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Block hitting spring

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A small block of mass m=1 kg and v0=6.32m/s hits a massless relaxed spring with spring constant k= 3 N/m, which starts to be compressed as the block continues to move horizontally. There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 0.7 m^−1. For simplicity that static and dynamic friction coefficients are the same.

    What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds)

    https://courses.edx.org/static/content-mit-801x~2013_SOND/html/final_p06.png [Broken]


    2. Relevant equations
    I chose x axis positive to the right. x0=0 at point of contact, x1=first momentarily stop
    Ffriction=-μmg=-αxmg
    Fspring=-kx

    Newton's Second Law
    ∑F=m*d2x/dt2

    3. The attempt at a solution

    -kx-αxmg=m*d2x/dt2
    -(k+αmg)x=m*d2x/dt2

    new konstant K=k+αmg
    -K/m=d2x/dt2

    This is SHO with general solution x(t)=Acos(ωt+θ)
    ω=√(K/m)=3.16

    To find A, and θ our initial conditions are:
    x0=x(t=0)=0
    v0=v(t=0)=6.32 m/s

    Thus, 0=x(t=0)=Acos(θ)
    This is true when θ=pi/2 or -3/2pi . Question: which angle should I use ???

    v0=v(t=0)=6.32/-ω=Asin(θ)

    A=√(x(t=o)^2+(v(t=0)/ω)^2)=√(0+(6.32/3.16)^2/)=2

    So, I know that my amplitude is 2, x1=2m.

    (Trying with 3/2pi. pi/2 can't feet because I am getting negative time.)

    2=x(t1)=2*cos(3.16t-3/2pi)

    1=cos(3.16t-3/2pi)
    This is true when (3.16t-3/2pi)=0
    t1=1.49 sec.

    Please, check my solution if it seemed to be right.
    And my t1 should be also half a period T=pi/ω=pi/3.16=0.99s. Isn't it? I am confused.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 12, 2014 #2
    This is not SHO because there is friction between the block and the surface.
     
  4. Jan 12, 2014 #3
    So, what approach should I take instead?
     
  5. Jan 12, 2014 #4
    Do you have any ideas?
     
  6. Jan 12, 2014 #5
    I have found how far it travels using work - energy theorem. May be using equations of motion: x(t)=x0+v0t+1/2at^2, v(t)=v0+at. and for acceleration put acceleration from N2L -(k+αmg)/m*x=d2x/dt2. May be then I can find my t?
     
  7. Jan 12, 2014 #6
    I have re-read the problem and your original solution, and I must say that I misunderstood what you were doing.

    While the system is not an SHO, it is not incorrect to treat it as an SHO until the block stops because the equation is the same.

    You have found the angular frequency ##\omega## of this system. How does that correspond to the period? What part of the period is spent by an SHO between max kinetic energy and max potential energy?
     
  8. Jan 12, 2014 #7
    T=2pi/ω. Between those two conditions block on spring spends half period T/2=pi/ω ?
     
  9. Jan 12, 2014 #8
    Correct. So what is your answer?
     
  10. Jan 12, 2014 #9
    T=pi/ω=pi/3.16=0.99 s.

    But I tried using equations of motion: v(t)=v0+at. for acceleration I put acceleration from N2L -(k+αmg)/m*x=d2x/dt2.

    0=v(t1)=vo-(k+αmg)/m*x*t
    x1=2
    t1=.316 s.

    Something wrong.
     
  11. Jan 12, 2014 #10
    Your equation of motion is $$ - {k + \alpha m g \over m} x = {d^2 x \over dt^2} .$$ This is a differential equation and its solution cannot be obtained by multiplying it by ##t##. In fact, its solution is given by ## x = A \cos (\omega t + \theta) ## as you wrote originally.
     
  12. Jan 12, 2014 #11
    But this equation v(t)=v0+at is also differential v(t)=v0+d2x/dt2*t . I have acceleration in both equations. I substitute one into another. Wanna understand why this is not right thing to do?
     
  13. Jan 12, 2014 #12
    ## v(t) = v_0 + at ## is only correct when ## a = \mathrm{const} ##. Which is not the case here. In this case, we have ## v(t) = -A \omega \sin (\omega t + \theta) ##, and ## a = -A \omega^2 \cos (\omega t + \theta) \ne \mathrm{const} ##.
     
  14. Jan 12, 2014 #13
    ahhh! sure, a is constant! that's why you can't. Thanks!
     
  15. Jan 12, 2014 #14


    Could either of you please explain why half of the period should be used?
     
  16. Jan 12, 2014 #15
    I found the max displacement for the mass via the energy work theorem, basically
    -> Potential Energy became Kinetic Energy at the end of the track
    -> in the collision no energy is lost as it is stated by the problem, so
    -> it is KE = Work F_spring + Work F_friction
    -> they both change with x so i took the integral from x0 to x1 for both the Work done
    that way I found max displacement.

    1) is that correct?
    2) is it possible to find the time without SHO equations?
     
    Last edited: Jan 12, 2014
  17. Jan 12, 2014 #16
    @ dummyano
    1) Yes, the way you found x is correct.
    2) It seems you can't find t without SHO

    I am very much in doubt about finding t from T/2 . It doesn't agree with t found from solving SHO general solution equation. See above. I am stuck.
     
  18. Jan 12, 2014 #17
    because it when object slides to the right it makes half a period.
     
  19. Jan 12, 2014 #18
    I am very much in doubt about finding t from T/2 . It doesn't agree with t found from solving SHO general solution equation. See above.

    Can you please comment?
     
  20. Jan 12, 2014 #19
    Thank you for answering.
    I will try SHO for t and post the result then
     
  21. Jan 12, 2014 #20

    TSny

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    Gold Member

    What fraction of a period does it take for a mass in SHM to travel from the equilibrium position to the first time it comes to rest?
     
    Last edited: Jan 12, 2014
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