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Speed of longitudinal wave 30 times the speed of a transverse wave?

  1. 1. The problem statement, all variables and given/known data
    What must be the stress (F/A) in a stretched wire of a material whose Young's modulus is Y for the speed of longitudinal waves to equal 30 times the speed of transverse waves?


    2. Relevant equations
    [itex]Y=\frac{Fl_0}{Al}[/itex]

    [itex]v_L=f\lambda = \sqrt{F/\mu}[/itex]

    [itex]v_T = \omega A sin(kx-\omega t)[/itex]


    3. The attempt at a solution
    I know that [itex]v_L=30v_T[/itex] but my main problem is that longitudinal velocity remains constant while transverse velocity is dependent on position and time, making it impossible for one to be a multiple of the other unless they are both equal to 0, which cannot be the case. I'm not sure what I'm missing ... thanks!
     
  2. jcsd
  3. those are irrevelant equations that u are using....the question says speed of transverse wave not transverse velocity of particle
     


  4. these equations are not written correct. You havent written the formula for speed of longitudnal wave. The formula written is for the speed of transverse wave instead of longitudnal wave. The formula written for speed of Transverse wave is the formulae for "transverse velocity of particle".
     
  5. Is [itex]v_T = \lambda f = \sqrt{F/\mu}[/itex] and [itex]v_L = \frac{Y}{\rho}[/itex] correct?
     

  6. Hint:
    You have missed a square root.
     
  7. [itex]v_L = \sqrt{\frac{Y}{\rho}}[/itex]??
     
  8. So if [itex]v_L=\sqrt{Y/\rho}[/itex] and [itex]v_t=\sqrt{F/\mu}[/itex] then

    [itex]\sqrt{Y/\rho}=30\sqrt{F/\mu}[/itex] which is equivalent to [itex]Y=\frac{900F\rho}{\mu}[/itex]

    [itex]Y=\frac{Fl_0}{Al}[/itex]

    so [itex]\frac{F}{A}=\frac{Yl}{l_0}=\frac{900Fl\rho}{l_0 \mu}[/itex]

    but [itex]l\rho = A[/itex] and [itex]\mu = \frac{mass}{length} ~~so~~\mu l_0=mass[/itex]

    so [itex]\frac{F}{A}=\frac{mass}{900AF} ~~so~~F^2=\frac{mass}{900}[/itex]

    which doesn't work ...

    I also tried it a different way, letting [itex]v_T=fλ[/itex] so that [itex]Y=900v^2\rho[/itex] so then [itex]\frac{F}{A}=\frac{Yl}{l_0} = \frac{900v^2\rho l}{l_0} = \frac{900v^2A^2}{l_0}[/itex] which also doesn't work. Any help would be appreciated! Thanks!!
     
  9. Skip the step where you have got in Y=FL/AL(0).

    You have no way of relating L and L(0).

    Also Lp=A is not right.

    p=M/V sp Lp=LM/V.
    But here L is change in length.
    So you can't write V=AL as L is change in length.So this doesn't help.


    Just focus on your relation
    Y/p=900F/u.

    Divide it by A on both sides

    Y/Ap=900F/Au.

    So F/A=Yu/900Ap

    Is there any way you can find out what u/Ap is ?
    ( Hint:You have applied a relation between u and l in your last post.That might help)
     
  10. I got it - thank you!!
     
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