1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...