Detailed steps to perform modal analysis are given below for a standard undamped two degrees of freedom system. The main advantage of solving a multidegree system using modal analysis is that it decouples the equations of motion (assuming they are coupled) making solving them much simpler.
In addition it shows the fundamental shapes that the system can vibrate in, which gives more insight into the system. Starting with standard 2 degrees of freedom system
In the above the generalized coordinates are \(x_{1}\) and \(x_{2}\). Hence the system requires two equations of motion (EOM’s).
The two EOM’s are found using any method such as Newton’s method or Lagrangian method. Using Newton’s method, free body diagram is made of each mass and then \(F=ma\) is written for each mass resulting in the equations of motion. In the following it is assumed that both masses are moving in the positive direction and that \(x_{2}\) is larger than \(x_{1}\) when these equations of equilibrium are written
Hence, from the above the equations of motion are\begin {align*} m_{1} x_{1}^{\prime \prime }+ k_{1} x_{1} - k_{2}(x_{2}-x_{1}) & = f_{1}(t)\\ m_{2} x_{2}^{\prime \prime }+ k_{2} (x_{2} - x_{1}) & = f_{2}(t) \end {align*}
or\begin {align*} m_{1} x_{1}^{\prime \prime }+ x_{1} (k_{1}+k_{2}) - k_{2} x_{2} & = f_{1}(t)\\ m_{2} x_{2}^{\prime \prime }+ k_{2} x_{2} - k_{2} x_{1} & = f_{2}(t) \end {align*}
In Matrix form \[\begin {bmatrix} m_{1} & 0\\ 0 & m_{2}\end {bmatrix} \begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\end {Bmatrix} + \begin {bmatrix} k_{1}+k_{2} & -k_{2}\\ -k_{2} & k_{2}\end {bmatrix} \begin {Bmatrix} x_{1}\\ x_{2}\end {Bmatrix} = \begin {Bmatrix} f_{1}(t)\\ f_{2}(t) \end {Bmatrix} \]
The above two EOM are coupled in stiffness, but not mass coupled. Using short notations, the above is written as\[ [M] \{x^{\prime \prime }\} + [K] \{x\} =\{f\} \] Modal analysis now starts with the goal to decouple the EOM and obtain the fundamental shape functions that the system can vibrate in. To make these derivations more general, the mass matrix and the stiffness matrix are written in general notations as follows
\[\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix} \begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\end {Bmatrix} + \begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix}\begin {Bmatrix} x_{1}\\ x_{2}\end {Bmatrix} = \begin {Bmatrix} f_{1}(t)\\ f_{2}(t) \end {Bmatrix} \]
The mass matrix \([M]\) and the stiffness matrix \([K]\) must always come out to be symmetric. If they are not symmetric, then a mistake was made in obtaining them. As a general rule, the mass matrix \([M]\) is PSD (positive definite matrix) and the \([K]\) matrix is positive semi-definite matrix. The reason the \([M]\) is PSD is that \({x}^{T} [M] \{x\}\) represents the kinetic energy of the system, which is typically positive and not zero. But reading some other references 1 it is possible that \([M]\) can be positive semi-definite. It depends on the application being modeled.
