Interactive Lecture
Understanding Two Degree of Freedom Undamped Vibration Systems
by Dr Arnaldo Delli Carri
A two-degree-of-freedom (2DOF) system requires two independent coordinates to completely describe its motion. These systems are fundamental in structural dynamics and mechanical vibrations, representing the next level of complexity after single degree of freedom systems.
Key Characteristics:
For a 2DOF undamped system with two masses \(m_1\) and \(m_2\) connected by springs \(k_1\) (connecting mass 1 to wall) and \(k_2\) (connecting masses 1 and 2):
\([M]\{\ddot{x}\} + [K]\{x\} = \{0\}\)
Where:
The natural frequencies are found by solving the eigenvalue problem:
\(\det([K] - \omega^2[M]) = 0\)
This yields two natural frequencies: \(\omega_1\) and \(\omega_2\) where \(\omega_1 < \omega_2\)
Each natural frequency has an associated mode shape that describes the relative motion of the two masses.
Mode Shapes: Each mode shape \(\{\Phi_i\}\) represents a pattern of motion where all parts of the system oscillate at the same frequency and pass through equilibrium simultaneously. The two masses move in-phase for the first mode and out-of-phase for the second mode.
Any free vibration is a superposition of the two modes. Collect the mode shapes as columns of the modal matrix \([\Phi] = [\{\Phi_1\}\ \{\Phi_2\}]\) and write the motion as \(\{x(t)\} = [\Phi]\{q(t)\}\), where \(q_1(t)\) and \(q_2(t)\) are the modal (principal) coordinates — the generalised coordinates of the system expressed in the eigenbasis. Each one obeys its own uncoupled equation, so
\(q_i(t) = A_i\cos(\omega_i t) + B_i\sin(\omega_i t)\)
The constants \(A_i, B_i\) follow from the initial conditions via \(\{q(0)\} = [\Phi]^{-1}\{x(0)\}\) and \(\{\dot{q}(0)\} = [\Phi]^{-1}\{\dot{x}(0)\}\). With \(B_i = \dot q_i(0)/\omega_i\), a system released from rest keeps only the cosine terms.
This simulation demonstrates the free vibration response of a 2DOF system. The visualisation shows three different scenarios:
Two degree of freedom systems are fundamental models in structural dynamics and mechanical engineering:
| Concept | Equation | Description |
|---|---|---|
| Equations of Motion | \([M]\{\ddot{x}\} + [K]\{x\} = \{0\}\) | Matrix form for 2DOF system |
| Eigenvalue Problem | \(([K] - \omega^2[M])\{\Phi\} = \{0\}\) | Solve for natural frequencies |
| Natural Frequencies | \(\omega_1, \omega_2\) | Two distinct natural frequencies |
| Mode Shapes | \(\{\Phi_1\}, \{\Phi_2\}\) | Relative motion patterns |
| General Solution | \(\{x(t)\} = \{\Phi_1\}(A_1\cos\omega_1 t + B_1\sin\omega_1 t) + \{\Phi_2\}(A_2\cos\omega_2 t + B_2\sin\omega_2 t)\) | Free response; constants \(A_i, B_i\) set by initial conditions |
A two-degree-of-freedom system has two masses \(m_1 = 2 \text{ kg}\) and \(m_2 = 3 \text{ kg}\) connected by springs with stiffnesses \(k_1 = 500 \text{ N/m}\) and \(k_2 = 800 \text{ N/m}\). Calculate the two natural frequencies of the system.
Show Solution
Step 1: Set up the mass and stiffness matrices
\([M] = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\), \([K] = \begin{bmatrix} 1300 & -800 \\ -800 & 800 \end{bmatrix}\)
Step 2: Formulate eigenvalue problem
The natural frequencies are found from: \(\det([K] - \omega^2[M]) = 0\)
\(\det \begin{bmatrix} 1300-2\omega^2 & -800 \\ -800 & 800-3\omega^2 \end{bmatrix} = 0\)
Step 3: Expand the determinant
\((1300-2\omega^2)(800-3\omega^2) - (-800)^2 = 0\)
\(1,040,000 - 3900\omega^2 - 1600\omega^2 + 6\omega^4 - 640,000 = 0\)
\(6\omega^4 - 5500\omega^2 + 400,000 = 0\)
Step 4: Solve the quadratic in \(\omega^2\)
Let \(x = \omega^2\): \(6x^2 - 5500x + 400,000 = 0\)
\(x = \frac{5500 \pm \sqrt{5500^2 - 4(6)(400,000)}}{12} = \frac{5500 \pm \sqrt{20,650,000}}{12} = \frac{5500 \pm 4544.2}{12}\)
\(\omega_1^2 = \frac{5500 - 4544.2}{12} = 79.65 \text{ rad}^2/\text{s}^2\), \(\omega_2^2 = \frac{5500 + 4544.2}{12} = 837.0 \text{ rad}^2/\text{s}^2\)
Natural frequencies:
\(\omega_1 = \sqrt{79.65} = 8.92 \text{ rad/s} = 1.42 \text{ Hz}\)
\(\omega_2 = \sqrt{837.0} = 28.93 \text{ rad/s} = 4.60 \text{ Hz}\)
Answer: \(\omega_1 = 8.92 \text{ rad/s}\) (1.42 Hz) and \(\omega_2 = 28.93 \text{ rad/s}\) (4.60 Hz).
