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HELP! Harmonic Motion

  1. Dec 9, 2004 #1
    Can you offer some help on the following question? I've been trying different things for a while but can't get the right answer. Thanks!

    At time t = 0, consider a 1/2 wavelength long section of the rope which is carrying the wave y = 0.06 cos(2.1 t - 2.4 x) between two points which have zero displacement (y = 0). Find the total force exerted by the rest of the rope on this section. Neglect any effects due to the weight of the rope. Use the small-angle approximation where q, sin(q), and tan(q) are all approximately equal to each other. A 5 meter length of this rope has a mass of 1.5 kg.

    What I have so far:
    I know the horizontal components of the Tensions cancel out, so the net force is due to the Y component of the one of the forces times 2. How do I find that?

    Thanks a lot!
     
  2. jcsd
  3. Dec 9, 2004 #2

    Andrew Mason

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    The force on the rope is the vector sum of the tensions at each end. Since horizontal tension T is constant,

    [tex]T_1 = T/cos\alpha_1 \text{ and } T_2 = T/cos\alpha2[/tex]

    You can determine the slope of the rope at each end (dy/dx) by partial differentiation of the wave function to find [itex]\alpha_1 \text{ and } \alpha2[/itex] where y = 0 (ie. [itex]2.1 t - 2.4 x = \pi/2, 3\pi/2[/itex])

    You should be able to determine the speed of the wave from the wave function using the general solution:
    [tex]y = Asin(2\pi/\lambda (x - \nu t))[/tex]

    to the wave equation for the ideal string:

    [tex]\frac{\partial ^2y}{\partial x^2} = \frac{\rho}{T}\frac{\partial ^2y}{\partial t^2}[/tex]

    and then use
    [itex]v = \sqrt{T/\rho} = \lambda \nu [/tex] to find the Tension T.

    AM
     
    Last edited: Dec 9, 2004
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