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12.2: Undamped Two-Mass-Two-Spring System

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    7699
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    Next, we analyze the two-degrees-of freedom (2-DOF) undamped mass-spring system of Figure \(\PageIndex{1}\). Dynamic translations \(y_{1}(t)\) and \(y_{2}(t)\) shown are relative to the static equilibrium positions. As usual for the purpose of drawing forces on dynamic freebody diagrams (DFBDs, as defined in Section 7.5), we let the translational springs be stretched at the instant depicted, so that, in particular, the dynamic tension in the lower spring \(k_{2}\left(y_{2}-y_{1}\right)\).

    clipboard_e18355122604f9ebb4e4ba9a71e6b928f.png
    Figure \(\PageIndex{1}\): Two-mass-twospring mechanical system

    From the DFBDs, Newton’s 2nd law gives the ODEs of motion:

    \[\begin{array}{c}
    m_{1} \ddot{y}_{1}=f_{1}(t)+k_{2}\left(y_{2}-y_{1}\right)-k_{1} y_{1} \\
    m_{2} \ddot{y}_{2}=f_{2}(t)-k_{2}\left(y_{2}-y_{1}\right)
    \end{array}\label{eqn:12.3} \]

    Transposing dependent-variable terms to the lefthand sides, and collecting terms gives

    \[\begin{array}{c}
    m_{1} \ddot{y}_{1}+\left(k_{1}+k_{2}\right) y_{1}-k_{2} y_{2}=f_{1}(t) \\
    m_{2} \ddot{y}_{2}-k_{2} y_{1}+k_{2} y_{2}=f_{2}(t)
    \end{array}\label{eqn:12.4} \]

    Now, expressing Equations \(\ref{eqn:12.4}\) in the same matrix form as Equation 11.3.16 gives

    \[\overbrace{\left[\begin{array}{cc}
    m_{1} & 0 \\
    0 & m_{2}
    \end{array}\right]}^\text{inertia matrix}\left[\begin{array}{c}
    \ddot{y}_{1} \\
    \ddot{y}_{2}
    \end{array}\right]+\overbrace{\left[\begin{array}{cc}
    k_{1}+k_{2} & -k_{2} \\
    -k_{2} & k_{2}
    \end{array}\right]}^\text{structural stiffness matrix}\left[\begin{array}{l}
    y_{1} \\
    y_{2}
    \end{array}\right]=\left[\begin{array}{c}
    f_{1}(t) \\
    f_{2}(t)
    \end{array}\right]\label{eqn:12.5} \]

    You can verify easily that, just as in Equation 11.3.16, the inertia and structural stiffness matrices of Equation \(\ref{eqn:12.5}\) are positive definite, and all diagonal elements of both matrices also are positive definite.

    Next, we want to solve matrix Equation \(\ref{eqn:12.5}\) for free vibrations, so we set to zero the applied forces, \(f_{1}(t)=f_{2}(t)=0\). Following the procedure in Section 12.1 for the 1-DOF system, let us seek response to non-zero initial translations, with zero initial velocities. Accordingly, we assume motion solutions of the unforced 2-DOF system in the form \(y_{1}(t)=Y_{1} \cos \omega t\) and \(y_{2}(t)=Y_{2} \cos \omega t\), in which \(\omega\), \(Y_{1}\), and \(Y_{2}\) are the unknown quantities. It is efficient and appropriate to express the assumed solution in matrix form:

    \[\left[\begin{array}{l}
    y_{1}(t) \\
    y_{2}(t)
    \end{array}\right]=\left[\begin{array}{l}
    Y_{1} \\
    Y_{2}
    \end{array}\right] \cos \omega t\label{eqn:12.6} \]

