From "Differential Equations" 3rd Edition, Blanchard/Devaney/Hall Section 2.3, Question #13 I have the final answer from the text, as well as my Professors take (class has 7 people). Looking for a more thorough explanation. Work has already been graded, but I still don't get it completely. From the text: We consider a mass sliding on a frictionless table between two walls that are 1-unit apart and connected to both walls with springs, as shown below. |-////-M-\\\\-| (M is the mass... hope you get the idea from this crude "drawing") Let k1 and k2 be the spring constants of the left and right spring, respectively, let m be the mass, and let b be the damping coefficient of the medium the spring is sliding through. Suppose L1 and L2 are the rest lengths of the left and right springs, respectively. Write a second-order differential equation for the position of the mass at time t. [Hint: The first step is to pick an origin, that is, a point where the position is 0. The left hand wall is a natural choice.] (Incidentally, our professor wants us to choose the left hand wall as the origin). My attempt, my professors explanation, and the answer. My first thought was that we don't know if the springs in the drawing are compressed, under tension, or whether their lengths at rest are in excess of 1-unit. My professor said that the text is written by mathematics professors, and we are to assume that L1+L2 are less than 1-unit in length. With that I proceeded to set up a force equation. m(dx2/dt2)+b(dx/dt)+k1x+k2x = "Something" Turns out the "Something" is k1L1-k2(L2+1) I understand that this is the force experienced by the stationary block when pressed against the left wall (when x=0), but what is this equation really stating? Reading it all, I'm interpreting it as "The force due to the accelerated mass plus the drag/frictional force (actually minus) plus the spring force of spring one plus (actually minus) the spring force of the other spring is equal to the sum of the spring forces". For some reason I'm not getting it conceptually. Please help if you can.