The first step in modal analysis is to solve the eigenvalue problem \(\det \left ( [K] - \omega ^{2} [M] \right ) =0\) in order to determine the natural frequencies of the system. This equations leads to a polynomial in \(\omega \) and the roots of this polynomial are the natural frequencies of the system. Since there are two degrees of freedom, there will be two natural frequencies \(\omega _{1},\omega _{2}\) for the system. \begin {align*} \det \left ( \left [ K\right ] -\omega ^{2}\left [ M\right ] \right ) & =0\\ \det \left ( \begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} -\omega ^{2}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix} \right ) & =0\\ \det \begin {bmatrix} k_{11}-\omega ^{2}m_{11} & k_{12}-\omega ^{2}m_{12}\\ k_{21}-\omega ^{2}m_{21} & k_{22}-\omega ^{2}m_{22}\end {bmatrix} & =0\\ \left ( k_{11}-\omega ^{2}m_{11}\right ) \left ( k_{22}-\omega ^{2}m_{22}\right ) -\left ( k_{12}-\omega ^{2}m_{12}\right ) \left ( k_{21}-\omega ^{2}m_{21}\right ) & =0\\ \omega ^{4}\left ( m_{11}m_{22}-m_{12}m_{21}\right ) +\omega ^{2}\left ( -k_{11}m_{22}+k_{12}m_{21}+k_{21}m_{12}-k_{22}m_{11}\right ) +k_{11}k_{22}-k_{12}k_{21} & =0 \end {align*}
The above is a polynomial in \(\omega ^{4}\). Let \(\omega ^{2}=\lambda \) it becomes\[ \lambda ^{2}\left ( m_{11}m_{22}-m_{12}m_{21}\right ) +\lambda \left ( -k_{11}m_{22}+k_{12}m_{21}+k_{21}m_{12}-k_{22}m_{11}\right ) +k_{11}k_{22}-k_{12} k_{21}=0 \]
This quadratic polynomial in \(\lambda \) which is now solved using the quadratic formula. Then the positive square root of each \(\lambda \) root to obtain \(\omega _{1}\) and \(\omega _{2}\) which are the roots of the original eigenvalue problem. Assuming from now that these roots are \(\omega _{1}\) and \(\omega _{2}\) the next step is to obtain the non-normalized shape vectors \(\boldsymbol {\varphi }_{1},\boldsymbol {\varphi }_{2}\) also called the eigenvectors associated with \(\omega _{1}\) and \(\omega _{2}\)
For each natural frequency \(\omega _{1}\) and \(\omega _{2}\) the corresponding shape function is found by solving the following two sets of equations for the vectors \(\varphi _{1},\varphi _{2}\)\[\begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} -\omega _{1}^{2}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \]
and\[\begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} -\omega _{2}^{2}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \]
For \(\omega _{1}\), let \(\varphi _{11}=1\) and solve for\begin {align*} \begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} -\omega _{1}^{2}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} 1\\ \varphi _{21}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \\\begin {bmatrix} k_{11}-\omega _{1}^{2}m_{11} & k_{12}-\omega _{1}^{2}m_{12}\\ k_{21}-\omega _{1}^{2}m_{21} & k_{22}-\omega _{1}^{2}m_{22}\end {bmatrix}\begin {Bmatrix} 1\\ \varphi _{21}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \end {align*}
Which gives one equation now to solve for \(\varphi _{21}\) (the first row equation is only used)\[ \left ( k_{11}-\omega _{1}^{2}m_{11}\right ) +\varphi _{21}\left ( k_{12}-\omega _{1}^{2}m_{12}\right ) =0 \]
Hence \[ \varphi _{21}=\frac {-\left ( k_{11}-\omega _{1}^{2}m_{11}\right ) }{\left ( k_{12}-\omega _{1}^{2}m_{12}\right ) } \]