    Substituting Equation \(\ref{eqn:12.6}\) into Equation \(\ref{eqn:12.5}\) with zero forcing gives

    \[\left[\begin{array}{cc}
    m_{1} & 0 \\
    0 & m_{2}
    \end{array}\right]\left[\begin{array}{c}
    Y_{1} \\
    Y_{2}
    \end{array}\right]\left(-\omega^{2}\right) \cos \omega t+\left[\begin{array}{cc}
    k_{1}+k_{2} & -k_{2} \\
    -k_{2} & k_{2}
    \end{array}\right]\left[\begin{array}{c}
    Y_{1} \\
    Y_{2}
    \end{array}\right] \cos \omega t=\left[\begin{array}{c}
    0 \\
    0
    \end{array}\right]\label{eqn:12.7} \]

    The common multiple \(\cos \omega t\) in Equation \(\ref{eqn:12.7}\) cannot be zero in general, so we cancel it out of the equation. Further, we consolidate all of the coefficients of \(Y_{1}\) and \(Y_{2}\) into a single matrix, giving

    \[\left[\begin{array}{cc}
    k_{1}+k_{2}-\omega^{2} m_{1} & -k_{2} \\
    -k_{2} & k_{2}-\omega^{2} m_{2}
    \end{array}\right]\left[\begin{array}{l}
    Y_{1} \\
    Y_{2}
    \end{array}\right]=\left[\begin{array}{l}
    0 \\
    0
    \end{array}\right]\label{eqn:12.8} \]

    Before dealing with Equation \(\ref{eqn:12.8}\), let us review a little from your mathematics background about solving two linear algebraic equations in two unknowns. In general, the equations are expressed in matrix notation as

    \[\left[\begin{array}{ll}
    a_{11} & a_{12} \\
    a_{21} & a_{22}
    \end{array}\right]\left[\begin{array}{l}
    x_{1} \\
    x_{2}
    \end{array}\right]=\left[\begin{array}{l}
    b_{1} \\
    b_{2}
    \end{array}\right], \text { with }\left[\begin{array}{ll}
    a_{11} & a_{12} \\
    a_{21} & a_{22}
    \end{array}\right] \equiv[\mathbf{a}], \text { and }\left[\begin{array}{l}
    x_{1} \\
    x_{2}
    \end{array}\right] \equiv\{\mathbf{x}\}, \text { and }\left[\begin{array}{l}
    b_{1} \\
    b_{2}
    \end{array}\right] \equiv\{\mathbf{b}\}\label{eqn:12.9} \]

    in which the \(a_{i j}\)’s are known constant coefficients, the \(x_{i}\)’s are the unknowns, and the \(b_{i}\)’s are known right-hand-side constants. It is useful to express the matrix equation symbolically by using bold fonts, brackets, and parentheses: \([\mathbf{a}]\{\mathbf{x}\}=\{\mathbf{b}\}\). Then the determinant and the adjoint matrix of the coefficient matrix are, respectively,

    \[\operatorname{det}[\mathbf{a}]=a_{11} a_{22}-a_{21} a_{12}\label{eqn:12.10} \]

    \[\operatorname{adj}[\mathbf{a}]=\left[\begin{array}{cc}
    a_{22} & -a_{12} \\
    -a_{21} & a_{11}
    \end{array}\right]\label{eqn:12.11} \]

    The solution of Equation \(\ref{eqn:12.9}\) for the unknowns involves the inverse \([\mathbf{a}]^{-1}\) of the coefficient matrix, which is defined in terms of the determinant and the adjoint matrix:

    \[\{\mathbf{x}\}=[\mathbf{a}]^{-1}\{\mathbf{b}\}=\frac{\operatorname{adj}[\mathbf{a}]}{\operatorname{det}[\mathbf{a}]}\{\mathbf{b}\}\label{eqn:12.12} \]

    If \(\{\mathbf{b}\}=0\), which is the case for Equation \(\ref{eqn:12.8}\), then there are two possible types of solutions:

    1. if \(\{\mathbf{b}\}=0\) and \(\operatorname{det}[\mathbf{a}] \neq 0\), then the solution must be the trivial result \(\{x\}=0\);
    2. however, if \(\{\mathbf{b}\}=0\) and \(\operatorname{det}[\mathbf{a}] = 0\), then the right-hand side of Equation \(\ref{eqn:12.12}\) has the indeterminate form \(0/0\), so that there exists, and we can solve for a non-trivial \(\{\mathbf{x}\} \neq \mathbf{0}\), but any constant multiple of \(\{\mathbf{x}\}\) is also a solution since \(\{\mathbf{b}\}=0\) in Equation \(\ref{eqn:12.9}\).

