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Homework Help: Block attached to spring on slope

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A block of mass M can slide freely (without friction) on a flat surface held at an angle [itex] \theta [/itex] to the horizontal. The block is attached to a spring of natural length l and spring constant k. The other end of the spring is fixed to a nail driven into the sloping surface.

    i) At equilibrium, by how much is the spring extended?

    ii) The block is set in motion directly up and down the slope (i.e. not across the slope). Show that the result is simple harmonic motion with an angular frequency which does not depend on [itex] \theta [/itex].

    2. Relevant equations
    [tex] F=-kx [/tex]

    3. The attempt at a solution
    i) To be in equilibrium the net force on the block must be zero. So using vector diagram, I got:
    [tex] F_{spring}=Mgsin(\theta) [/tex]
    [tex] F_{normal}=Mgcos(\theta) [/tex]


    [tex] F_{spring}=Mgsin(\theta) =-kx [/tex]
    [tex] x=\frac{-Mgsin(\theta)}{k} [/tex]

    I know to show simple harmonic motion need to show that:
    [tex] a=-\omega^2x [/tex]
    I was thinking taking the second derivative of the x found in the above part to find a, but isn't that x the distance the spring is extended when at equilibrium, not a formula for the x position?
  2. jcsd
  3. Nov 12, 2014 #2

    Simon Bridge

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    ...You mean you don't know what the "x" in the formula you have to demonstrate means?
    You should ask then.

    But does it matter?
    Your process is the same: derive the formula for acceleration as a function of x and see if it matches. If you like you can keep track of both the overall extension and the extension at equilibrium and see which one matches the formula you have to show.
  4. Nov 12, 2014 #3
    Thanks for the reply

    So letting [itex] x_0 [/itex] be the distance from equilibrium.
    [tex] F=kx [/tex]
    [tex] ma=kx [/tex]
    [tex] a=\frac{k}{m}x [/tex]
    And since [itex] \omega =\sqrt{\frac{k}{m}}[/itex]
    [tex] a=\omega^2x [/tex]
    And the x found in part i) would be the distance from the natural length of the spring to equilibrium.
    Does this make sense?
  5. Nov 12, 2014 #4

    Simon Bridge

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    No - if you just let the box rest at equilibrium, what is it's acceleration?
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