Impulse force can be defined as a large magnitude force that acts for a short time.

An example of an Impulse force is a blast load applied on top of a building. For this experiment, a building with two floors is considered. A blast load is a load applied on the structure which comes immediately after an explosion. It combines overpressure and impulse which lasts for a short duration of time.

The response of the structure due to the impulse load is influenced by the stiffness and mass of the structure. However, the stiffness and mass of the structures are influenced by the nature of construction and materials used.

The magnitude of the impulse force can be calculated by the impulse momentum principle.

$\text{Impulse} = F \Delta T = m \dot{x}_2 - m \dot{x}_1$

The initial conditions are given by

$x(t=0) = x_0 = 0$
$\dot{x}(t=0) = \dot{x}_0 = \frac{1}{m}$

The governing differential equation are,

$m_1 \ddot{x}_1(t) + (k_1 + k_2) x_1(t) - k_2 x_2(t) = 0$
$m_2 \ddot{x}_2(t) - k_2 x_1(t) + k_2 x_2(t) = 0$

Writing in matrix form,

$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{Bmatrix} \ddot{x}_1(t) \\ \ddot{x}_2(t) \end{Bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \begin{Bmatrix} x_1(t) \\ x_2(t) \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}$

Where,

$x_1(t) = X_1 \cos(\omega t + \phi)$
$x_2(t) = X_2 \cos(\omega t + \phi)$

Substituting this and writing in the matrix form,

$\begin{bmatrix} -m_1 \omega^2 + k_1 + k_2 & -k_2 \\ -k_2 & -m_2 \omega^2 + k_2 \end{bmatrix} \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}$

To obtain the solution the determminant of the coefficient of $X_1$ and $X_2$ must be zero,

$(m_1 m_2) \omega^4 - \{(k_1 + k_2) m_2 + k_2 m_1\} \omega^2 + (k_1 + k_2) k_2 - k_2^2 = 0$

where, $\omega_1$ and $\omega_2$ are roots of the equations,

$\omega_1^2, \omega_2^2 = \frac{1}{2} \left\{ \frac{(k_1 + k_2) m_2 + k_2 m_1}{m_1 m_2} \right\} \mp \frac{1}{2} \left\{ \left( \frac{(k_1 + k_2) m_2 + k_2 m_1}{m_1 m_2} \right)^2 - 4 \frac{(k_1 + k_2) k_2 - k_2^2}{m_1 m_2} \right\}^{1/2}$

The mode shape,

For the first mode $X_1 = \begin{Bmatrix} X_1^{(1)} \\ X_2^{(1)} \end{Bmatrix} = X_1^{(1)} \begin{Bmatrix} 1 \\ r_1 \end{Bmatrix} \quad \text{where } r_1 = \frac{X_2^{(1)}}{X_1^{(1)}}$

Fot the second mode $X_2 = \begin{Bmatrix} X_1^{(2)} \\ X_2^{(2)} \end{Bmatrix} = X_1^{(2)} \begin{Bmatrix} 1 \\ r_2 \end{Bmatrix} \quad \text{where } r_2 = \frac{X_2^{(2)}}{X_1^{(2)}}$

The general form of equation of motion of $m_1$ and $m_2$ are,

$x_1 = X_1^{(1)} \cos(\omega_1 t + \phi_1) + X_1^{(2)} \cos(\omega_2 t + \phi_2)$
$x_2 = r_1 X_1^{(1)} \cos(\omega_1 t + \phi_1) + r_2 X_1^{(2)} \cos(\omega_2 t + \phi_2)$

The initial conditions (at $t = 0$) are,

$x_1 = 0 \quad \dot{x}_1 = 0$
$x_2 = 0 \quad \dot{x}_2 = 0$

Based on the initial conditions, $X_1^{(1)}$, $X_1^{(2)}$, $\phi_1$ and $\phi_2$ can be obtained.