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Amplitude of standing wave.

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.386 m. The maximum transverse acceleration of a point at the middle of the segment is 9000 m/s^2 and the maximum transverse velocity is 3.00 m/s.


    2. Relevant equations
    y(x,t) = A*sin(kx)*sin(wt)

    dy(x,t)/dt = A*w*sin(kx)*sin(wt)

    d^2y(x,t)/dt^2 = -A*w^2*sin(kx)*sin(wt)


    3. The attempt at a solution

    I equated the max velocity with the secound derivative of the stading wave function.

    3.00 = A*w*sin(kx)cos(wt)

    Quick question: The x in the equation refers to the location on the string, not the
    displacement of the string from equilibrium correct?

    2nd question: is max velocity still at displacement = 0, and max acceleration at x =
    1*amplitude?

    --------------------------------------------------------------------------------

    From here I tried to solve for k, first by solving for lambda.

    lambda = 2L/n

    lambda = 2L (because n = 1, because (sorry) f = fundemental frequency

    lambda = 0.772 m

    from there I solved for k

    k = 2pi/lambda

    k = 8.1388 /m

    ------------------------------------------------------------------------------

    using this I solved for w (omega) using w = v*k

    I used max velocity for this equation

    w = 24.417/s

    now, does the fact this is MAX velocity effect the caluculation?

    -------------------------------------------------------------------------------

    I failed at solving for t, and I don't want to waste your time, so I won't repeat my
    attempts here.

    -------------------------------------------------------------------------------

    From there I subbed the values into the equation and attempted to solve for A
    (amplitude).
     
  2. jcsd
  3. Oct 13, 2009 #2
    I've thought about this question a little more. If velocity and acceleration are maxed, then the cos/sin values must be equal to one. So, I've done this work:





    1: A*w = 3.00
    2: A*w^2 = 9000

    1: w = 3.00/A

    subbing into 2

    A*(3.00/A)^2 = 9000

    then:

    9.00/A = 9000

    A = 9.00/9000

    A = 0.00100m = 1.00*10^-3 m

    Does this look correct to you learned folk? I only ask because there is no answer in the textbook.
     
  4. Oct 14, 2009 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct, as both the maximum acceleration and maximum speed occurs at the same place, at the middle of the string. Was it the maximum amplitude the question?

    ehild
     
  5. Oct 15, 2009 #4
    Oh, yes it was, sorry. Thanks for the help.
     
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