Therefore the first shape vector is\[ \boldsymbol {\varphi }_{1}=\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} =\begin {Bmatrix} 1\\ \frac {-\left ( k_{11}-\omega _{1}^{2}m_{11}\right ) }{\left ( k_{12}-\omega _{1}^{2}m_{12}\right ) }\end {Bmatrix} \]
Similarly the second shape function is obtained. For \(\omega _{2}\), let \(\varphi _{12}=1\) and solve for\begin {align*} \begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} -\omega _{2}^{2}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} 1\\ \varphi _{22}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \\\begin {bmatrix} k_{11}-\omega _{2}^{2}m_{11} & k_{12}-\omega _{2}^{2}m_{12}\\ k_{21}-\omega _{2}^{2}m_{21} & k_{22}-\omega _{2}^{2}m_{22}\end {bmatrix}\begin {Bmatrix} 1\\ \varphi _{22}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \end {align*}
Which gives one equation now to solve for \(\varphi _{22}\) (the first row equation is only used)\[ \left ( k_{11}-\omega _{2}^{2}m_{11}\right ) +\varphi _{22}\left ( k_{12}-\omega _{2}^{2}m_{12}\right ) =0 \]
Hence \[ \varphi _{22}=\frac {-\left ( k_{11}-\omega _{2}^{2}m_{11}\right ) }{\left ( k_{12}-\omega _{2}^{2}m_{12}\right ) } \]
Therefore the second shape vector is\[ \boldsymbol {\varphi }_{2}=\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} =\begin {Bmatrix} 1\\ \frac {-\left ( k_{11}-\omega _{2}^{2}m_{11}\right ) }{\left ( k_{12}-\omega _{2}^{2}m_{12}\right ) }\end {Bmatrix} \]
Now that the two non-normalized shape vectors are found, the next step is to perform mass normalization
Let \[ \mu _{1}=\boldsymbol {\varphi }_{1}^{T}\left [ M\right ] \boldsymbol {\varphi }_{1} \]
This results in a scalar value \(\mu _{1}\), which is later used to normalize \(\boldsymbol {\varphi }_{1}\). Similarly\[ \mu _{2}=\boldsymbol {\varphi }_{2}^{T}\left [ M\right ] \boldsymbol {\varphi }_{2} \]
For example, to find \(\mu _{1}\)\begin {align*} \mu _{1} & =\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} \\ & =\begin {Bmatrix} \varphi _{11} & \varphi _{21}\end {Bmatrix}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} \\ & =\begin {Bmatrix} \varphi _{11}m_{11}+\varphi _{21}m_{21} & \varphi _{11}m_{12}+\varphi _{21}m_{22}\end {Bmatrix}\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} \\ & =\varphi _{11}\left ( \varphi _{11}m_{11}+\varphi _{21}m_{21}\right ) +\varphi _{21}\left ( \varphi _{11}m_{12}+\varphi _{21}m_{22}\right ) \end {align*}
Similarly, \(\mu _{2}\) is found\begin {align*} \mu _{2} & =\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} \\ & =\begin {Bmatrix} \varphi _{12} & \varphi _{22}\end {Bmatrix}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} \\ & =\begin {Bmatrix} \varphi _{12}m_{11}+\varphi _{22}m_{21} & \varphi _{12}m_{12}+\varphi _{22}m_{22}\end {Bmatrix}\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} \\ & =\varphi _{12}\left ( \varphi _{12}m_{11}+\varphi _{22}m_{21}\right ) +\varphi _{22}\left ( \varphi _{12}m_{12}+\varphi _{22}m_{22}\right ) \end {align*}
Now that \(\mu _{1},\mu _{2}\) are obtained, the mass normalized shape vectors are found. They are called \(\boldsymbol {\Phi }_{1},\boldsymbol {\Phi }_{2}\)\[ \boldsymbol {\Phi }_{1}=\frac {\boldsymbol {\varphi }_{1}}{\sqrt {\mu _{1}}}=\frac {\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} }{\sqrt {\mu _{1}}}=\begin {Bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}}\end {Bmatrix} \]
