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Simple harmonic motion and oscillation

  1. Jun 14, 2008 #1
    A 780-g mass is attached to a vertical spring and lowered slowly until the mass rests at its equilibrium position 30 cm below the original length of the spring. It is then set into simple harmonic motion.

    (a) What is the frequency and period of the oscillation?
    (b) What is the maximum velocity of the oscillating mass? Where does the maximum velocity occur?
    (c) What is the maximum acceleration of the oscillating mass? Where does the maximum acceleration occur?
    (d) Draw a diagram showing how the total energy of the oscillating system changes with time.

    Can someone get me off to a start here? I don't understand how to find the frequency without knowing Lamda or angular frequence (w)..
  2. jcsd
  3. Jun 14, 2008 #2
    What are the equations that you know? We have to know what you know, so that we can pick out what you are struggling with. Maybe to get you started

    [tex]m\ddot{x} + kx=0[/tex]
    leads to
    [tex]x(t)=A_0 cos(\omega t + \delta)[/tex]

    where [itex]\omega = \sqrt{k/m}[/itex]. So now you know the angular frequency.
  4. Jun 15, 2008 #3

    Yeah, but what is k? The spring constant? or does it have another meaning?
  5. Jun 18, 2008 #4
    I just don't get the k part..
  6. Jun 18, 2008 #5
    k is the spring constant make a free body diagram and you'll see!
  7. Jun 18, 2008 #6
    Large value of k means that the spring resists being stretched, a very hard spring. A small value of k means the spring is very soft and can be stretched quite easily. Restoring force of a spring is F= -k(delta s). Opposite in direction to the displacement.
  8. Jun 18, 2008 #7

    I still don't see.. I only know the weight, and the distance from it's equillibrium to rest position.
  9. Jun 18, 2008 #8
    Tell me, if you had to associate familiar physics term with weight, what would it be?
  10. Jun 19, 2008 #9
    Huh? W=ma?
  11. Jun 20, 2008 #10
    Weight is a force. Are spring constants associated with a force at all? What does it mean when something is in equilibrium?
  12. Jun 20, 2008 #11

    Oh, i see, well the weight acting down is .78*9.8=7.644 (mg)
    that means there must be an equal force acting up which must be 7.644 in the opposite direction. So is this equal force the spring constant? (k)
  13. Jun 20, 2008 #12
    A force constant like k expresses the force generated per unit of some variable (whether it be displacement or acceleration)
  14. Jun 21, 2008 #13

    in other words
  15. Jun 21, 2008 #14
    okay, so k(-.3)=7.644

    T=2pi*sqrt(m/k) (period of a loaded spring)

  16. Jun 22, 2008 #15
    just like that?
  17. Jun 22, 2008 #16
    Just like that.
  18. Jun 22, 2008 #17

    for part b, the max velocity is:

    well, i know two equations

    vmax=wA (w being angular velocity)
    however, i also know that velocity is greatest at the equilibrium position
    so how do i find this? do i need to know k?

    time out, i know k..k=25.48
    so vmax=sqrt(25.48/.78)A
    vmax=5.715(A) but is A (the amplitude) the 30cm?
  19. Jun 22, 2008 #18
    You won't be able to find a numerical answer for the amplitude with the information provided. You would need to know it's initial position and initial velocity when it was first set into harmonic motion.

    An algebraic answer should be fine.
  20. Jun 24, 2008 #19
    so just leave it at:
  21. Jun 26, 2008 #20
    Yep, I don't see anyway around the lack of initial conditions.
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