 in the second line. It should be a plus sign.

    Basically (after substitution) you wrote x- 91.99* 3*10^-3 = 5 -x or
    2x= 5 + 91.99*3*10^-3 or

    2x= 5 + 2.76

    But it should be 2x= 5 - 2.76.

    (it's easy to see. In the first 30 ms, the first pulse travels 2.76 m. There is still a distance of 5-2.76 to travel for both pulses when the second pulse will be emitted. They will obviously meet halfway through this remaining distance, therefore the answer is x= 1/2(5-2.76).

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