Similarly\[ \boldsymbol {\Phi }_{2}=\frac {\boldsymbol {\varphi }_{2}}{\sqrt {\mu _{2}}}=\frac {\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} }{\sqrt {\mu _{2}}}=\begin {Bmatrix} \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {Bmatrix} \]
The modal transformation matrix is the \(2\times 2\) matrix made of of \(\boldsymbol {\Phi }_{1},\boldsymbol {\Phi }_{2}\) in each of its columns \begin {align*} \left [ \Phi \right ] & =\left [ \boldsymbol {\Phi }_{1}\boldsymbol {\Phi }_{2}\right ] \\ & =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix} \end {align*}
Now the \(\left [ \Phi \right ] \) is found, the transformation from the normal coordinates \(\left \{ x\right \} \) to modal coordinates, which is called \(\left \{ \eta \right \} \) is found
\begin {align*} \left \{ x\right \} & =\left [ \Phi \right ] \left \{ \eta \right \} \\\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix}\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} \end {align*}
The transformation from modal coordinates back to normal coordinates is\begin {align*} \left \{ \eta \right \} & =\left [ \Phi \right ] ^{-1}\left \{ x\right \} \\\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix} ^{-1}\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} \end {align*}
However, \(\left [ \Phi \right ] ^{-1}=\left [ \Phi \right ] ^{T}\left [ M\right ] \) therefore\begin {align*} \left \{ \eta \right \} & =\left [ \Phi \right ] ^{T}\left [ M\right ] \left \{ x\right \} \\\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix} ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} \end {align*}
The next step is to apply this transformation to the original equations of motion in order to decouple them
The EOM in normal coordinates is\[\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix}\begin {Bmatrix} x_{1}\\ x_{2}\end {Bmatrix} =\begin {Bmatrix} f_{1}\left ( t\right ) \\ f_{2}\left ( t\right ) \end {Bmatrix} \]
Applying the above modal transformation \(\left \{ x\right \} =\left [ \Phi \right ] \left \{ \eta \right \} \) on the above results in \[\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix} \left [ \Phi \right ] \begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} \left [ \Phi \right ] \begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\begin {Bmatrix} f_{1}\left ( t\right ) \\ f_{2}\left ( t\right ) \end {Bmatrix} \]
pre-multiplying by \(\left [ \Phi \right ] ^{T}\) results in\[ \left [ \Phi \right ] ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix} \left [ \Phi \right ] \begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\left [ \Phi \right ] ^{T}\begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} \left [ \Phi \right ] \begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\left [ \Phi \right ] ^{T}\begin {Bmatrix} f_{1}\left ( t\right ) \\ f_{2}\left ( t\right ) \end {Bmatrix} \]
The result of \(\left [ \Phi \right ] ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix} \left [ \Phi \right ] \) will always be \(\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix} \). This is because mass normalized shape vectors are used. If the shape functions were not mass normalized, then the diagonal values will not be \(1\) as shown.
