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Two blocks and a spring system - Finding equation of motion

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A system is composed of two blocks of mass ##m_1## and ##m_2## connected by a massless spring with spring constant k. The blocks slide on a frictionless plane. The unstretched length of the spring is ##l##. Initially ##m_2## is held so that the spring is compressed to ##l/2## and ##m_1## is forced against a stop, as shown. ##m_2## is released at ##t=0##.

    Find the motion of the center of mass of the system as a function of time.


    2. Relevant equations



    3. The attempt at a solution
    At any time t, let the distance of block ##m_1## from the wall be ##x_1## and that of ##m_2## be ##x_2##. The extension in the spring length is ##x_2-x_1-l##.
    Applying Newton's second law on ##m_1##,
    $$k(x_2-x_1-l)=m_1\ddot{x_1} \Rightarrow \ddot{x_1}=\frac{k(x_2-x_1-l)}{m_1} (*)$$
    Similarly,
    $$\ddot{x_2}=-\frac{k(x_2-x_1-l)}{m_2} (**)$$
    Subtracting (*) from (**) and substituting ##z=x_2-x_1-l## and ##\mu=m_1m_2/(m_1+m_2)##,
    $$\ddot{z}=-\frac{k}{\mu}z$$
    Solution of the above differential equation is of the form,
    $$z(t)=A\sin(\omega t)+B\cos(\omega t)$$
    where ##\omega=k/\mu##.
    At t=0, z(0)=-l/2, z'(0)=0 hence, ##A=0## and ##B=-l/2##. The equation of motion is
    $$z(t)=-\frac{l}{2}\cos(\omega t)$$
    Is the above equation correct? The question asks about the motion of center of mass and I am in a dilemma if the equation I have reached represents that. If not, what does the above equation represent then? :confused:

    Any help is appreciated. Thanks!
     

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  3. Aug 20, 2013 #2

    TSny

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    Block 1 does not move for a while after the system is released - it remains pushed against the wall by the spring until the spring extends enough to pull the block to the right.

    Since you are only interested in the motion of the center of mass, you can try using the famous theorem that the center of mass of a system moves like a point mass having a mass equal to the total mass of the system and acted on by a single force equal to the net external force acting on the system.
     
  4. Aug 20, 2013 #3
    The net force acting on the system is the normal force on block ##m_1##. The normal force varies as ##m_2## moves, I am unsure how to write it in equations. :confused:

    Let the distance of ##m_2## be ##x## from the wall. Then change in length of spring is ##l-x##. The normal force on ##m_1## is ##k(l-x)## to the right. Hence,
    $$k(l-x)=(m_1+m_2)\ddot{x}_{CM}$$
    ##x_{CM}=m_2x/(m_1+m_2) \Rightarrow \ddot{x}_{CM}=m_2\ddot{x}/(m_1+m_2)##
    $$\Rightarrow k(l-x)=m_2\ddot{x}$$.
    This doesn't look right to me. :confused:
     
  5. Aug 20, 2013 #4

    TSny

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    That's right.

    If the left end of the spring were just attached directly to the wall, could you describe the motion of m2 after it is released? With the left end attached to m1 against the wall, would the motion of m2 be any different up until the time m1 leaves the wall?

    Does that help to see how the spring force acting on m1 will change with time? Will that help in finding the time dependence of the normal force acting on m1?
     
  6. Aug 20, 2013 #5
    Yes, it will be SHM.
    No.
    Won't the normal force be equal to the spring force for the time ##m_1## is attached to wall?
     
  7. Aug 20, 2013 #6

    TSny

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    Yes, that's correct.
     
  8. Aug 20, 2013 #7
    Can you please check my equations in post #3?
     
  9. Aug 20, 2013 #8

    TSny

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    Looks like the equations are correct to me. But I don't really like using ##x## for the position of mass 2 relative to the wall. Would you mind using ##x_2## instead? That way, you can let ##x## represent the amount of stretch of the spring from it's natural length. Then you can just write Hooke's law in the usual form ##F = -kx##. Note that ##x_2 = l+x##.

