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More SHM

  1. Jan 4, 2006 #1

    G01

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    A .1 kg ball oscillates horizontally on a spring on a frictionless Table. k is 2.5 N/m. Its velocity is .2m/s when x= -.05 m What is :

    a. The Amplitude
    b. The Max Acceleration
    c. What is the balls position when [tex] a = a_{max} [/tex]

    This one I know. Here x=the amplitude because a will be greatest when the spring is at its extremes.

    d. What is the speed of the ball at x = .03m

    So I know: [tex] \omega = \sqrt{\frac{k}{m}} = 5 rad/sec [/tex]

    [tex] T = 2\pi \sqrt{\frac{m}{k}} = 2/5\pi [/tex]

    [tex] x=A\cos (\omega t + \phi_0) [/tex]

    [tex] v_x = -\omega A\sin (\omega t + \phi_0) [/tex]

    So:

    [tex]x=A \cos (\frac{2\pi t}{T} +\phi_0)[/tex]

    I'm lost at how to use this info to solve the problem. Any hints?
     
    Last edited: Jan 4, 2006
  2. jcsd
  3. Jan 4, 2006 #2
    At certain time 't' velocity and position are given. Use this on the following to get 2 equations that you can solve.


    [tex] x=A\cos (\omega t + \phi_0) [/tex]

    [tex] v_x = -\omega A\sin (\omega t + \phi_0) [/tex]

    Use the fact that

    [itex] sin^2 \theta + cos^2 \theta = 1 [/itex]


    Accelaration can be found by differentiating [itex] v_x [/itex] with respect to 't' and then proceed to find [itex] a_{max} [/itex] from there.
     
    Last edited: Jan 4, 2006
  4. Jan 4, 2006 #3

    G01

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    OK I've tried adding the two equations together and I get:

    [tex] x+v_x=A\cos(\omega t + \phi_0) - \omega A\sin (\omega t +\phi_0) [/tex]

    I still don't see where to go from here. Sorry, I'm trying to teach SHM to myself and i guess i didn't do as great of a job as I thought. :)
     
  5. Jan 4, 2006 #4
    That is not what I said. Look at the following trig expression.

    [itex] sin^2 \theta + cos^2 \theta = 1 [/itex]

    I said use the above fact to some how get rid of sin and cosine from both of your equations

    So you need to first plug in the given values for x and vx.


    At time t, x = -0.5 m and vx = 0.2 m/s. After you plug in these values, do the following to eliminate the trig parts from your equations. See what you get . You should be able to find the Amphlitude.

    [itex] (\frac{x}{A}) ^2 + (\frac{v_x}{-wA}) ^2 = ?[/itex]
     
  6. Jan 5, 2006 #5

    G01

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    Ok I got the answer. Thank you very much.
     
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