    Therefore, for our current application, Equation \(\ref{eqn:12.8}\), we must seek a type 2 solution.

    To find the same type 2 result as in the last paragraph, but without using matrices or theory from linear algebra, let us write Equation \(\ref{eqn:12.8}\) as two separate scalar equations, and then attempt to solve them algebraically for \(Y_{1}\) and \(Y_{2}\):

    \[\left(k_{1}+k_{2}-\omega^{2} m_{1}\right) Y_{1}-k_{2} Y_{2}=0 \Rightarrow Y_{2}=\frac{1}{k_{2}}\left(k_{1}+k_{2}-\omega^{2} m_{1}\right) Y_{1}\label{eqn:12.13a} \]

    \[-k_{2} Y_{1}+\left(k_{2}-\omega^{2} m_{2}\right) Y_{2}=0=-k_{2} Y_{1}+\left(k_{2}-\omega^{2} m_{2}\right) \frac{1}{k_{2}}\left(k_{1}+k_{2}-\omega^{2} m_{1}\right) Y_{1}\label{eqn:12.13b} \]

    Re-arranging the last part of Equation \(\ref{eqn:12.13b}\) a bit differently gives

    \[\left[\left(k_{1}+k_{2}-\omega^{2} m_{1}\right)\left(k_{2}-\omega^{2} m_{2}\right)-k_{2}^{2}\right] \frac{Y_{1}}{k_{2}}=0\label{eqn:12.14} \]

    We want to solve Equation \(\ref{eqn:12.14}\) for a non-zero , which leads us to conclude that:

    1. the bracketed term multiplying \(Y_{1} / k_{2}\) must be zero [Note that the bracketed term is the determinant of the coefficient matrix in Equation \(\ref{eqn:12.8}\).]; and
    2. we cannot yet assign any specific value to \(Y_{1}\) (but we will be able to do so later if we chose to regard this problem as being time response to specific initial translations).

    The bracketed term in Equation \(\ref{eqn:12.14}\) must be zero; upon carrying out the multiplications, we see that it leads to a quadratic equation in the unknown \(\omega^{2}\):

    \[m_{1} m_{2}\left(\omega^{2}\right)^{2}+\left[-m_{1} k_{2}-m_{2}\left(k_{1}+k_{2}\right)\right] \omega^{2}+k_{1} k_{2}=0\label{eqn:12.15} \]

    This type of polynomial equation is generally called the characteristic equation of the free-vibration problem. For any realistic values of the mass and stiffness constants, Equation \(\ref{eqn:12.15}\) has two real, positive roots, which we denote as \(\omega_{1}^{2}\) and \(\omega_{2}^{2}\).

    Let us consider a numerically simplified case for which the masses are equal, \(m_{1}=m_{2} \equiv m\), and the stiffness constants also are equal, \(k_{1}=k_{2} \equiv k\). In this case, the quadratic equation Equation \(\ref{eqn:12.15}\) becomes

    \[m^{2}\left(\omega^{2}\right)^{2}-3 m k \omega^{2}+k^{2}=0\label{eqn:12.16} \]

    Using the standard solution of quadratic equations, Equation 2.1.2, we find the roots of Equation \(\ref{eqn:12.16}\) to be

    \[\omega^{2}=\frac{3 \pm \sqrt{5}}{2} \frac{k}{m}=(0.382,2.618) \frac{k}{m}\label{eqn:12.17} \]

