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Finding the Amplitude of a spring (Simple Harmonic Motion)

  1. Dec 9, 2008 #1
    [SOLVED] Finding the Amplitude of a spring (Simple Harmonic Motion)

    First post here at PF, so forgive me if I make a faux pas. I'm trying to study for an upcoming Physics test and I'm having a bit of trouble with this.

    1. The problem statement, all variables and given/known data
    A massless spring with spring constant 19 N/m hangs vertically. A body of mass 0.20 kg is attached to its free end and then released. Assume that the spring was un-stretched before the body was released. Find

    a. How far below the initial position the body descends, and the
    b. Frequency of the resulting SHM.
    c. Amplitude of the resulting SHM.


    2. Relevant equations
    [tex]a = -\omega^2x = \frac{d^2x}{dt^2} = -\frac{kx}{m}[/tex]

    [tex]x = A\cos(\omega t+\phi)[/tex]

    [tex]f = \frac{1}{T} = \frac{\omega}{2\pi}[/tex]

    [tex]F = -kx[/tex]

    [tex]F = ma[/tex]

    3. The attempt at a solution
    I found (b), the frequency, since
    [tex]f = \frac{\omega}{2\pi} = \frac{\sqrt{k/m}}{2\pi} = \frac{\sqrt{(19 N/m)/(0.2 kg)}}{2\pi} = \frac{\sqrt{95/s^2}}{2\pi} = \frac{\sqrt{95}}{2\pi s} \approx 1.551250 Hz[/tex]
    However, I am having a hard time finding how far it will initially fall before coming back up. Once I have that, I could find the amplitude (or vice versa), I think. I know it will have to do with gravity, but I'm not sure how to set up this part of the problem.

    Any help is appreciated. Thanks.
     
    Last edited: Dec 10, 2008
  2. jcsd
  3. Dec 9, 2008 #2

    alphysicist

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    Hi Maxx573,

    Try using conservation of energy to find how far it descends. What do you get?
     
  4. Dec 9, 2008 #3
    Looking at the conservation of energy:

    [tex]K = \frac{1}{2} kA^2\sin^2(\omega_0 t+\phi) = \frac{1}{2}mv^2[/tex]

    [tex]U = \frac{1}{2} kA^2\cos^2(\omega_0 t+\phi) = \frac{1}{2}kx^2[/tex]

    [tex]E = \frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]

    OK, obviously something here should help me, but I just keep running into the same things. When [tex]U = E[/tex], then [tex]x = A[/tex] and [tex]v = 0[/tex]. When [tex]K = E[/tex], then [tex]a = 0[/tex] and [tex]t = 0[/tex]. Acceleration is at its maximum when all the energy is potential energy (and velocity is 0), and velocity is at its maximum when the spring is at equilibrium (and acceleration is 0).

    All of this makes perfect sense to me, but I haven't been able to apply it to solving for [tex]A[/tex] or [tex]x[/tex] without already knowing one of them. I feel like the answer is staring me in the face, but I just can't find it.
     
  5. Dec 10, 2008 #4

    alphysicist

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    In setting up this problem I would say there are three types of energies to keep track of: kinetic, gravitational potential, and spring potential. The two points you are interested in are the initial position, and the lowest point in the motion.

    What is the kinetic, gravitational potential, and spring potential energies at the intial point? at the final point? Setting the total energy at the initial point equal to the total energy at the final point should give you the answer.
     
  6. Dec 10, 2008 #5
    Ah, I think I've got it now.

    So for the system's initial energy:
    [tex]K = 0[/tex] (everything is at rest)
    [tex]U_{spring} = \frac{1}{2}(19 N/m)(0)^2 = 0[/tex] (the initial position is 0)
    [tex]U_{gravitational} = (0.2 kg)(9.8 \frac{m}{s^2})(x)[/tex]
    (for this problem, I will refer to the lowest point as h=0, so initially h=x)

    And the system's energy at the bottom of the bottom of the oscillation:
    [tex]K = \frac{1}{2}m(0)^2 = 0[/tex] (velocity is 0)
    [tex]U_{spring} = \frac{1}{2}(19 N/m)x^2[/tex]
    [tex]U_{gravitational} = (0.2 kg)(9.8 \frac{m}{s^2})(0) = 0[/tex] (height is now 0)

    [tex]E_0 = E_1[/tex]
    [tex](0.2 kg)(9.8 \frac{m}{s^2})x = \frac{1}{2}(19 N/m)x^2[/tex]
    [tex]1.96 N x = 9.5 (N/m) x^2[/tex]
    [tex]1.96 N = 9.5 (N/m) x[/tex]
    [tex]1.96m = 9.5x[/tex]

    [tex]x \approx 0.2063158 m[/tex]

    And since the mass starts from the point where the spring is neither compressed nor stretched, [tex]x = A[/tex]

    That right?
     
  7. Dec 10, 2008 #6

    alphysicist

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    That looks right to me.

    This isn't quite right. Remember that x=0.206m is the distance from the highest to the lowest point of the oscillation. But where is the amplitude measured from?
     
  8. Dec 10, 2008 #7
    Oh right. Amplitude is measured from the center of the oscillation to the highest (or lowest) point. So [tex]A = \frac{x}{2} \approx 0.103m[/tex]

    Thanks a ton :)
     
  9. Dec 10, 2008 #8

    alphysicist

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    Glad to help!
     
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