Find The Natural Frequencies And Mode Shapes Of The System Shown In Fig For M1=m2=1kg

Find The Natural Frequencies And Mode Shapes Of The System Shown In Fig For M1=m2=1kg

Understanding the dynamic behavior of mechanical systems is crucial in designing structures and machinery that can withstand vibrations without failure. When analyzing a system's vibrational characteristics, two key concepts are the natural frequencies and the corresponding mode shapes. These are intrinsic properties of a system, indicating the frequencies at which the system tends to oscillate naturally and the deformation patterns associated with these oscillations. In this article, we will explore the process of determining the natural frequencies and mode shapes of a specific two-mass system, where the masses M1 and M2 are both 1 kg, as depicted in the hypothetical figure.

System Description and Assumptions

Overview of the System

Although the figure is not provided here, typical two-mass systems consist of two masses connected by springs and/or dampers. For the purpose of this analysis, we assume the system to be a simplified, idealized model with the following characteristics:

• Two point masses: M1 and M2, each with mass 1 kg.


• Connecting elements: Springs (with stiffnesses k1 and k2) connect the masses to each other and possibly to fixed supports.


• No damping: To focus solely on natural frequencies and mode shapes, damping is neglected.


• Linear, elastic system: The springs obey Hooke’s law, and displacements are small.

Typical Configuration Assumed

A common configuration involves:

• M1 connected to a fixed support via spring k1.


• M2 connected to M1 via spring k2.


• Both masses free to oscillate in one dimension (say, horizontal).

This configuration leads to coupled equations of motion, which can be analyzed to find the system's natural frequencies and mode shapes.

Mathematical Formulation of the System

Deriving Equations of Motion

Let’s denote:

• \( x_1(t) \): displacement of mass M1 from equilibrium.


• \( x_2(t) \): displacement of mass M2 from equilibrium.

Assuming the springs obey linear elasticity, the forces on each mass are:

• For M1: \( -k1 x1 \) from the fixed support and \( -k2 (x1 - x_2) \) from the spring connecting M1 and M2.


• For M2: \( -k2 (x2 - x_1) \).

Applying Newton’s second law:

\[
M1 \ddot{x}1 + (k1 + k2) x1 - k2 x_2 = 0
\]
\[
M2 \ddot{x}2 + k2 x2 - k2 x1 = 0
\]

Given \( M1 = M2 = 1\, \text{kg} \), the equations simplify to:

\[
\ddot{x}1 + (k1 + k2) x1 - k2 x2 = 0
\]
\[
\ddot{x}2 + k2 x2 - k2 x_1 = 0
\]

Assuming Harmonic Solutions

To find the natural frequencies, assume solutions of the form:

\[
x1(t) = X1 e^{i \omega t}
\]
\[
x2(t) = X2 e^{i \omega t}
\]

where \( \omega \) is the angular frequency to be determined, and \( X1, X2 \) are the mode shape amplitudes.

Substituting into the equations:

\[

• \omega^2 X1 + (k1 + k2) X1 - k2 X2 = 0

\]
\[

• \omega^2 X2 + k2 X2 - k2 X_1 = 0

\]

This yields a homogeneous system:

\[
[(k1 + k2) - \omega^2] X1 - k2 X_2 = 0
\]
\[

• k2 X1 + (k2 - \omega^2) X2 = 0

\]

Expressed in matrix form:

\[
\begin{bmatrix}
(k1 + k2) - \omega^2 & -k_2 \\

• k2 & k2 - \omega^2

\end{bmatrix}
\begin{bmatrix}
X_1 \\
X_2
\end{bmatrix}
= \mathbf{0}
\]

Non-trivial solutions exist only when the determinant of the coefficient matrix is zero:

\[
\det
\begin{bmatrix}
(k1 + k2) - \omega^2 & -k_2 \\

• k2 & k2 - \omega^2

\end{bmatrix}
= 0
\]

Calculating the determinant:

\[
[(k1 + k2) - \omega^2](k2 - \omega^2) - (k2)^2 = 0
\]

This characteristic equation allows us to solve for \( \omega^2 \), the squared natural frequencies.

Calculating the Natural Frequencies

Deriving the Characteristic Equation

Expanding the determinant:

\[
[(k1 + k2) - \omega^2](k2 - \omega^2) - k2^2 = 0
\]

\[
[(k1 + k2)k2 - (k1 + k2)\omega^2 - k2 \omega^2 + \omega^4] - k_2^2 = 0
\]

Simplify:

\[
(k1 + k2)k2 - (k1 + 2k2)\omega^2 + \omega^4 - k2^2 = 0
\]

Rearranged as:

\[
\omega^4 - (k1 + 2k2) \omega^2 + [(k1 + k2)k2 - k2^2] = 0
\]

Note that:

\[
(k1 + k2)k2 - k2^2 = k1 k2
\]

Thus, the quadratic in \( \omega^2 \) is:

\[
\omega^4 - (k1 + 2k2) \omega^2 + k1 k2 = 0
\]

Let \( y = \omega^2 \):

\[
y^2 - (k1 + 2k2) y + k1 k2 = 0
\]

The solutions:

\[
y{1,2} = \frac{(k1 + 2k2) \pm \sqrt{(k1 + 2k2)^2 - 4 k1 k_2}}{2}
\]

Corresponding natural frequencies:

\[
\boxed{
\omega{1,2} = \sqrt{y{1,2}}
}
\]

Numerical Example: Assigning Spring Constants

Suppose:

• \( k_1 = 100\, \text{N/m} \)


• \( k_2 = 50\, \text{N/m} \)

Compute:

\[
(k1 + 2k2) = 100 + 100 = 200
\]
\[
\Delta = \sqrt{200^2 - 4 \times 100 \times 50} = \sqrt{40,000 - 20,000} = \sqrt{20,000} \approx 141.42
\]

Then:

\[
y_1 = \frac{200 + 141.42}{2} = \frac{341.42}{2} \approx 170.71
\]
\[
y_2 = \frac{200 - 141.42}{2} = \frac{58.58}{2} \approx 29.29
\]

Thus:

\[
\omega_1 = \sqrt{170.71} \approx 13.06\, \text{rad/sec}
\]
\[
\omega_2 = \sqrt{29.29} \approx 5.41\, \text{rad/sec}
\]

The corresponding natural frequencies in Hz:

\[
f{1,2} = \frac{\omega{1,2}}{2\pi} \approx \frac{13.06}{6.283} \approx 2.08\, \text{Hz}
\]
\[
f_{2,2} = \frac{5.41}{6.283} \approx 0.86\, \text{Hz}
\]

Determining Mode Shapes

Eigenvector Computation

For each natural frequency, the mode shape corresponds to the eigenvector \( \mathbf{X} = [X1, X2]^T \). To find