The result of \(\left [ \Phi \right ] ^{T}\begin {bmatrix} k_{11} & k_{12}\\ k_{21} & k_{22}\end {bmatrix} \left [ \Phi \right ] \) will be \(\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix} \).
Let the result of \(\left [ \Phi \right ] ^{T}\begin {Bmatrix} f_{1}\left ( t\right ) \\ f_{2}\left ( t\right ) \end {Bmatrix} \) be \(\begin {Bmatrix} \tilde {f}_{1}\left ( t\right ) \\ \tilde {f}_{2}\left ( t\right ) \end {Bmatrix} ,\)Therefore, in modal coordinates the original EOM becomes\[\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} =\begin {Bmatrix} \tilde {f}_{1}\left ( t\right ) \\ \tilde {f}_{2}\left ( t\right ) \end {Bmatrix} \]
The EOM are now decouples and each can be solved as follows\begin {align*} \eta _{1}^{\prime \prime }\left ( t\right ) +\omega _{1}^{2}\eta _{1}\left ( t\right ) & =\tilde {f}_{1}\left ( t\right ) \\ \eta _{2}^{\prime \prime }\left ( t\right ) +\omega _{2}^{2}\eta _{2}\left ( t\right ) & =\tilde {f}_{2}\left ( t\right ) \end {align*}
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules
\begin {align*} \left \{ \eta \left ( 0\right ) \right \} & =\left [ \Phi \right ] ^{T}\left [ M\right ] \left \{ x\left ( 0\right ) \right \} \\ \left \{ \eta ^{\prime }\left ( 0\right ) \right \} & =\left [ \Phi \right ] ^{T}\left [ M\right ] \left \{ x^{\prime }\left ( 0\right ) \right \} \end {align*}
Or in full form\[\begin {Bmatrix} \eta _{1}\left ( 0\right ) \\ \eta _{2}\left ( 0\right ) \end {Bmatrix} =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix} ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} x_{1}\left ( 0\right ) \\ x_{2}\left ( 0\right ) \end {Bmatrix} \]
and\[\begin {Bmatrix} \eta _{1}^{\prime }\left ( 0\right ) \\ \eta _{2}^{\prime }\left ( 0\right ) \end {Bmatrix} =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix} ^{T}\begin {bmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end {bmatrix}\begin {Bmatrix} x_{1}^{\prime }\left ( 0\right ) \\ x_{2}^{\prime }\left ( 0\right ) \end {Bmatrix} \]
Each of these EOM are solved using any of the standard methods. This will result is solutions \(\eta _{1}\left ( t\right ) \) and \(\eta _{2}\left ( t\right ) \)
The solutions found above are in modal coordinates \(\eta _{1}\left ( t\right ) ,\eta _{2}\left ( t\right ) \). The solution needed is \(x_{1}\left ( t\right ) \,,x_{2}\left ( t\right ) \). Therefore, the transformation \(\left \{ x\right \} =\left [ \Phi \right ] \left \{ \eta \right \} \) is now applied to convert the solution to normal coordinates\begin {align*} \begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}} & \frac {\varphi _{12}}{\sqrt {\mu _{2}}}\\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}} & \frac {\varphi _{22}}{\sqrt {\mu _{2}}}\end {bmatrix}\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} \\ & =\begin {Bmatrix} \frac {\varphi _{11}}{\sqrt {\mu _{1}}}\eta _{1}\left ( t\right ) +\frac {\varphi _{12}}{\sqrt {\mu _{2}}}\eta _{2}\left ( t\right ) \\ \frac {\varphi _{21}}{\sqrt {\mu _{1}}}\eta _{1}\left ( t\right ) +\frac {\varphi _{22}}{\sqrt {\mu _{2}}}\eta _{2}\left ( t\right ) \end {Bmatrix} \end {align*}
Hence\[ x_{1}\left ( t\right ) =\frac {\varphi _{11}}{\sqrt {\mu _{1}}}\eta _{1}\left ( t\right ) +\frac {\varphi _{12}}{\sqrt {\mu _{2}}}\eta _{2}\left ( t\right ) \]
and\[ x_{2}\left ( t\right ) =\frac {\varphi _{21}}{\sqrt {\mu _{1}}}\eta _{1}\left ( t\right ) +\frac {\varphi _{22}}{\sqrt {\mu _{2}}}\eta _{2}\left ( t\right ) \]
