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Cauchy Boundary Conditions on a Wave

  1. May 3, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    So using the D'Alembert solution, I know the solution of the wave equation is of the form:

    y(x,t) = f(x-ct) + g(x+ct)

    I'm told that at t=0 the displacement of an infinitely long string is defined as y(x,t) = sin (pi x/a) in the range -a<= x <= a

    and y =0 otherwise.

    The string is initially at rest.

    I'm told that the waves move along the string with speed c and told to sketch the displacement of the string at t=0, t=a/2c and t=a/c

    2. Relevant equations



    3. The attempt at a solution

    So substituting t=0 into the d'alembert solution gives

    f(x) + g(x) = sin pix/a

    similarly since the string is initially at rest, we can calculate that f(x) - g(x) = const. therefore f(x) = 1/2sin pix/a + k where k is some const. and g(x) = 1/2 sin pix/a - k

    So is the full solution y(x,t) = 1/2 [ sin (pi(x-ct)/a) + sin(pi(x+ct)/a) ] ? Isn't this a stationary wave..? Im not sure how to sketch for t = a/2c etc...thanks :)
     
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  3. May 3, 2010 #2

    ideasrule

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    I think you're overcomplicating this. You can easily sketch the wave at t=0; the equation's given to you. For t=a/2c, the wave travels at speed c, so the pattern at t=0 would just be shifted by a/2c*c=a/2.
     
  4. May 3, 2010 #3

    bon

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    Thanks..I do see that, but I'm trying to understand the general method so that I can apply it to situations where the boundary conditions aren't so simple...

    any guidance would be great
    thanks
     
  5. May 4, 2010 #4

    ehild

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    No, you do not get standing waves on an infinite string.
    Remember that f(x)=g(x)= sin(pix/a) is valid only in the interval between [-a, a ], and both functions are 0 everywhere else.
    Letting time run, the argument of f(x) is replaced by x-ct and that of g(x) by x+ct. The condition above means that

    f(x) = 1/2 sin(pi/a(x-ct)) if -a<=x-ct <=a that is -a+ct<=x<=a+ct

    and g(x)= 1/2 sin(pi/a(x+ct)) if -a<=x+ct <=a that is -a-ct<=x<=a-ct ,

    and y=0 everywhere else.

    Can you sketch y(x,t) now?

    ehild
     
    Last edited: May 4, 2010
  6. May 4, 2010 #5

    bon

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    Sorry I'm not quite sure I agree with what you've written here :S

    I thought it was that f(x) + g(x) = sin pix/a (rather than f(x) = g(x) = ...)

    and f(x) - g(x) = constant.

    So f(x) = 1/2 sin pix/a + k
    and g(x) = 1/2 sin pix/a - k

    so f(x) + g(x) = 1/2 sin pi(x-ct)/a + 1/2 sin pi (x+ct)/a...? Over the limits on x that you have spelt out..

    which seem to be two waves of equal magnitude, phase etc moving in opposite directions? i.e. a standing wave?
     
  7. May 4, 2010 #6

    ehild

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    OK, you are right, but that constant is arbitrary, so choose it zero for start.
    The sum of two waves of the same frequency and amplitude moving in opposite direction in a closed interval would produce a standing wave, but it is not the case here.
    Your function is not y(x.t)=1/2 sin (pi(x-ct)/a)+ 1/2 sin (pi(x+ct)/a) in the whole interval from minus infinity to infinity, but you need a piece-wise definition:

    y(x,t) =f((x-ct) + g(x+ct)

    where f(x) = 1/2 sin(pi/a(x-ct)) if ct-a<=x<=ct +a , zero otherwise

    and g(x)= 1/2 sin(pi/a(x+ct)) if -ct-a<=x<=-ct +a, zero otherwise.

    y(x,t) = 0 outside [-ct-a, -ct+a] U[ct-a, ct+a].

    Try to sketch y for t = 4a/c, you will understand what I mean.

    If you want to include an arbitrary constant, you can do it, by subtracting it form g and adding to f, it is just shifting them up or down.

    ehild
     
  8. May 4, 2010 #7

    bon

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    Ahh great help. I see now. Thanks ehild.
     
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