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Simple Harmonic

  1. Apr 28, 2008 #1
    [SOLVED] Simple Harmonic

    1. The problem statement, all variables and given/known data
    A 1.00 kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 19.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest with an initial position of xi = 0.200 m, and it subsequently undergoes simple harmonic oscillations.

    (f) Find the speed of the object when its position is equal to one third of the maximum value.
    (g) Find the acceleration of the object when its position is equal to one third of the maximum value.

    3. The attempt at a solution
    I have solved for w, and to find F) I used the conservation of energy to find the velocity. My question then is how to I find part G)?

    Can I do: a = w^2(1/3*A) which gives me a correct answer.
    However if I use the same method to find v: v= w(1/3*A) I get a wrong answer. Can anyone tell me why this is and if I need to use another method to solve for a.
     
  2. jcsd
  3. Apr 28, 2008 #2

    alphysicist

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    It has to do with the relative phases of x, v, and a. x and a are 180 degrees out of phase, so when one is at a positive maximum, then other is at a negative maximum. So at the time when the position is 1/3 of the amplitude, the acceleration magnitude is 1/3 of its maximum value.

    At the time when the position is 1/3 of its maximum value, the velocity is not at 1/3 its value, so you cannot do the same thing with the velocity.

    (The two easiest points to see are at the middle and the endpoint. When the particle is at the amplitude, the acceleration magnitude is a maximum but the velocity is not. When the particle is at the equilibrium position, so x=0, the acceleration is 0 but the velocity is not.)

    (Also, when you did a=w^2 (1/3A), since w^2=k/m, you were also doing ma=kx, so this becomes Newton's law for the acceleration. The velocity equation you mentioned as an incorrect example does not.)
     
  4. Apr 28, 2008 #3
    Ok, so that is a legitimate way for finding the acceleration, and will work all the time?
     
  5. Apr 28, 2008 #4

    alphysicist

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    If you compare the time dependent formulas for x and a (the ones with sines or cosines), you can see that:

    [tex]
    a = - \omega^2 x
    [/tex]

    so your way will work for a; also it is equivalent to newton's law [itex]\vec F=m\vec a[/itex].
     
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