    For future reference, note that the lower of the \(\pm\) signs gives the smaller (the “first”) of the two \(\omega^{2}\) roots, and that the upper sign gives the larger (the “second”) of the two \(\omega^{2}\) roots. Only positive circular frequencies are physically meaningful, and they are:

    \[\omega=(0.618,1.618) \sqrt{k / m} \Rightarrow \omega_{1}=0.618 \sqrt{k / m} \text { and } \omega_{2}=1.618 \sqrt{k / m}\label{eqn:12.18} \]

    Of the two circular frequencies in Equations \(\ref{eqn:12.18}\), the smaller value \(\omega_{1}\) is called the first or fundamental natural frequency, and the larger value \(\omega_{2}\) is called the second natural frequency. By convention, natural frequencies are always numbered in ascending order for all higher-order systems that have two or more natural frequencies. It is mathematically systematic to define an integer subscript \(n\) that is either 1 or 2, and to identify the natural frequencies more generally as \(\omega_n\), \(n\) = 1, 2. (In this case, the symbol \(n\) performs double-duty, because it both implies natural frequency of vibration and takes the values 1 or 2 to distinguish the two natural frequencies from each other.)

    Next, let us seek values of \(Y_1\) and \(Y_2\) associated with each of the natural frequencies. It is appropriate and efficient to extend use of the subscript \(n\) to these motion magnitude values by labeling as \(Y_{1n\) and \(Y_{2n}\) the values associated with frequency \(\omega_n\), \(n\) = 1, 2. Let us re-write Equations \(\ref{eqn:12.13a}\) and \(\ref{eqn:12.13b}\) using this notation and the numerically simplified case of equal masses and equal stiffness constants:

    \[\left(2 k-\omega_{n}^{2} m\right) Y_{1 n}-k Y_{2 n}=0 \Rightarrow Y_{2 n}=\frac{1}{k}\left(2 k-\omega_{n}^{2} m\right) Y_{1 n}\label{eqn:12.19a} \]

    \[-k Y_{1 n}+\left(k-\omega_{n}^{2} m\right) Y_{2 n}=0 \Rightarrow Y_{2 n}=\frac{k}{k-\omega_{n}^{2} m} Y_{1 n}\label{eqn:12.19b} \]

    As is stated in the discussion of Equation \(\ref{eqn:12.14}\), we cannot yet assign any specific value to \(Y_{1n}\). If we leave \(Y_{1n}\) arbitrary (but not zero) for now, then Equations \(\ref{eqn:12.19a}\) and \(\ref{eqn:12.19b}\) appear to give two different ways of solving for \(Y_{2n}\) in term of \(Y_{1n}\). Let us use natural frequency Equation \(\ref{eqn:12.17}\) to evaluate both equations:

    \[Y_{2 n}=\frac{1}{k}\left(2 k-\omega_{n}^{2} m\right) Y_{1 n}=\left(2-\omega_{n}^{2} \frac{m}{k}\right) Y_{1 n}=\left(2-\frac{3 \pm \sqrt{5}}{2}\right) Y_{1 n}=\frac{1 \mp \sqrt{5}}{2} Y_{1 n}\label{eqn:12.20a} \]

    \[Y_{2 n}=\frac{k}{k-\omega_{n}^{2} m} Y_{1 n}=\frac{1}{1-\omega_{n}^{2} \frac{m}{k}} Y_{1 n}=\frac{1}{1-\frac{3 \pm \sqrt{5}}{2}} Y_{1 n}=\frac{2}{-1 \mp \sqrt{5}} Y_{1 n} \nonumber \]

    \[Y_{2 n}=\frac{2}{-1 \mp \sqrt{5}} \times \frac{1 \mp \sqrt{5}}{1 \mp \sqrt{5}} Y_{1 n}=\frac{1 \mp \sqrt{5}}{2} Y_{1 n}\label{eqn:12.20b} \]

    Both equations Equations \(\ref{eqn:12.20a}\) and \(\ref{eqn:12.20b}\) give the same answer! Mathematically, this result is related to the zero of the determinant of the coefficient matrix in Equation \(\ref{eqn:12.8}\), but the detailed theory is best left to a more advanced book (for example, Craig, 1981, Section 13.1). For our purposes, it is enough to recognize from this example that, in this type of problem, both equations have the same solution, so it is necessary to solve only one of them.