Notice that the solution in normal coordinates is a linear combination of the modal solutions. The terms \(\frac {\varphi _{ij}}{\sqrt {\mu }}\) are just scaling factors that represent the contribution of each modal solution to the final solution. This completes modal analysis
This is a numerical example that implements the above steps using a numerical values for \(\left [ K\right ] \) and \(\left [ M\right ] \). Let \(k_{1}=1,k_{2}=2,m_{1}=1,m_{2}=3\) and let \(f_{1}\left ( t\right ) =0\) and \(f_{2}\left ( t\right ) =\sin \left ( 5t\right ) \). Let initial conditions be \(x_{1}\left ( 0\right ) =0,x_{1}^{\prime }\left ( 0\right ) =1,x_{2}\left ( 0\right ) =1.5,x_{2}^{\prime }\left ( 0\right ) =3\), hence
\[\begin {Bmatrix} x_{1}\left ( 0\right ) \\ x_{2}\left ( 0\right ) \end {Bmatrix} =\begin {Bmatrix} 0\\ 1 \end {Bmatrix} \]
and
\[\begin {Bmatrix} x_{1}^{\prime }\left ( 0\right ) \\ x_{2}^{\prime }\left ( 0\right ) \end {Bmatrix} =\begin {Bmatrix} 1.5\\ 3 \end {Bmatrix} \]
In normal coordinates, the EOM are
\begin {align*} \begin {bmatrix} m_{1} & 0\\ 0 & m_{2}\end {bmatrix}\begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\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}\left ( t\right ) \\ f_{2}\left ( t\right ) \end {Bmatrix} \\\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix}\begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 3 & -2\\ -2 & 2 \end {bmatrix}\begin {Bmatrix} x_{1}\\ x_{2}\end {Bmatrix} & =\begin {Bmatrix} 0\\ \sin \left ( 5t\right ) \end {Bmatrix} \end {align*}
In this example \(m_{11}=1,m_{12}=0,m_{21}=0,m_{22}=3\) and \(k_{11}=3,k_{12}=-2,k_{21}=-2,k_{22}=2\) and \(f_{1}\left ( t\right ) =0\) and \(f_{2}\left ( t\right ) =\sin \left ( 5t\right ) \)
step 2 is now applied which solves the eigenvalue problem in order to find the two natural frequencies
\begin {align*} \det \left ( \left [ K\right ] -\omega ^{2}\left [ M\right ] \right ) & =0\\ \det \left ( \begin {bmatrix} 3 & -2\\ -2 & 2 \end {bmatrix} -\omega ^{2}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix} \right ) & =0\\ \det \begin {bmatrix} 3-\omega ^{2} & -2\\ -2 & 2-3\omega ^{2}\end {bmatrix} & =0\\ \left ( 3-\omega ^{2}\right ) \left ( 2-3\omega ^{2}\right ) -\left ( -2\right ) \left ( -2\right ) & =0\\ 3\omega ^{4}-11\omega ^{2}+2 & =0 \end {align*}
Let \(\omega ^{2}=\lambda \) hence\[ 3\lambda ^{2}-11\lambda +2=0 \] The solution is \(\lambda _{1}=3.\,\allowbreak 475\) and \(\lambda _{2}=0.192\), therefore\[ \omega _{1}=\sqrt {3.475}=1.864 \] And \[ \omega _{2}=\sqrt {0.192}=0.438 \] step 3 is now applied which finds the non-normalized eigenvectors. For each natural frequency \(\omega _{1}\) and \(\omega _{2}\) the corresponding shape function is found by solving the following two sets of equations for the eigen vectors \(\varphi _{1},\varphi _{2}\)\[ \left ( \begin {bmatrix} 3 & -2\\ -2 & 2 \end {bmatrix} -\omega _{1}^{2}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix} \right ) \begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \]
For \(\omega _{1}=\allowbreak 1.\,\allowbreak 864\)\begin {align*} \left ( \begin {bmatrix} 3 & -2\\ -2 & 2 \end {bmatrix} -1.864^{2}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix} \right ) \begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \\\begin {bmatrix} -0.475 & -2\\ -2 & -8.424 \end {bmatrix}\begin {Bmatrix} 1\\ \varphi _{21}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \end {align*}
