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What is the frequency of the oscillation?

  1. Dec 11, 2004 #1
    A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially
    held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and
    oscillates up and down, with its lowest position being 10 cm below yi.
    (a) What is the frequency of the oscillation?
    (b) What is the speed of the object when it is 8.0 cm below the
    initial position?
    (c) An object of mass 300 g is attached to the first object, after which the system oscillates
    with half the original frequency. What is the mass of the first object?
    (d) Relative to yi, where is the new
    equilibrium (rest) position with both objects attached to the spring?
     
  2. jcsd
  3. Dec 11, 2004 #2

    Pyrrhus

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    Homework Helper

    Apply Newton's 2nd Law to get the equation of motion for this system.

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    [tex] mg - kx = m \ddot{x} [/tex]

    [tex] g - \frac{k}{m}x = \ddot{x} [/tex]

    this gives a second order differential equation with the following solution:

    [tex] x = A \sin (\omega t - \phi) + B [/tex]
     
  4. Dec 11, 2004 #3
    For finding the frequency, I did mg=kx, and k=mg/x. since f=1/2pi*sqrt(k/m) I substitute mg/x with k. But it was wrong.
     
    Last edited: Dec 11, 2004
  5. Dec 11, 2004 #4
    the frequency is:

    [tex]
    f = \frac{\omega}{2\pi}
    [/tex]

    and

    [tex]
    \omega = \sqrt{\frac{k}{m}}
    [/tex]
     
    Last edited by a moderator: Dec 12, 2004
  6. Dec 11, 2004 #5
    Hmm.. Did they tell you what the spring constant is?
     
  7. Dec 11, 2004 #6
  8. Dec 12, 2004 #7
    Actually, if it has the spring got a gravitational acceleration then you can say that.

    [tex]
    g = \omega ^2x
    [/tex]

    But i am still not sure about this.
     
  9. Dec 12, 2004 #8
    If you have omega then:

    for b) you can let x(t) = 8 and find t.
    then take the derivative of x(t) and find the velocity.

    I think that should work.
     
  10. Dec 12, 2004 #9
    how did you come up with the equation g=w^2*x?
     
  11. Dec 12, 2004 #10
    well - if u take the second derivative of x(t)
     
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