    Next, let us collect together from Equations \(\ref{eqn:12.8}\), \(\ref{eqn:12.20a}\), and \(\ref{eqn:12.20b}\) all the numerical results for each value of the index \(n\) into modes of vibration, the physical significance of which will become clear later. We have observed already that \(Y_{1n}\) is some arbitrary non-zero value, so nothing is lost if we set \(Y_{1n}=9\) (with dimension of translation, i.e., length) for both modes of vibration.

    \(n\) = 1, first (fundamental) mode of vibration: [the lower sign in Equations \(\ref{eqn:12.20a}\) and \(\ref{eqn:12.20b}\)]

    \[\omega_{1}=0.618 \sqrt{\frac{k}{m}}, \quad Y_{21}=\frac{1+\sqrt{5}}{2} Y_{11}=1.618 \Rightarrow\left[\begin{array}{l}
    Y_{11} \\
    Y_{21}
    \end{array}\right]=\left[\begin{array}{c}
    1 \\
    1.618
    \end{array}\right]\label{eqn:12.21a} \]

    \(n\) = 2, second mode of vibration: [the upper sign in \(\ref{eqn:12.20a}\) and \(\ref{eqn:12.20b}\)]

    \[\omega_{2}=1.618 \sqrt{\frac{k}{m}}, \quad Y_{22}=\frac{1-\sqrt{5}}{2} Y_{12}=-0.618 \Rightarrow\left[\begin{array}{c}
    Y_{12} \\
    Y_{22}
    \end{array}\right]=\left[\begin{array}{c}
    1 \\
    -0.618
    \end{array}\right]\label{eqn:12.21b} \]

    For reasons that will become clear later, the column matrices \(\left[\begin{array}{l} Y_{11} \\ Y_{21} \end{array}\right]\) and \(\left[\begin{array}{l} Y_{12} \\ Y_{22} \end{array}\right]\) are called the mode shapes of, respectively, the first and second modes of vibration. NOTE: In Equations \(\ref{eqn:12.21a}\) and \(\ref{eqn:12.21b}\) we set \(Y_{11}=Y_{12}=1\) with dimension of length, but that does not necessarily imply that the magnitudes of physical responses \(y_1(t)\) and \(y_2(t)\) are one meter or one foot or one inch (depending upon the unit system in use). The actual physical response magnitudes are determined by initial conditions, as is discussed in the next paragraph.

    Recall that we are seeking response to non-zero initial deformation, with zero initial velocity, and so we chose the solution in the form \(\left[\begin{array}{l}
    y_{1}(t) \\
    y_{2}(t)
    \end{array}\right]=\left[\begin{array}{l}
    Y_{1} \\
    Y_{2}
    \end{array}\right] \cos \omega t\), Equation \(\ref{eqn:12.6}\). In view of the free-vibration results Equations \(\ref{eqn:12.21a}\) and \(\ref{eqn:12.21b}\), it is quite plausible physically (and it can be proved rigorously) that the general initial-translation response can be written as the linear sum of two terms like Equation \(\ref{eqn:12.6}\), one for each of the vibration modes:

    \[\left[\begin{array}{l}
    y_{1}(t) \\
    y_{2}(t)
    \end{array}\right]=C_{1}\left[\begin{array}{l}
    Y_{11} \\
    Y_{21}
    \end{array}\right] \cos \omega_{1} t+C_{2}\left[\begin{array}{l}
    Y_{12} \\
    Y_{22}
    \end{array}\right] \cos \omega_{2} t\label{eqn:12.22} \]

