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Oscillatory motion (vertical spring-mass system)

  1. Dec 4, 2009 #1
    "Resonance in an un-damped spring-mass system"

    If I have a force that is pushing upwards on a spring-mass system, and I basically have to find an equation that will give me the velocity and y-position for any given t, how much does that differ from the general form of Asin ([tex]\omega[/tex]t +[tex]\phi[/tex])?

    This is what I know:
    m= 2
    k= 32

    Vertical position y of mass: dy/dt = v

    Velocity of the mass: dv/dt = f(t)/M - (k/M) * y

    I know the external force is f(t) is 10sin([tex]\omega[/tex]t)

    I know that y(0) and v(0) are zero.

    I have to find the position and velocities when [tex]\omega[/tex]= 2, 3, 3.5, 4, 4.5, 5 rad sec and I have to use 40,000 time steps (totaling 25 seconds) where each step is 0.000625.

    I know that the "preferred" frequency of the system will be at 4 rad/ sec, using [tex]\omega[/tex]= [tex]\sqrt{\frac{k}{m}}[/tex]

    This is where I am having a bit of trouble: I found that when I'm examining 2 rad/sec, that v'' = 5cos2t - 16y', or that v'' = 5cos2t-16v. What I'm trying to do, is find the velocity and position of this system for any given value of t.

    I was told that I can use the "general" solution of this type: v= A cos 2t + B cos 4t + C sin 2t + D sin 4t.

    Am I in the right direction? Any insight would be appreciated. Thanks in advance!
  2. jcsd
  3. Dec 5, 2009 #2
    Is this being driven by an oscillitary force, such as F(t) = F0sin(ωt) ?
    Is ω0=sqrt(k/m), where ω0 ≠ ω ?

    Is this homework??

    Bob S
  4. Dec 5, 2009 #3
    No, this isn't homework.

    Yes, the external force that is applied is given by 10sin(wt) where w will vary.

    I am first examining what happens when w = 2 rad/sec.

    I've been working on this and came up with the following at 2 rad/ sec (don't know if it is correct)

    dv/ dt = 5sin(2t) - 16y

    since y' = v, y= v^2/2

    dv/ dt = 5sin(2t)- 8v^2

    dv/dt +8v^2 = 5sin (2t)
    8v^2 dv = 5sin (2t)dt

    (8/3)v^3 = (-5/2)cos(2t)
    v = cubed root of (-15/16 cos (2t))
  5. Dec 5, 2009 #4
    Isn't sqrt(k/m) fixed at sqrt(32/2) = w0 = 4 radians per second?

    Isn't your differential equation

    y'' +w02y = (10/m) sin(wt) [STRIKE]- g[/STRIKE] where the right hand term is the driving term?

    where w0= sqrt(k/m)

    Note that the units on the left hand and right hand sides have to match.

    Bob S
    Last edited: Dec 5, 2009
  6. Dec 5, 2009 #5
    Hi gramentz-

    Look at the thumbnail and see if it fits all conditions, including driving force and two initial conditions.

    [added] Here is a thumbnail plot of the solution using w=3.5 radians per second, and w0 = 4 radians per second. Both the amplitude and derivative are zero at t=0.

    Bob S

    Attached Files:

    Last edited: Dec 5, 2009
  7. Dec 6, 2009 #6
    When we assume the general solution of the form y= csin(w0t) + dcos(w0t)....is this because all of the derivatives will take such form? Since dy/dt = v in the original equation I was given, that implies that step 7 in the thumbnail could also be v(t)? Thank you for helping me with this.
  8. Dec 6, 2009 #7
    Hi gramentz-

    The particular solution provides the long-term (steady state) solution to match the driving force, especially when there is damping. But the particular solution does not match the initial conditions; specifically y(0) = y'(0) = 0. To do this, we need a solution to [4] with two arbitrary constants that can match the particular solution [1-3] to the two initial conditions. So that is why a general solution of the form [5] was chosen. The final solution [8] matches the two initial conditions at t=0, AND the particular equation [1] at all times.

    Bob S
    Last edited: Dec 6, 2009
  9. Dec 8, 2009 #8
    It took me quite some time, but I finally understand what you wrote and how you arrived at your answers. Thanks for taking the time to help me out, I appreciate it.
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