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Simple Harmonic Motion-Mass

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A 0.2 kg object is suspended from a spring with a spring constant of k=10 N/m and is undergoing simple harmonic motion. What is its acceleration of the object at the instant when it is -0.05 m away from equilibrium?

    A. 1000 m/s2
    B. 40 m/s2
    C. 0.1 m/s2
    D. 2.5 m/s2


    2. Relevant equations

    F=ma
    F=-kx
    F=mg

    3. The attempt at a solution

    F=F(gravity)+F(spring)
    ma=mg+(-kx)
    0.20kg*a=(0.20kg*9.81m/s^2)+(-(10N/m*-0.05m))
    a=12.31m/s^2

    The answer is supposed to be D
     
  2. jcsd
  3. Nov 3, 2011 #2

    BruceW

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    Homework Helper

    Yeah, this is a tricky question if you don't already know the answer.

    The equilibrium point will shift due to the gravity force, so when the object is -0.05m away from equilibrium, this will not be the value of x.

    To calculate the new equilibrium point, just solve for a=0. And then you can get the value of x which is -0.05m from the equilibrium point.
     
  4. Nov 3, 2011 #3

    gneill

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    Staff: Mentor

    Sure it will, if x is the displacement from equilibrium.
     
  5. Nov 3, 2011 #4

    gneill

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    Staff: Mentor

    A straightforward way to approach this problem is to use write an expression for the position versus time (use a sine or cosine function). Find the corresponding acceleration expression (a touch of simple calculus). Take the position expression and rearrange for an expression for t in terms of x. Use it to find acceleration in terms of x.
     
  6. Nov 3, 2011 #5

    BruceW

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    Homework Helper

    I was referring to her/his? equations, where x is not the displacement from equilibrium. Your post isn't very helpful.

    Yep, that's another way of doing it.
     
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