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Standing Wave Question?

  1. Dec 4, 2004 #1
    Adjacent antinodes of a standing wave on a string are a distance 14.5 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 8.00×10-2 s . The string lies along the + x - axis and is fixed at x = 0.

    Find the displacement of a point on the string as a function of position and time.
    Find the speed of propagation of a transverse wave in the string.
    Find the amplitude at a point a distance 3.20 to the right of an antinode.


    - the first question was pretty easy easy i just used the formula :
    A*cos(2*pi*t/T)*sin(2*pi*x/(2*s)) T = period A = amplitude
    and i got the result: 3.25*10^(-3)m. which is correct, now can anybody give me any clues on how to calculate the speed of the propagation and the amplitude at 3.20 to the right of the antinode?
     
  2. jcsd
  3. Dec 4, 2004 #2

    Pyrrhus

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    Homework Helper

    if i remember correctly

    [tex] v = \lambda f [/tex]

    [tex] v = \frac{w}{k} [/tex]

    For the third question you know an atinode is located at a distance of [itex] \frac{\lambda}{4} [/itex] so [itex] x = \frac{\lambda}{4} + 3.2 [/itex]

    Also
    [tex] k = \frac{2 \pi}{\lambda} [/tex]

    Remember Stationary waves formula

    [tex] y = (2A \sin kx) \cos \omega t [/tex]
     
    Last edited: Dec 4, 2004
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