# Harmonic Oscillation problem

by Dracovich
Tags: harmonic, oscillation
 Sci Advisor HW Helper P: 3,147 Does this look familiar? $$\frac {d^2x}{dt^2} + \frac {k}{m} x = -g$$
 P: 87 Yeah i at least know the first part, although now i am a bit confused (i should've caught and asked about this in class). Since the textbook says: $$\frac {d^2x}{dt^2} = a = -kx$$ But isn't -kx suppose to give F not a ? And i'm assuming you took $$\frac {d^2x}{dt^2} + kx= mg$$ And got $$\frac {d^2x}{dt^2} + \frac {k}{m} x = -g$$ But i'm afraid i don't see how those two amount to the same, since $$\frac {d^2x}{dt^2} + kx= 0$$ but mg=F I just hope i got all the latex correct o_O not very used to that
 P: 87 Harmonic Oscillation problem Oh wait i think i just got it. So you got $$\frac {d^2x}{dt^2} + kx = 0$$ As the formula for simple harmonic oscillation, and then the resulting forces are mg+kx=0 which leads to what you just said. Although i'm afraid i don't see where that leads me :/
 Sci Advisor HW Helper P: 3,147 Let's back up. Newton's law of motion says force = mass times acceleration. In your situation the two forces are the spring force (Hooke's Law) and gravity. So $$m \frac {d^2x}{dt^2} = -kx -mg$$ The k here is the spring constant. Now divide both sides by m to get my previous equation. $$\frac {d^2x}{dt^2} + \frac {k}{m} x = -g$$ If the g weren't there, could you solve the equation? (Hint: It's a simple harmonic oscilator!) The presence of the constant g on the right side does alter the nature of the differential equation. For example, using the simple change of variables $$y = x + \frac {mg}{k}$$ gives $$\frac {d^2y}{dt^2} + \frac {k}{m} y = 0$$ This actually answers your original question!
 Sci Advisor HW Helper P: 3,147 Just replace x with $$y - \frac {mg}{k}$$ in the DE and remember that the second part is a constant!