    In Equation \(\ref{eqn:12.22}\), the dimensionless multiplying constants \(C_1\) and \(C_2\) show the contribution to the total response due to each mode of vibration; they are found from given initial translations \(y_1(0)\) and \(y_2(0)\) by writing Equation \(\ref{eqn:12.22}\) for \(t\) = 0:

    \[\left[\begin{array}{l}
    y_{1}(0) \\
    y_{2}(0)
    \end{array}\right]=C_{1}\left[\begin{array}{l}
    Y_{11} \\
    Y_{21}
    \end{array}\right]+C_{2}\left[\begin{array}{l}
    Y_{12} \\
    Y_{22}
    \end{array}\right]=\left[\begin{array}{ll}
    Y_{11} & Y_{12} \\
    Y_{21} & Y_{22}
    \end{array}\right]\left[\begin{array}{l}
    C_{1} \\
    C_{2}
    \end{array}\right]\label{eqn:12.23} \]

    The determinant of the \(2\times 2\) mode-shape matrix [\(\mathbf{Y}\)] in Equation \(\ref{eqn:12.23}\) is non-zero, so that matrix can be inverted and the equation can be solved by application of Equation \(\ref{eqn:12.12}\), for arbitrary known \(y_1(0)\) and \(y_2(0)\), to give mode contribution constants \(C_1\) and \(C_2\):

    \[\left[\begin{array}{c}
    C_{1} \\
    C_{2}
    \end{array}\right]=\frac{1}{\operatorname{det}[\mathbf{Y}]}\left[\begin{array}{cc}
    Y_{22} & -Y_{12} \\
    -Y_{21} & Y_{11}
    \end{array}\right]\left[\begin{array}{l}
    y_{1}(0) \\
    y_{2}(0)
    \end{array}\right]\label{eqn:12.24} \]

    Now, suppose that the initial translation is chosen to have the shape of the first mode, in other words, to be proportional to \(\left[\begin{array}{l}
    Y_{11} \\
    Y_{21}
    \end{array}\right]\) with dimensionless constant of proportionality \(D_1\): \(\left[\begin{array}{l}
    y_{1}(0) \\
    y_{2}(0)
    \end{array}\right]=D_{1}\left[\begin{array}{l}
    Y_{11} \\
    Y_{21}
    \end{array}\right]\). Substituting this initial translation into Equation \(\ref{eqn:12.24}\) gives

    \[\left[\begin{array}{c}
    C_{1} \\
    C_{2}
    \end{array}\right]=\frac{1}{\operatorname{det}[\mathbf{Y}]}\left[\begin{array}{cc}
    Y_{22} & -Y_{12} \\
    -Y_{21} & Y_{11}
    \end{array}\right] D_{1}\left[\begin{array}{c}
    Y_{11} \\
    Y_{21}
    \end{array}\right]=\frac{D_{1}}{\operatorname{det}[\mathbf{Y}]}\left[\begin{array}{c}
    Y_{22} Y_{11}-Y_{12} Y_{21} \\
    -Y_{21} Y_{11}+Y_{11} Y_{21}
    \end{array}\right]=\left[\begin{array}{c}
    D_{1} \\
    0
    \end{array}\right] \nonumber \]

    Therefore, the total response Equation \(\ref{eqn:12.22}\) is \(\left[\begin{array}{l}
    y_{1}(t) \\
    y_{2}(t)
    \end{array}\right]=D_{1}\left[\begin{array}{l}
    Y_{11} \\
    Y_{21}
    \end{array}\right] \cos \omega_{1} t\) for \(t \geq 0\). Let us express this mathematical result in words: if we impose the initial translation in the shape of the first mode of vibration, then the subsequent free vibration of the system will be pure sinusoidal motion at the first natural frequency, and the relative positions of the two masses at each instant of time will remain in the first mode shape. In other words, both masses will vibrate with frequency \(\omega_1\), and at each instant \(t \geq 0\) the ratio \(y_{2}(t) / y_{1}(t)\) will have the same value, \(Y_{21} / Y_{11}\), which is 1.618 for the numerical example of Equations \(\ref{eqn:12.16}\), \(\ref{eqn:12.17}\), and \(\ref{eqn:12.18}\) and Equation \(\ref{eqn:12.20a}\), \(\ref{eqn:12.20b}\), \(\ref{eqn:12.21a}\), and \(\ref{eqn:12.21b}\). In this case, the two masses move in the same direction at each instant, so they are said to move in phase with each other.