This gives one equation to solve for \(\varphi _{21}\) (the first row equation is only used)\[ -0.475-2\varphi _{21}=0 \]
Hence \[ \varphi _{21}=\frac {0.475}{-2}=-0.237\, \]
The first eigen vector is\[ \boldsymbol {\varphi }_{1}=\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} =\begin {Bmatrix} 1\\ -0.237\, \end {Bmatrix} \]
Similarly for \(\omega _{2}=0.438\)\begin {align*} \left ( \begin {bmatrix} 3 & -2\\ -2 & 2 \end {bmatrix} -0.438^{2}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix} \right ) \begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \\\begin {bmatrix} 2.808 & -2\\ -2 & 1.425 \end {bmatrix} \allowbreak \begin {Bmatrix} 1\\ \varphi _{22}\end {Bmatrix} & =\begin {Bmatrix} 0\\ 0 \end {Bmatrix} \end {align*}
This gives one equation to solve for \(\varphi _{22}\) (the first row equation is only used)\[ 2.808-2\varphi _{22}=0 \]
Hence \[ \varphi _{22}=\frac {-2.808}{-2}=1.404 \]
The second eigen vector is\[ \boldsymbol {\varphi }_{2}=\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} =\begin {Bmatrix} 1\\ 1.404 \end {Bmatrix} \]
Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors) \[ \mu _{1}=\boldsymbol {\varphi }_{1}^{T}\left [ M\right ] \boldsymbol {\varphi }_{1}\]
Hence\begin {align*} \mu _{1} & =\begin {Bmatrix} 1\\ -0.237\, \end {Bmatrix} ^{T}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix}\begin {Bmatrix} 1\\ -0.237\, \end {Bmatrix} \\ & =1.\,\allowbreak 169 \end {align*}
Similarly, \(\mu _{2}\) is found\[ \mu _{2}=\boldsymbol {\varphi }_{2}^{T}\left [ M\right ] \boldsymbol {\varphi }_{2} \]
Hence\begin {align*} \mu _{2} & =\begin {Bmatrix} 1\\ 1.404 \end {Bmatrix} ^{T}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix}\begin {Bmatrix} 1\\ 1.404 \end {Bmatrix} \\ & =6.914 \end {align*}
Now that \(\mu _{1},\mu _{2}\) are found, the mass normalized eigen vectors are found. They are called \(\boldsymbol {\Phi }_{1},\boldsymbol {\Phi }_{2}\)\begin {align*} \boldsymbol {\Phi }_{1} & =\frac {\boldsymbol {\varphi }_{1}}{\sqrt {\mu _{1}}}=\frac {\begin {Bmatrix} \varphi _{11}\\ \varphi _{21}\end {Bmatrix} }{\sqrt {\mu _{1}}}=\frac {\begin {Bmatrix} 1\\ -0.237\, \end {Bmatrix} }{\sqrt {1.169}}\\ & =\begin {Bmatrix} 0.925\\ -0.219\, \end {Bmatrix} \allowbreak \end {align*}
Similarly\begin {align*} \boldsymbol {\Phi }_{2} & =\frac {\boldsymbol {\varphi }_{2}}{\sqrt {\mu _{2}}}=\frac {\begin {Bmatrix} \varphi _{12}\\ \varphi _{22}\end {Bmatrix} }{\sqrt {\mu _{2}}}=\frac {\begin {Bmatrix} 1\\ 1.404 \end {Bmatrix} }{\sqrt {6.\allowbreak 914}}\\ & =\begin {Bmatrix} 0.380\\ 0.534\, \end {Bmatrix} \end {align*}
Therefore, the modal transformation matrix is \begin {align*} \left [ \Phi \right ] & =\left [ \boldsymbol {\Phi }_{1}\boldsymbol {\Phi }_{2}\right ] \\ & =\begin {bmatrix} 0.925 & 0.380\\ -0.219 & 0.534 \end {bmatrix} \end {align*}
This result can be verified using Matlab’s eig function as follows
K=[3 -2;-2 2]; M=[1 0;0 3]; [phi,lam]=eig(K,M) phi = -0.3803 -0.9249 -0.5340 0.2196 diag(sqrt(lam)) 0.4380 1.8641
Matlab result agrees with the result obtained above. The sign difference is not important.
Now step 5 is applied. Matlab generates mass normalized eigenvectors by default.
