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Oscillatory motion

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data
    mass of A=5,0kg
    mass of B=2,0kg
    friction coefficient between A and the plan=0,30

    The system is moving with an acceleration of 7,5 m /s^2 and the angle theta is constant and equals to 37º.

    Then, a mechanism makes the body A and the body B stops. B stop when it is in the position of the figure and it begins to oscillate in an simple harmonic motion.

    lenght of the wire that connects A to B = 0,85 m

    1) Calc the value of the frequency of pendulum motion.

    2)Write the equation of the elongation of the motion of body B.

    3) Calc the value of the maximum velocity of the motion and the first instant in which that velocity is reached.
    2. Relevant equations



    3. The attempt at a solution

    1)it gaves me 1,83 but it's wrong

    2) x(t)=Asin(wt+Q)
    A=0,51
    f=1,83
    Q= -pi/2 (this is wrong but I don't know why =S )
    x(t)=0,51sin(11,5t-pi/2)

    3) v(t)=[x(t)] '
    a(t)= [v(t)] '

    v_max => a(t) = 0 <=> v_max=5,9 m/s

    I haven't done the instant because this was already wrong
     

    Attached Files:

  2. jcsd
  3. Dec 26, 2009 #2

    AEM

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    For a simple pendulum, the period is given by [itex] 2 \pi \sqrt{ \frac{L}{g} } [/itex] where L is the length of the pendulum. The period and the frequency are related by T = 1/f .

    The purpose of this complicated arrangement is essentially to give the pendulum bob an initial velocity when it starts its oscillation. You will have to calculate this initial velocity in order to calculate Q. The way to do that will be to use the equation for the position of the pendulum bob and the equation for its velocity at the start of the oscillation (t = 0).

    That should get you started.
     
  4. Dec 27, 2009 #3
    But why Q= -pi/2 it's wrong? I dont understand :S
     
    Last edited: Dec 28, 2009
  5. Dec 30, 2009 #4

    AEM

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    I'm sorry I couldn't respond earlier. I was away.

    How did you calculate Q?
     
  6. Dec 31, 2009 #5
    x(t)=Asin(wt+Q)

    x(0)= -A
    -A=Asin(w.0+Q)
    -1=sin Q
    sin (-pi/2)=sinQ
    Q= -pi/2
     
  7. Dec 31, 2009 #6

    AEM

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    I just started to solve your problem, but I find that there is some information missing. As I read the problem statement and look at the diagram you included, here's what I think is happening: The masses A, and B, are accelerating to the right while C is accelerating downwards. This means that when block A is stopped, the mass B will have a velocity in the X direction. This velocity should be taken into account when you try to solve the equations for oscillatory motion. Therefore, you will need to calculate the motion of the system as a whole before you worry about the pendulum. You will need to know either how long the system accelerated, OR for how far it moved while it was accelerated. Then you can calculate the velocity of B in the X direction.

    Once you have the velocity of B horizontally, you will need to convert it into an initial angular velocity, call it [itex] \theta_0 [/itex]

    The frequency of oscillation is easy to calculate, as I mentioned above. However, you will have to use two equations simultaneously to find the amplitude and the phase angle (your Q).

    These equations are

    [tex] \theta(t) = A cos(\omega t + \phi) [/tex]

    and

    [tex] \frac{d \theta}{dt} = - \omega A sin(\omega t + \phi) [/tex]


    In your work, you have chosen to use [itex] x(t) = A sin( \omega t + Q) [/itex]. I think it would be better to work with angles than linear variables. Also, one can use either sine or cosine for the function representing the motion, but you should be aware that you will get different values for the angle [itex] \phi [/itex] ( your Q ). This angle reflects both which function you choose to express your answer in and how the motion begins. For example, if you start a pendulum from rest by pulling it back, holding it for an instant and then letting go, AND use the cosine expression above, the angle [itex] \phi [/itex] will be zero.

    You solve the two equations I wrote above by putting in the values that you have for the initial value of theta and initial angular velocity, and t = 0. It is also better for you to work with radian measure, rather than expressing your angles in degrees.

    If you can tell me how long the acceleration lasted, or for how far block A moved, I'll solve the problem and see if there are any other hints I can give you.
     
  8. Jan 9, 2010 #7
    i'm heading a problem in mathematical equation about the initial condition, what methode should we used if there is an addition force, such as damping force. This equation will be applied to performed oscillatory system in fortran language programme .
     
  9. Jan 10, 2010 #8

    AEM

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    Could you give us the differential equation that you've come up with? It should be a second order linear DE (unless you haven't made the small angle approximation) with constant coefficients. Such DEs can be solve exactly without much difficulty and the solution method can be found in any "Introduction to Differential Equations" textbook.
     
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