    Suppose next that the initial translation is chosen to have the shape of the second mode, to be proportional to \(\left[\begin{array}{l}
    Y_{12} \\
    Y_{22}
    \end{array}\right]\) with dimensionless constant of proportionality \(D_2\). By following the procedure of the last paragraph, we can find that the total response Equation \(\ref{eqn:12.22}\) is \(\left[\begin{array}{l}
    y_{1}(t) \\
    y_{2}(t)
    \end{array}\right]=D_{2}\left[\begin{array}{l}
    Y_{12} \\
    Y_{22}
    \end{array}\right] \cos \omega_{2} t\) for \(t \geq 0\). In words, the subsequent free vibration of the system will be pure sinusoidal motion at the second natural frequency, and the relative positions of the two masses at each instant of time will remain in the second mode shape: both masses will vibrate with frequency \(\omega_2\), and at each instant \(t \geq 0\) the ratio \(y_{2}(t) / y_{1}(t)\) will have the same value, \(Y_{22} / Y_{12}\), which is −0.618 for the numerical example of Equations \(\ref{eqn:12.16}\), \(\ref{eqn:12.17}\), and \(\ref{eqn:12.18}\) and Equation \(\ref{eqn:12.20a}\), \(\ref{eqn:12.20b}\), \(\ref{eqn:12.21a}\), and \(\ref{eqn:12.21b}\). In this case, the two masses move in opposite directions at each instant, so they are said to move out of phase.

    The modes of vibration of this 2-DOF mass-spring system, the numerical example of Equations \(\ref{eqn:12.16}\), \(\ref{eqn:12.17}\), and \(\ref{eqn:12.18}\) and Equation \(\ref{eqn:12.20a}\), \(\ref{eqn:12.20b}\), \(\ref{eqn:12.21a}\), and \(\ref{eqn:12.21b}\), are depicted graphically in Figure \(\PageIndex{2}\). The system is shown in its static equilibrium position for reference, and in deformed positions corresponding to each of the modes of vibration. We can not animate on the printed page the dynamic motion of a vibration mode, so we imagine that we are viewing a snapshot of the masses taken at one instant of time when the deformations are non-zero during any cycle of vibration.

    clipboard_ea230295484624ccd612ca3bfc643b2bf.png
    Figure \(\PageIndex{2}\): Vibration modes of the 2-DOF mass-spring system that has equal masses and equal springs

    The object in this chapter is to derive and illustrate the physical character of modes of vibration of undamped 2-DOF systems. The clearest and most direct way to do this is to restrict attention to response resulting from simple initial deformation, with zero initial velocity and zero input action. We will not go further in this book, but you should be informed that the theory started here can be extended to account for arbitrary non-zero initial conditions (both deformation and velocity) and arbitrary non-zero input actions imposed on damped 2-DOF systems. Indeed, the theory can be extended to damped higher-order systems having any number of degrees of freedom, which are often called N-DOF systems, with \(N\geq 2\). The general method of analysis is called modal analysis, and it is used extensively in engineering practice and research for both mathematical and experimental studies. This book just scratches the surface of modal analysis; the theory is extended to general response of LTI mechanical systems by more advanced books on linear-systems analysis and structural dynamics, two examples being Meirovitch, 2001, Chapter 7 and Craig, 1981, Chapters 13-15.


    This page titled 12.2: Undamped Two-Mass-Two-Spring System is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by William L. Hallauer Jr. (Virginia Tech Libraries' Open Education Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.