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    Here is the question and the answer

    Question
    A travelling wave has displacement given by:
    y = 0.1 sin(!t − kx)

    Sketch the initial profile between x = − and x = +; labelling the x
    axis in units of .

    Answer
    Sketch should show a graph of y as a function of x with shape like -sin
    (with y = 0 and decreasing at x = 0) covering 2 wavelengths moving between
    y = ±0.1.

    However surely it is to be drawn as a +sin wave as its 0.1sin and not -0.1sin?
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    (Original post by LowRider)
    Here is the question and the answer

    Question
    A travelling wave has displacement given by:
    y = 0.1 sin(!t − kx)

    Sketch the initial profile between x = − and x = +; labelling the x
    axis in units of .

    Answer
    Sketch should show a graph of y as a function of x with shape like -sin
    (with y = 0 and decreasing at x = 0) covering 2 wavelengths moving between
    y = ±0.1.

    However surely it is to be drawn as a +sin wave as its 0.1sin and not -0.1sin?
    I am trying to help but atm I cant read what some of it is ..could you edit it?
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    I didnt think the whole thing was needed but here it si

    Question
    A travelling wave has displacement given by:
    y = 0.1 sin((lambda)t − kx)

    Sketch the initial profile between x = −lambda and x = +lambda; labelling the x
    axis in units of lambda.

    Answer
    Sketch should show a graph of y as a function of x with shape like -sin
    (with y = 0 and decreasing at x = 0) covering 2 wavelengths moving between
    y = ±0.1.
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    Look for the key points.

    1) Initial profile => starting wave, i.e. t = 0. Therefore reduce equation to

    y = 0.1 sin(-kx)

    2) Use that k = 2Pi/lambda to get

    y= 0.1sin(-2pi*x/lambda)

    2) Now look for key points. Setting x = +lambda gives y = 0.1sin(-2*pi) = 0, so can now plot this point (x = lambda, y = 0)

    Set x = -lambda, get 0 again (y = 0.1sin(2*pi) = 0)

    For inbetween points set x = +-lambda/2 and see what happens...

    Hope that helps!
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    It would look like this correct?

    Sorry its messy

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    Thats almost it. Got the x intercepts right: zero at  x = \pm \frac{\lambda}{2} and  \pm \lambda and is max or min at  x = \pm \frac{\lambda}{4} and  x = \pm \frac{3\lambda}{4} Only mistake is the whole thing is upside down - must have forgotten the minus sign in the sin expression!
 
 
 
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