1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Composite wave

  1. Jul 25, 2014 #1
    1. The problem statement, all variables and given/known data

    The figure below shows a snapshot of a standing wave on a composite string. It is a node in the composite point. If the aluminium string has a diameter of 1.50 mm, which diameter does the steel string have ?

    http://s716.photobucket.com/user/Pitoraq/media/Fys22222_zps61d0eb53.png.html


    2. Relevant equations

    V = √(F/ρ(length)) = √(F/ρA)


    3. The attempt at a solution

    Al: Node in both endpoints

    y = f(ωt)*sin(kx) <==> sin(k1L1) = 0 <==>

    2π/λ1*L1 = P1π <==>

    λ1 = 2/p1*L1

    Steel: Node in both endpoints


    sin(K2L2) = 0 <==> 2π/λ2*L2 = P2π <==>

    λ2 = 2/p2*L2

    λ = v/f

    λ1 = v/f1 = 2/p1*L1 <==> vp1/2L1 = f1 <==>

    vp2/2L2 = f2

    Same frequency

    vp2/2L2 = vp1/2L1 <==>

    √(F/ρ(al)A1)*P1/2L1 = √(F/ρ(Steel)A2)*P2/2L2 <==>


    √(F/ρ(al)4πd1)*P1/2L1 = √(F/ρ(Steel)4πd2)*P2/2L2 , breaking out d2 <==>

    d2 = ((p2)^2*(L1)^2*(ρ(Al))*d1)/((p1)^2*(L2)^2*ρ(Steel))

    Where d1 = 0.0015
    ρ(Al) = 2.7+ * 10^3/m^3
    ρ(Steel) = 7.87*10^3/m^3
    L1 = 0.3 m
    L2 = 0.25 m

    p2 = 5
    p1 = 3

    Are the nodes from the figure

    when i put does values into the equation above i got 2.058 * 10^-3 m

    In the answer it should be 1.76 mm
     
  2. jcsd
  3. Jul 25, 2014 #2

    olivermsun

    User Avatar
    Science Advisor

    I think if you are more careful writing the equations (especially the grouping under the square roots) you'll can see more clearly where the problem is. I'll rewrite some of it

    Starting from:
    $$
    \begin{align}
    \frac{vp_2}{L_2} &= \frac{vp_1}{L_1} \\
    \frac{p_2}{L_2} \sqrt{\frac{F}{\mu_2}} &= \frac{p_1}{L_1} \sqrt{\frac{F}{\mu_1}} \\
    \frac{p_2}{L_2} \sqrt{\frac{F}{\pi(d_2/2)^2\rho_\mathrm{Steel}}}
    &= \frac{p_1}{L_1} \sqrt{\frac{F}{\pi(d_1/2)^2\rho_\mathrm{Al}}} \\
    \frac{p_2/L_2}{p_1/L_1} \sqrt{\frac{\rho_\mathrm{Al}}{\rho_\mathrm{Steel}}}
    &= \frac{d_2}{d_1}
    \end{align}
    $$

    So it seems as if you forgot d1, d2 were inside the √ and as a result your p1, L1, p2, L2 got squared by accident.
     
    Last edited: Jul 25, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Composite wave
  1. Composite Systems - QM (Replies: 3)

Loading...