Why is the Second Derivative Negative in SHM?

[Pseudo Statistic]

Say you had a spring hanging vertically-- spring constant k. A mass m is attached to it and it's at its equilibrium position, b.
Say, I decide to extend it (downward) such that its total extension distance from how it would be normally (without the mass) is x. Say I take the upward direction as positive. The moment I let go of it, the sum of the forces should be:
kx - mg = m \dfrac{d^{2}x}{dt^{2}}
Now, at some point I'd be asked a question (on a test) that says "Prove SHM".
To do so I have to get my force equation to the form \dfrac{d^{2}x}{dt^{2}} = - \omega^{2} (x-b), yet I can only do this if I had gotten:
mg - kx = m \dfrac{d^{2}x}{dt^{2}}
As my force equation.
Can someone tell me what I did wrong in forming my original force equation? I'd like to clear this up before my exam tomorrow.
Thanks alot.

 

[Doc Al]

Say, I decide to extend it (downward) such that its total extension distance from how it would be normally (without the mass) is x. Say I take the upward direction as positive.

Even though you took upward as positive, you still defined your "x" as being downward. You have the wrong sign for \dfrac{d^{2}x}{dt^{2}}. An upward acceleration would be - \dfrac{d^{2}x}{dt^{2}}.

 

[Pseudo Statistic]

Exactly why is \dfrac{d^{2}x}{dt^{2}} negative? Can you elaborate?
Thanks for the reply.

 

[Doc Al]

Exactly why is \dfrac{d^{2}x}{dt^{2}} negative? Can you elaborate?

When \dfrac{d^{2}x}{dt^{2}} is positive it means that the acceleration is in the +x direction. You defined x to be + downward. (Otherwise the restoring force would be -kx, not kx.)

You can stick with your definition of x, and call down to be postive. In which case the forces would be -kx & mg; set that sum equal to m \dfrac{d^{2}x}{dt^{2}}.