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Energy of a transverse wave

  1. Jul 20, 2017 #1
    1. The problem statement, all variables and given/known data
    There's a string with tension T & mass density μ that has a transverse wave with ψ(x,t) = f(x±vt). f(x) is an even function & goes to zero as x→±∞
    Show that the total energy in the string is given by ∫dw*T*((f'(w))2; limits of integration are ±∞

    2. Relevant equations
    The kinetic energy of an infinitesimal part of string is KE = 0.5*μ*dx*(dψ/dt)2.

    Its potential energy is 0.5*T*dx*(dψ/dx)2.

    Note that w=ck. c = √(T/μ)

    3. The attempt at a solution
    Total energy = KE + PE = μ/2[(dψ/dt)2 + v2*(dψ/dx)2]. I add them & integrate over ±∞. However, where does the f(w) come from. What does it have to do with f(x±vt)?
     
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  3. Jul 20, 2017 #2

    haruspex

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    You have gone a bit wrong by eliminating T and keeping v. The form of answer has T but not v.

    Also, you have omitted the dx terms from the total energy. Please post your working for the integration step.
     
  4. Jul 20, 2017 #3
    Ok, so I figured out that dψ/dx = dψ/dt. So infinitesimal energy = μ/2(2V2*(dψ/dx)2)dx. Since v2 = T/μ, infinitesimal energy = T*((dψ/dx)2)dx. Now I integrate to get total energy for whole string: total energy = ∫T*((dψ/dx)2)*dx. Limits of integration are ±∞. I don't see how they go from this to the answer.
     
  5. Jul 20, 2017 #4

    haruspex

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    That is not even dimensionally correct.
    Differentiate f(x+vt) wrt each of x, t.
     
  6. Jul 21, 2017 #5
    df/dx = f(x+vt)*dx;
    df/dt = f(x+vt)*v*dt
    infinitesimal energy = KE + PE = μ/2[(f2(x+vt)*dx+v2*f2(x+vt)*dt] = 0.5*μ*(1+v2)*f2(x+vt)*(dx+dt)

    Total energy = ∫0.5*μ*(1+v2)*f2(x+vt)*dx I'm guessing dt goes to zero since we're integrating with respect to space/over entire spring & not time. Is this right?
     
  7. Jul 21, 2017 #6
    Oh I forgot something:

    df/dx = f'(x+vt)*dx;
    df/dt = f'(x+vt)*v*dt

    So KE + PE = μ/2[(f'2(x+vt)*dx+v2*f'2(x+vt)*dt] = 0.5*μ*(1+v2)*f'2(x+vt)*(dx+dt)

    Forgot to add primes in front of f's to indicate derivative.
     
  8. Jul 21, 2017 #7

    haruspex

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    You do not want dx in there twice. Delete one of them. Similarly dt in the next line.
    For the integral, you are interested in the total energy at any instant, so what is the variable of integration?
     
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