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Hooke's law and wave velocity related problem

  1. Jan 30, 2017 #1
    1. The problem statement, all variables and given/known data
    The extension in a string, obeying hooke’s law is Y when wave velocity in it is V. if extension is increased to 1.5Y, then wave velocity V’ becomes?

    1) V' =V. 2)V'= 1.22V . 3)V'=1.5V. 4) V'=0.75V.

    2. Relevant equations
    wave velocity= frequency*wave length.

    3. The attempt at a solution
    the frequency will be unchanged in both the cases, so v/v' = Y/1.5Y , which gives V'= 1.5V. but this is not the answer. where did i go wrong?
     
  2. jcsd
  3. Jan 30, 2017 #2

    DrClaude

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    Staff: Mentor

    Do you mean tension?
     
  4. Jan 30, 2017 #3
    nope. it's the extension and not tension.
     
  5. Jan 30, 2017 #4

    DrClaude

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    So your are simply changing the length of the string.
    Have you ever played a guitar (or seen someone playing one)?
     
  6. Jan 30, 2017 #5
    except in movies no.
     
  7. Jan 30, 2017 #6

    DrClaude

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    Think about how someone changes the notes on a guitar.

    By the way, it would be helpful if you came up with more "relevant equations."
     
  8. Jan 30, 2017 #7
    in particularly? there is this hooke's law which states F(force) = -K(spring constant)*x (extension). how to relate this with wave velocity?
     
  9. Jan 30, 2017 #8

    DrClaude

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  10. Jan 30, 2017 #9
  11. Jan 30, 2017 #10

    DrClaude

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    I realize now that the problem statement is not clear. How is the extension changed? By taking a longer length of the same string, or stretching it? I was assuming the former.
     
  12. Jan 30, 2017 #11
    Do you know the equation for the wave velocity of a string as a function of the string tension? The equation has to take into account the fact that the string has mass.
     
  13. Feb 1, 2017 #12
    Yup. V=√(T/mass per unit length). Substituting T for Y*area*extension/original length. Since the other components like mass, length,Young's modulus,area are unchanged, I'm equating V/V' =√(y/1.5y). Which gives V/V' = 0.82. thus I'm getting V' = 1.21V. Is this method correct?
     
  14. Feb 1, 2017 #13
    Yes.
     
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