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Maximum force on rope and wavefunction

  1. Jul 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A sinusoidal wave on a rope with linear density ##\mu=0.012 kg/m## is described in SI units by $$\xi(x,t)=A sin (kx-\omega t)= 0.15sin(0.8x-50t)$$
    a) find the maximum acceleration of a rope element
    b) find the maximum transverse force on a piece of rope ##1 cm## long
    c) Show how the force depends on the tension of the rope
    2. Relevant equations
    Wave equation

    3. The attempt at a solution
    a) This is simply ##a_{max}=\omega^2 A=375 m/s^2##
    b) This should be ##F_{max, y}=m a_{max}=\mu \cdot 0.01 \cdot \omega^2 A = 0.045 N##
    c) Tension should be related to such force by ##F_{y}=-T \frac{\partial \xi}{\partial x}## but here is the problem, since
    $$\frac{\partial \xi}{\partial x}=-0.12 sin(0.8x-50t) $$

    So the maximum value is ##0.12 ## and the force should be

    $$F_{max,y}= 0.12 T= 0.12 \cdot \mu \cdot v^2$$

    Now ##v=\frac{\omega}{k}=\frac{50}{0.8}=62.5 m/s## so it should be

    $$F_{max,y}=0.09 N$$

    Different from the value found in b), which is strange, did I get something wrong?
     
    Last edited: Jul 2, 2016
  2. jcsd
  3. Jul 2, 2016 #2
    I think I have a clue but I'm not sure... when you say that Fy= -T de/dx ...

    Isn't it Fy= -2T de/dx at the maximum point ? Since the tension acts on both sides of the rope element

    And the derivative of the sine is the cosine
     
    Last edited: Jul 2, 2016
  4. Jul 5, 2016 #3

    haruspex

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    In the equation F=ma, how exactly is F defined? Constrast that with tension in a rope.
     
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