
We're going to start by considering a simplified version of the SMD model. We'll assume that there is no damping in the system and that the external force is zero. So, the only way the system will oscillate is if it has some non-zero initial displacement and/or velocity. We refer to this as _undamped free vibration_.

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/undamped-free-vibration/img1.jpg"
	alt="Spring-mass model with no external force. | EngineeringSkills.com"
	width="500"
	height="313"
	caption="Fig 1. Spring-mass model with no external force."
/>{" "}

With these simplifications, the differential equation of motion reduces to,

$$
m\ddot{u} + ku = 0
\tag{1}
$$

The typical method of analytically solving a differential equation is to assume a solution and substitute that solution back into the original equation. This case is no different. Looking again at the differential equation we note that we need a candidate solution for $u$ that we can add to its second derivative. A function that satisfies this requirement is $e^{\lambda t}$, where $\lambda$ is an arbitrary constant. So, we start by assuming a candidate solution,

$$
u = e^{\lambda t}
\tag{2}
$$

We now differentiate our candidate solution and substitute it back into our original equation,

$$
\ddot{u} = \lambda^2 e^{\lambda t}
\tag{3}
$$

Substituting into the original equation,

$$
m\lambda^2 e^{\lambda t} + k e^{\lambda t} = 0
\tag{4}
$$

We now rearrange to isolate the <em>characteristic equation</em>,

$$
e^{\lambda t}\:\underbrace{(m\lambda^2  + k)}_\text{characteristic equation}  = 0
\tag{5}
$$

Since we know, $e^{\lambda t}\neq 0$,

$$
\begin{align}
& m\lambda^2 + k = 0\\
& \lambda =\pm \sqrt{\frac{-k}{m}}\\
& \lambda =\pm \sqrt{-1}\sqrt{\frac{k}{m}}\\
& \lambda =\pm i \sqrt{\frac{k}{m}}
\end{align}
\tag{6}
$$

$\lambda$ represents the two roots of the characteristic equation.

### Root 1: $\lambda_1$

$$
\lambda_1 = i\:\sqrt{\frac{k}{m}}
\tag{7}
$$

This yields one solution for $u$,

$$
u_1 = e^{\lambda_1 t} = \exp\left[i\sqrt{\frac{k}{m}}\:t\right]
\tag{8}
$$

### Root 2: $\lambda_2$

The other solution is,

$$
u_2 = e^{\lambda_2 t} = \exp\left[-i\sqrt{\frac{k}{m}}\:t\right]
\tag{9}
$$

Therefore the **general solution** is any linear combination of these two solutions,

$$
u = C_1\:\exp\left[i\sqrt{\frac{k}{m}}\:t\right] + C_2\:  \exp\left[-i\sqrt{\frac{k}{m}}\:t\right]
\tag{10}
$$

where $C_1$ and $C_2$ are constants. Using Euler's formula,

$$
\begin{align}
e^{ix} &= \cos(x) + i\sin(x)\\
e^{-ix} &= \cos(x) - i\sin(x)
\end{align}
\tag{11}
$$

the general solution can be represented as,

$$
\boxed{u = A\cos\left(\sqrt{\frac{k}{m}}t\right) + B\sin\left(\sqrt{\frac{k}{m}}t \right)}
\tag{12}
$$

Remember, this is a general solution to the equation that still contains the unknown constants $A$ and $B$. These are determined using two initial conditions for the SDoF system. The initial conditions are typically the position and velocity at some time $t$. We'll see how to use the initial conditions to determine $A$ and $B$ soon.

In the next lecture we'll introduce an important property of the undamped spring-mass system, the natural frequency.


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