    The reason for harping on the notation is that you are going to be interested in the external force from the wall which, as you already said, is determined by the spring force which is most simply expressed in terms of ##x##.

    Also, you should be able to write down the time dependence of ##x## immediately from your knowledge of simple harmonic motion. (At least up to the time that m1 leaves the wall.) No need to set up and solve a differential equation, unless you just want to.
     
  10. Aug 20, 2013 #9

    TSny

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    Just realized that since you can write down the stretch of the spring ##x(t)## from knowledge of SHM, you can then easily get ##x_2(t)## and therefore ##x_{CM}(t)## up until m1 leaves the wall. After that, the motion of the CM is simple.

    I solved it originally by working with the external force from the wall. But that's not necessary. Sorry for not seeing that earlier.
     
  11. Aug 21, 2013 #10
    The position of ##m_2## from the wall for the time ##m_1## is attached to wall can be represented by
    $$x_2(t)=l\left(1-\frac{\cos(\omega t)}{2}\right)$$
    where ##\omega=\sqrt{k/m_2}##.
    Hence, the position of CM of system,
    $$x_{CM}=\frac{m_2l}{m_1+m_2}\left(1-\frac{\cos(\omega t)}{2}\right)$$
    I am confused as to how I would represent the motion of CM after ##m_1## detaches from the wall? Do I follow the same approach as I did in my post #1? But that would give me the extension of spring as a function of time.
     
  12. Aug 21, 2013 #11
    What happens with the CM when there are no external forces?
     
  13. Aug 21, 2013 #12

    TSny

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    That looks correct.

    This is where the theorem about the motion of the center of mass that I cited earlier will help.
     
  14. Aug 21, 2013 #13
    The net external force on the system is zero. This means that the velocity of CM stays constant.

    When ##m_1## is about to move, the velocity of ##m_2## is ##l\omega/2##. The velocity of CM at that instant is ##\displaystyle \frac{m_2l\omega}{2(m_1+m_2)}## and this stays constant for the rest of the motion. Hence, the position of CM can be represented by
    $$x_{CM}=\frac{m_2l\omega}{2(m_1+m_2)}\left(t+\frac{\pi}{2\omega}\right)$$

    Correct?

    EDIT: I think its better to write
    $$x_{CM}=\frac{m_2l\omega}{2(m_1+m_2)}t$$

    for ##t \geq \pi/(2\omega)##.
     
    Last edited: Aug 21, 2013
  15. Aug 21, 2013 #14

    TSny

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    That's now quite correct if you are measuring the position of the CM from the wall. Look at where that puts the CM at time ##t = \pi/(2\omega)##
     
  16. Aug 21, 2013 #15
    At ##t=\pi/(2\omega)##,
    $$x_{CM}=\frac{ml\pi}{4(m_1+m_2)}$$
    This is wrong. :confused:

    Does this mean the above applies only for ##t>\pi/(2\omega)##?
     
  17. Aug 21, 2013 #16

    TSny

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    What is the length of the spring when m1 leaves the wall? Where does that put the CM at that instant?
     
  18. Aug 21, 2013 #17
    It is unstretched when ##m_1## leaves the wall.
    At a distance ##m_2l/(m_1+m_2)## from the wall.
     
  19. Aug 21, 2013 #18

    TSny

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    Right. But that doesn't agree with what you wrote for the location of the CM at ##t = \pi/(2\omega)##.
     
  20. Aug 22, 2013 #19
    Yes. I modify my answer.
    $$x_{CM}(t)=\frac{m_2l}{(m_1+m_2)}\left(1-\frac{\cos(\omega t)}{2}\right) \, \, \text{for} \, \, t\leq \frac{\pi}{2\omega}$$
    $$x_{CM}(t)=\frac{m_2l\omega}{2(m_1+m_2)}t\, \, \text{for}\, \, t>\frac{\pi}{2\omega}$$
    Is this correct?
     
  21. Aug 22, 2013 #20

    TSny

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    Note that at ##t =\frac{\pi}{2\omega}## your solution for ##t \leq \frac{\pi}{2\omega}## does not agree with your solution for ##t > \frac{\pi}{2\omega}##. Decide which one is correct, and then modify the other.
     
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