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Review of single and multi-degree of freedom (mdof) systems: Vibration and Applications of MDOF Systems

For more than a SDOF system the analysis requires a more generalized approach. However, overall it doesn’t matter if there are 2 or 22 DOFs. Consider first a special 2 DOF free vibration system:

    \[ \begin{bmatrix} m & 0 \\  0 & m \end{bmatrix} \begin{Bmatrix} \ddot{x_1} \\ \ddot{x_2} \end{Bmatrix} + \begin{bmatrix} k + k_c & -k_c \\ -k_c & k + k_c \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}\]

The simplest approach is to look for Simultaneous Simple Harmonic Motion (SSHM). That is:

    \[ \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} \sin(pt + \phi) \]

Which is a solution (not the general one) if:

    \[ \begin{bmatrix} k+k_c - mp^2 & -k_c \\ -k_c & k+k_c-mp^2 \end{bmatrix} \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} \]

This can be true if and only if (iff), \text{det}[] = 0. Therefore:

    \[ p^4-p^2\Big[\frac{(k+k_c)}{m} + \frac{(k+k_c)}{m}\Big] + \frac{k^2+2k_c k}{m^2} = 0 \]

There are two soloutions for p^2 \Rightarrow p_1^2 \ \text{and} \ p_2^2.

    \[ \begin{split} p_{1,2}^2 &= \frac{k+k_c}{m} \pm \sqrt{\frac{k^2+2kk_c + k_c^2 - (k^2 + 2kk_c)}{m^2}} \\&= \frac{k+k_c}{m} \pm \frac{k_c}{m} \\&= \frac{k}{m}, \  \frac{k+2k_c}{m} \end{split}\]

Therefore:

    \[p_1 = \sqrt{\frac{k}{m}}, \ p_2 = \sqrt{\frac{k+2k_c}{m}} \]

As a result, there are 2 frequencies at which our assumption is true. NOTE: It turns out that the general solution can be determined from these p_1, p_2, and the ratio of the amplitudes between the two masses during each of the two SSHMs. Therefore:

    \[ (k + k_c - mp^2)X_1 -k_cX_2 = 0 \]

Or:

    \[ -k_cX_1 + (k + k_c - mp^2)X_2 = 0 \]

Therefore, for either p_1^2 or p_2^2:

    \[ \begin{split} \frac{X_2}{X_1} &= \frac{k +k_c -mp^2}{k_c} \\&= \frac{k_c}{k+k_c-mp^2} \end{split} \]

If we put p_1^2 into either we get:

    \[ \Big(\frac{X_2}{X_1}\Big)_1 = \frac{1}{1} \equiv X_1 \]

And if we put p_2^2 into either we get:

    \[ \Big(\frac{X_2}{X_1}\Big)_2 = \frac{-1}{1} \equiv X_2 \]

Thus, we define the mode shapes corresponding to each ratio:

    \[ \{u\}_1 = \begin{Bmatrix} 1 \\ X_1 \end{Bmatrix} \]

    \[ \{u\}_2 = \begin{Bmatrix} 1 \\ X_2 \end{Bmatrix} \]

Therefore, we have a mode shape corresponding to each of the natural frequencies. \{u\}_1 and \{u\}_2 are “normalized” vectors where the first component is set arbitrarily to unity. The general solution to the free vibration problem is then:

    \[ \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = c_1 \{u\}_1 \sin(p_1t +\phi_1) + c_2\{u\}_2 \sin(p_2t+\phi_2) \]

There are 4 constants c_1, c_2, \phi_1, \phi_2 for the 4 initial conditions for x_1, x_2, \dot{x_1}, \dot{x_2}. If the initial conditions are selected then the solution will include both modes in general but only one mode for certain cases. For example, when x_1(0) = 1, x_2(0) = 1, \dot{x}_1(0) = 0, and \dot{x}_2(0) = 0, \Rightarrow mode 1!

Consider the following initial conditions:

    \[x_1(0) = 1,\ x_2(0) = 1,\ \dot{x}_1(0) = 0,\ \dot{x}_2(0) = 0\]

    \[x_1(t) = c_1\sin(p_1t+\phi_1) + c_2\sin(p_2t+\phi_2) \]

    \[x_2(t) = c_1\sin(p_1t+\phi_1) - c_2\sin(p_2t+\phi_2) \]

At t = 0:

    \[1 = c_1\sin\phi_1 + c_2\sin\phi_2 \quad (1) \]

    \[0 = c_1\sin\phi_1 - c_2\sin\phi_2 \quad (2) \]

    \[0 = p_1\cos\phi_1 + p_2\cos\phi_2 \quad (3) \]

    \[0 = p_1\cos\phi_1 -  p_2\cos\phi_2 \quad (4) \]

Therefore:

    \[ \cos\phi_1 = 0 \quad \text{(sum (3) + (4))} \]

    \[ \cos\phi_2 = 0 \quad \text{((3) - (4))} \]

Therefore:

    \[ \phi_1 = \frac{\pi}{2},\ \phi_2 = \frac{\pi}{2}\]

From (1) and (2) (+):

    \[2c_1\sin\phi_1 = 1\]

    \[c_1 = \frac{1}{2}\]

From (1) and (2) (-):

    \[c_2 = \frac{1}{2}\]

Therefore:

    \[\begin{split} x_1 &= \frac{1}{2}\sin\Big(p_1t + \frac{\pi}{2}\Big) + \frac{1}{2}\sin\Big(p_2t + \frac{\pi}{2}\Big) \\&= \frac{1}{2}\cos p_1t + \frac{1}{2}\cos p_2t \end{split} \]

    \[x_2 = \frac{1}{2}\cos p_1t - \frac{1}{2}\cos p_2t \]

Using identities:

    \[x_1 = \Big[\cos(p_2-p_1)\frac{t}{2} \bullet \cos(p_1+p_2)\frac{t}{2}\Big] \]

    \[x_2 = \Big[\sin(p_2-p_1)\frac{t}{2} \bullet \sin(p_2+p_1)\frac{t}{2}\Big] \]

These may be interpreted as a higher frequency oscillation at the average of the two natural frequencies with a variable amplitude, given by a lower frequency given by \frac{1}{2} the difference in natural frequencies.

If p_1 and p_2 are close to each relative to their magnitude then (p_2-p_1) << p_1, the motion becomes:

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