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Reflection and Transmission

  1. Dec 2, 2006 #1
    1. The problem statement, all variables and given/known data
    Consider an infinitely long continuous string with tension [tex]\tau[/tex]. A mass [tex]M[/tex] is attached to the string at x=0. If a wave train with velocity [tex]\frac{\omega}{k}[/tex] is incident from the left, show that reflection and transmission occur at x=0 and that the coefficients R and T are given.
    Consider carefully the boundary condition on the derivatives of the wave functions at x=0. What are the phase changes for the reflected and transmitted waves?

    2. Relevant equations
    i. [tex]R = sin^2\theta[/tex]
    ii. [tex]T = cos^2\theta[/tex]
    iii. [tex]tan\theta = \frac{M\omega^2}{2k\tau}[/tex]
    iv. [tex]\psi_1(x,t) = \psi_i + \psi_r = Ae^{i(\omega t - kx)} + Be^{i(\omega t + kx)}[/tex]
    v. [tex]\psi_2(x,t) = \psi_t[/tex], however, I do not know what this is.

    ** note ** [tex]\psi_i[/tex] is the incident wave, [tex]\psi_r[/tex] is the reflected wave, and [tex]\psi_t[/tex] is the transmitted wave

    3. The attempt at a solution
    I am used to dealing with situations where the string is of 2 different densities, therefore, [tex]\psi_t[/tex] will have a different value for k than [tex]\psi_i[/tex]. However, in this case, the densities are the same on either side of the mass, and the only obstruction is the mass. If I knew how to find an equation for [tex]\psi_2(x,t)[/tex], then I could potentially solve the rest of the problem.
    Thank you.
  2. jcsd
  3. Dec 3, 2006 #2
    I still haven't reached any solution to my problem.

    Any help is appreciated.
  4. Dec 4, 2006 #3
    EDIT: Ignore this post. The result leads nowhere.

    According to my calculations, this is true: [tex]\frac{d^2\psi_1}{dt^2}(0,t) = \frac{d^2\psi_2}{dt^2}(0,t)[/tex]

    Do you see why?

    Hint: Apply Newton's (2nd?) Law to the central mass and find an expression for the net force on [tex]M[/tex]
    Last edited: Dec 4, 2006
  5. Dec 4, 2006 #4
    I realise this is due in about ~1/2 hour, but the second boundary condition is given by:
    [tex]M\frac{d^2\psi_1}{dt^2} = M\frac{d^2\psi_2}{dt^2} = \tau \left( \frac{d\psi_1}{dx} - \frac{d\psi_2}{dx}\right) (0,t)[/tex]
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