INSTRUCTOR: A 2 kilogram mass is attached to a spring with spring constant 24 newtons per meter. The system is then immersed in a medium imparting a damping force equal to 16 times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point 40 centimeters below equilibrium.

I'm just going to go ahead and note that our mass is equal to 2. So m is equal to 2. Our spring constant is 24 newtons per meter, so k is equal to 24. And b, our damping coefficient, is 16 [INAUDIBLE] it's 16 times the instantaneous velocity of the mass. So this means that the equation we are attempting to solve is 2 times the second derivative of x plus 16 times the first derivative of x plus 24x equals 0.

And also, a few other things we should note. It is released from rest at a point 40 centimeters below equilibrium. This means, for our solution, x of 0 is equal to 0.4, because 40 centimeters is 0.4 meters.

Also, because it is released from rest, this means that x prime of 0 is equal to 0. It is not pushed off at initial speed or anything like that. In finding our exact equation of motion, we will need those values.

All right. So let's next consider our characteristic equation. That is 2 lambda squared plus 16 lambda plus 24 equals 0, which factors.

We can factor a 2 out. So 2, parentheses, lambda squared plus 8 lambda plus 12, close parentheses, equals 0, which means that lambda is equal to negative 2 or negative 6. And actually, I'm going to treat those the other way. So lambda is equal to negative 6 or negative 2, just to be consistent with my other work. So our general solution is x of t is equal to c1 times e to the negative 6t, end exponent, plus c2 times e to the negative 2t, end exponent.

All right. Well, using the fact that x of 0 equals 0.4, I have a system of equations to find c1 and c2. Using the x of 0 equals 0.4, that means that c1 plus c2 equals 0.4. And using the fact that x prime of 0 equals 0 tells me that negative 6 times c1 minus 2 times c2 equals 0.

Combining these two equations together, using elimination or substitution or some other way of solving a system of equations, we see that c1 equals negative 0.2, and c sub 2 equals 0.6. This means that our equation of motion is x of t equals negative 0.2 times e to the negative 6t, end exponent, plus 0.6 times e to the negative 2t, end exponent.