Vibration of Two Degree of Freedom Systems

Data: $m_1 = 2$ kg, $m_2 = 1$ kg, $k_1 = 3000$ N/m, $k_2 = 2000$ N/m.

 wall |--k1--[ m1 ]--k2--[ m2 ]
          x1       x2

Equations of motion

$$m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 = 0, \qquad m_2\ddot{x}_2 - k_2x_1 + k_2x_2 = 0$$

In matrix form with $x = X\sin\omega t$:

$$\begin{bmatrix} 5000 - 2\omega^2 & -2000 \\ -2000 & 2000 - \omega^2 \end{bmatrix}\begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}$$

Frequency equation

$$(5000 - 2\omega^2)(2000 - \omega^2) - (2000)^2 = 0$$ $$2\omega^4 - 9000\omega^2 + 6\times10^6 = 0 \;\Rightarrow\; \omega^4 - 4500\omega^2 + 3\times10^6 = 0$$ $$\omega^2 = \frac{4500 \pm \sqrt{4500^2 - 4(3\times10^6)}}{2} = \frac{4500 \pm 2872.28}{2}$$ $$\omega_1^2 = 813.86 \Rightarrow \omega_1 = 28.53\ \text{rad/s}\ (f_1 = 4.54\ \text{Hz})$$ $$\omega_2^2 = 3686.14 \Rightarrow \omega_2 = 60.71\ \text{rad/s}\ (f_2 = 9.66\ \text{Hz})$$

Mode shapes

From the second row, $\dfrac{X_2}{X_1} = \dfrac{2000}{2000 - \omega^2}$ (equivalently $\dfrac{5000 - 2\omega^2}{2000}$).

 Mode 1: m1 -> , m2 ---> (same direction)
 Mode 2: m1 -> , m2 <-   (node between m1 and m2)

Answer: $\omega_1 = 28.53$ rad/s, $\omega_2 = 60.71$ rad/s; mode shapes $\{1,\ 1.686\}$ and $\{1,\ -1.186\}$.

(a) Principle

A small auxiliary spring-mass system ($m_2$, $k_2$) is attached to the main system ($m_1$, $k_1$) which is excited by $F_0\sin\omega t$. When $\sqrt{k_2/m_2} = \omega$, the absorber's force on the main mass cancels the exciting force and the main mass stands still. The absorber mass vibrates instead.

  F0 sin(wt)
     |
   [ m1 ]--k2--[ m2 ]
     |
     k1
   ///

Equations of motion:

$$m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 = F_0\sin\omega t, \qquad m_2\ddot{x}_2 + k_2(x_2 - x_1) = 0$$

Put $x = X\sin\omega t$:

$$\begin{bmatrix} k_1 + k_2 - m_1\omega^2 & -k_2 \\ -k_2 & k_2 - m_2\omega^2 \end{bmatrix}\begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} F_0 \\ 0 \end{Bmatrix}$$ $$X_1 = \frac{(k_2 - m_2\omega^2)F_0}{(k_1+k_2-m_1\omega^2)(k_2 - m_2\omega^2) - k_2^2}, \qquad X_2 = \frac{k_2F_0}{(\ldots)}$$

$X_1 = 0$ when $k_2 - m_2\omega^2 = 0$, i.e. $\omega = \sqrt{k_2/m_2}$. Then $X_2 = -\dfrac{F_0}{k_2}$, so the absorber spring force $k_2X_2 = -F_0$ exactly cancels the exciting force. The absorber works only at one tuned frequency; away from it, the combined system has two resonances.

(b) Design

Operating frequency: $\Omega = \dfrac{2\pi \times 1500}{60} = 157.08$ rad/s. The main system is at resonance, so $k_1 = m_1\Omega^2 = 100 \times 157.08^2 = 2467401$ N/m.

Tuning: $k_2 = m_2\Omega^2 = 10 \times 157.08^2 = 246740\ \text{N/m} = 246.7\ \text{kN/m}$.

New natural frequencies: mass ratio $\mu = m_2/m_1 = 0.1$. With $\omega_a = \omega_1 = \Omega$ and $r = \omega/\Omega$, the frequency equation is

$$r^4 - (2+\mu)r^2 + 1 = 0 \;\Rightarrow\; r^2 = \frac{2.1 \pm \sqrt{2.1^2 - 4}}{2} = 0.7298,\ 1.3702$$ $$\omega_a = \Omega\sqrt{0.7298} = 134.19\ \text{rad/s}\ (1281\ \text{rpm}), \qquad \omega_b = \Omega\sqrt{1.3702} = 183.87\ \text{rad/s}\ (1756\ \text{rpm})$$

The two new resonant speeds lie about 15% below and 17% above the operating speed, so the machine runs between them, safely away from both.

Answer: Absorber stiffness $= 246.7$ kN/m; new natural speeds $\approx 1281$ rpm and $1756$ rpm.

Equations of motion

Take masses $m_1, m_2$, springs $k_1$ (wall to $m_1$), $k_2$ (between masses), dampers $c_1, c_2$ in parallel with them, and force $F_1\sin\omega t$ on $m_1$.

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix}\begin{Bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{Bmatrix} + \begin{bmatrix} c_1+c_2 & -c_2 \\ -c_2 & c_2 \end{bmatrix}\begin{Bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{Bmatrix} + \begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} F_1\sin\omega t \\ 0 \end{Bmatrix}$$

or $[M]\{\ddot{x}\} + [C]\{\dot{x}\} + [K]\{x\} = \{F\}$.

Steady-state amplitudes

Write the force as $F_1e^{i\omega t}$ and the response as $\{x\} = \{X\}e^{i\omega t}$. Substituting:

$$\left[-\omega^2[M] + i\omega[C] + [K]\right]\{X\} = \{F\}$$

The bracket is the mechanical impedance matrix $[Z(\omega)]$ with entries $Z_{rs} = -\omega^2 m_{rs} + i\omega c_{rs} + k_{rs}$. Then

$$\{X\} = [Z(\omega)]^{-1}\{F\}, \qquad X_1 = \frac{Z_{22}F_1}{Z_{11}Z_{22} - Z_{12}^2}, \quad X_2 = \frac{-Z_{12}F_1}{Z_{11}Z_{22} - Z_{12}^2}$$

The amplitudes are complex: $|X_r|$ is the amplitude, and the argument gives the phase lag behind the force. Damping makes the denominator never vanish, so the resonance peaks are finite.

Coupling

• Static (elastic) coupling: the stiffness matrix has off-diagonal terms ($k_{12} \ne 0$); a static force on one coordinate displaces the other.
• Dynamic (inertia) coupling: the mass matrix has off-diagonal terms ($m_{12} \ne 0$), e.g. a vehicle body with the c.g. not midway between the axles.
• The coupling depends on the choice of coordinates, not on the system.

Principal (normal) coordinates

They are special coordinates in which both $[M]$ and $[K]$ become diagonal, so the equations are uncoupled and each is a single-degree-of-freedom equation:

$$\ddot{q}_r + \omega_r^2 q_r = 0, \qquad \{x\} = [\Phi]\{q\}$$

where $[\Phi]$ is the matrix of mode shapes. Each principal coordinate vibrates in one mode only.