Spring Motion

$$8x’’ + 1.4x’ + 10x = 0$$ $$x(0) = -4$$ $$x’(0) = 0$$

We start solving this model of harmonic motion by taking a look at the differential equation itself. It's a second-order equation with the mass $(m = 8)$, damping $(d = 1.4)$, and spring constant $(k = 10)$ as the coefficients. The external forcing term is $0$. The initial time, $t_0$, is 0, the inital position, $x(0)$, is $-4$, and the initial velocity, $x’(0)$, is $0$.

$$p(r) = 8r^2 + 1.4r + 10$$

As this equation is homogenous, $b(t) = 0$, it is easy to solve. First, we will find the roots of the corresponding characteristic polynomial in order to determine what type of solution we are going to get.

$$r = \frac{-1.4 \pm \sqrt{(1.4)^2 - 4(8)(10)}}{2(8)}$$ $$r =-0.0875 \pm 1.1146047505730452i$$

The roots of the characteristic polynomial can easily be found using the quadratic formula. In this case, our roots are complex as the discriminant ($d^2 - 4(m)(k)$) is less than $0$. This means that the spring's motion is under damped. We will now use this information to get the fundamental set of solutions for this equation.

$$\{ e^{-0.09t}\cos(1.11t), e^{-0.09t}\sin(1.11t) \}$$

This is the fundamental set of solutions for this equation. If we just wanted the general solution to the equation, we could use the set of solutions to find it and be done. However, we want the particular solution to this differential equation given the initial conditions (otherwise, we wouldn't be able to graph it or model its motion). We thus will need to find the coefficients.

$$x(t) = Ae^{-0.09t}\cos(-0.09t) + Be^{-0.09t}\sin(-0.09t)$$

$$x’(t) = A(-0.09e^{-0.09t}\cos(1.11t) - 1.11e^{-0.09t}\sin(1.11t))$$ $$+ B(-0.09e^{-0.09t}\sin(1.11t) - 1.11e^{-0.09t}\cos(1.11t))$$

Now that we have the general solution, we can solve for the particular solution to the differential equation by solving for the coefficients $A$ and $B$. To do so, we will also need to find the the derivative of the general solution. Unfortunately, because this solution has exponentials multiplying sines and cosines, we will need to use product rule. You should be able to find this derivative yourself, but we have provided it here for reference.

$$x(0) = -4 = A$$ $$x’(0) = 0 = 1.11B + 0.35$$ $$B = -0.31$$

Fortunately because $t_0 = 0$, $\sin(1.11(0)) = 0$. This makes $A$ easy to find, as the exponential and the cosine simplify to $1$, meaning that $A$ is equal to $x_0$. We can then easily find $B$ by plugging in $A$'s value to the derivative of the general solution at $v_0$ and solving for $B$.

$$x(t) = -4e^{-0.09t}\cos(1.11t) - 0.31e^{-0.09t}\sin(1.11t)$$

Plugging the values we found for $A$ and $B$ into the general solution gives us the particular solution you see when you first click solve. This is what we were looking for, so we are done with the problem.