Vibration Analysis

There are multiple ways of solving this problem. The first step is to derive the equations of motion of the system. Let the x coordinates of the masses from left to right be $x_1$ , $x_2$ and $x_3$ . Let the natural lengths of the springs be $L_1$ and $L_2$ . The equations of motion of the system are found either using the 2nd law of motion or Lagrangian mechanics. The derivation is left out of this solution.

$M\ddot{x}_1 -2K(x_2-x_1-L_1) =0 \ \dots \ (1)$ $M\ddot{x}_2 +2K(x_2-x_1-L_1) -K(x_3-x_2-L_2)=0 \ \dots \ (2)$ $M\ddot{x}_3 + K(x_3-x_2-L_2)=0 \ \dots \ (3)$

Let:

$x_2 - x_1 - L_1 = y_1$ $x_3 - x_2 - L_2 = y_2$

Now, performing $(3)-(2)$ and $(2)-(1)$ and simplifying the expressions, and using $M=K=1$ leads to:

$\ddot{y}_1 = -4y_1 + y_2 \ \dots \ (4)$ $\ddot{y}_2 = 2y_1 - 2y_2 \ \dots \ (5)$

Manipulating (4) and (5) to obtain a differential equation purely in terms of $y_1$ leads to:

$\frac{d^4 y_1}{dt^4} + 6\frac{d^2 y_1}{dt^2} + 6y_1 =0$

The same governing differential equation is found for $y_2$ . Now, to find the frequencies of oscillation, let us assume a solution of the form:

$y_1 = A\sin(\omega t + \phi)$

Plugging this into the differential equation leads to:

$(\omega^4 - 6\omega^2 + 6)y_1=0$ $\implies \omega^4 - 6\omega^2 + 6=0$

The above equation is a quadratic equation in $\omega^2$ . The sum of the roots of this equation are:

$\boxed{\omega_1^2 + \omega_2^2 = 6}$

The natural frequencies of the system are also known as the eigenfrequencies of the system. The term 'eigen' is used here as the above result can also be obtained by solving an eigenvalue problem. The reader may try to derive and solve this eigenvalue problem by him/her self. The corresponding eigenvectors to each eigenfrequency are known as natural modes of oscillation or eigenmodes.