For a 2-DOF system with equal masses \(m_1 = m_2 = 1 \text{ kg}\) and spring constants \(k_1 = k_2 = k = 1000 \text{ N/m}\), determine:
Show Solution
Step 1: Natural frequencies
\([M] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), \([K] = \begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} = \begin{bmatrix} 2000 & -1000 \\ -1000 & 1000 \end{bmatrix}\)
Solve \(\det([K] - \omega^2[M]) = 0\):
\((2000-\omega^2)(1000-\omega^2) - (-1000)^2 = 0\)
\(\omega^4 - 3000\omega^2 + 10^6 = 0\)
\(\omega^2 = \dfrac{3000 \pm \sqrt{3000^2 - 4(10^6)}}{2} = \dfrac{3000 \pm 2236.07}{2}\)
\(\omega_1^2 = 381.97 \text{ rad}^2/\text{s}^2,\quad \omega_2^2 = 2618.03 \text{ rad}^2/\text{s}^2\)
\(\omega_1 = \sqrt{381.97} = 19.54 \text{ rad/s}\) (lower frequency)
\(\omega_2 = \sqrt{2618.03} = 51.17 \text{ rad/s}\) (higher frequency)
Step 2: Mode shapes
Mode 1: Substitute \(\omega_1^2 = 381.97\) into \(([K] - \omega^2[M])\{\Phi\} = 0\). Row 1 gives \((2000 - 381.97)\Phi_1 - 1000\,\Phi_2 = 0\), so \(\Phi_2 = 1.618\,\Phi_1\):
Mode shape: \(\{\Phi_1\} = \begin{bmatrix} 1 \\ 1.618 \end{bmatrix}\)
Mode 2: Substitute \(\omega_2^2 = 2618.03\). Row 1 gives \((2000 - 2618.03)\Phi_1 - 1000\,\Phi_2 = 0\), so \(\Phi_2 = -0.618\,\Phi_1\):
Mode shape: \(\{\Phi_2\} = \begin{bmatrix} 1 \\ -0.618 \end{bmatrix}\)
Step 3: Physical interpretation
Mode 1 (in-phase): Both masses move in the same direction, with mass 2 swinging 1.618 times as far as mass 1. The spring \(k_2\) stretches by \(\Phi_2 - \Phi_1 = 0.618\) per unit of \(\Phi_1\).
Mode 2 (out-of-phase): The masses move in opposite directions, with mass 2 moving 0.618 times as far as mass 1 and against it. The spring \(k_2\) now stretches by \(\Phi_2 - \Phi_1 = -1.618\) — the larger relative deformation of the two modes.
Answer: \(\omega_1 = 19.54 \text{ rad/s}\) with \(\{\Phi_1\} = [1,\ 1.618]^T\); \(\omega_2 = 51.17 \text{ rad/s}\) with \(\{\Phi_2\} = [1,\ -0.618]^T\).
A 2-DOF system has natural frequencies \(\omega_1 = 10 \text{ rad/s}\) and \(\omega_2 = 30 \text{ rad/s}\) with mode shapes \(\{\Phi_1\} = [1, 1]^T\) and \(\{\Phi_2\} = [1, -1]^T\). If the system is released from rest with initial displacements \(x_1(0) = 0\) and \(x_2(0) = 2 \text{ cm}\), determine the response equations for both masses.
Show Solution
Step 1: Transform initial conditions to modal coordinates
The modal (principal) coordinates \(q_1, q_2\) weight each mode's contribution, with \(\{x\} = [\Phi]\{q\}\).
Modal matrix: \([\Phi] = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\)
Initial modal displacements: \(\{q(0)\} = [\Phi]^{-1}\{x(0)\} = \frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) cm
Initial modal velocities: \(\{\dot{q}(0)\} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) (released from rest)
Step 2: Modal responses
\(q_1(t) = 1 \cos(10t)\) cm
\(q_2(t) = -1 \cos(30t)\) cm
Step 3: Physical coordinates
\(\{x(t)\} = [\Phi]\{q(t)\} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} \cos(10t) \\ -\cos(30t) \end{bmatrix}\)
Answer:
\(x_1(t) = \cos(10t) - \cos(30t)\) cm
\(x_2(t) = \cos(10t) + \cos(30t)\) cm
The response is a superposition of both natural modes, so the two frequencies beat against each other.