Now that \(\left [ \Phi \right ] \) is found, the transformation from the normal coordinates \(\left \{ x\right \} \) to modal coordinates, called \(\left \{ \eta \right \} ,\) is obtained\begin {align*} \left \{ x\right \} & =\left [ \Phi \right ] \left \{ \eta \right \} \\\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925 & 0.380\\ -0.219 & 0.534 \end {bmatrix}\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} \end {align*}
The transformation from modal coordinates back to normal coordinates is\begin {align*} \left \{ \eta \right \} & =\left [ \Phi \right ] ^{-1}\left \{ x\right \} \\\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925 & 0.380\,\\ -0.219\, & 0.534\, \end {bmatrix} ^{-1}\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} \end {align*}
However, \(\left [ \Phi \right ] ^{-1}=\left [ \Phi \right ] ^{T}\left [ M\right ] \) therefore\begin {align*} \left \{ \eta \right \} & =\left [ \Phi \right ] ^{T}\left [ M\right ] \left \{ x\right \} \\\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925 & 0.380\,\\ -0.219\, & 0.534\, \end {bmatrix} ^{T}\begin {bmatrix} 1 & 0\\ 0 & 3 \end {bmatrix}\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} \\ & =\begin {bmatrix} 0.925\, & -0.657\,\\ 0.38 & 1.6 \end {bmatrix}\begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} \end {align*}
The next step is to apply this transformation to the original equations of motion in order to decouple them.
Applying step 6 results in\begin {align*} \begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} \omega _{1}^{2} & 0\\ 0 & \omega _{2}^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} & =\left [ \Phi \right ] ^{T}\begin {Bmatrix} 0\\ \sin \left ( 5t\right ) \end {Bmatrix} \\\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 1.864^{2} & 0\\ 0 & 0.438^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} & =\begin {bmatrix} 0.925 & 0.380\\ -0.219 & 0.534 \end {bmatrix} ^{T}\begin {Bmatrix} 0\\ \sin \left ( 5t\right ) \end {Bmatrix} \\\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 3.\,\allowbreak 47 & 0\\ 0 & 0.192\, \end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\end {Bmatrix} & =\begin {Bmatrix} -0.219\,\sin \left ( 5t\right ) \\ 0.534\sin \left ( 5t\right ) \end {Bmatrix} \allowbreak \end {align*}
The EOM are now decoupled and each EOM can be solved easily as follows\begin {align*} \eta _{1}^{\prime \prime }\left ( t\right ) +3.47\eta _{1}\left ( t\right ) & =-0.219\,\sin \left ( 5t\right ) \\ \eta _{2}^{\prime \prime }\left ( t\right ) +0.192\eta _{2}\left ( t\right ) & =0.534\sin \left ( 5t\right ) \end {align*}
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules\begin {align*} \begin {Bmatrix} \eta _{1}\left ( 0\right ) \\ \eta _{2}\left ( 0\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925\, & -0.657\,\\ 0.38 & 1.6 \end {bmatrix}\begin {Bmatrix} x_{1}\left ( 0\right ) \\ x_{2}\left ( 0\right ) \end {Bmatrix} \\\begin {Bmatrix} \eta _{1}\left ( 0\right ) \\ \eta _{2}\left ( 0\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925\, & -0.657\,\\ 0.38 & 1.6 \end {bmatrix}\begin {Bmatrix} 0\\ 1 \end {Bmatrix} \\ & =\begin {Bmatrix} -0.657\,\\ 1.6 \end {Bmatrix} \end {align*}
and\begin {align*} \begin {Bmatrix} \eta _{1}^{\prime }\left ( 0\right ) \\ \eta _{2}^{\prime }\left ( 0\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925\, & -0.657\,\\ 0.38 & 1.6 \end {bmatrix}\begin {Bmatrix} x_{1}^{\prime }\left ( 0\right ) \\ x_{2}^{\prime }\left ( 0\right ) \end {Bmatrix} \\ & =\begin {bmatrix} 0.925\, & -0.657\,\\ 0.38 & 1.6 \end {bmatrix}\begin {Bmatrix} 1.5\\ 3 \end {Bmatrix} \\ & =\begin {Bmatrix} -0.584\,\\ 5.37 \end {Bmatrix} \end {align*}
Each of these EOM are solved using any of the standard methods. This results in solutions \(\eta _{1}\left ( t\right ) \) and \(\eta _{2}\left ( t\right ) .\) Hence the following EOM’s are solved \begin {align*} \eta _{1}^{\prime \prime }\left ( t\right ) +3.47\eta _{1}\left ( t\right ) & =-0.219\,\sin \left ( 5t\right ) \\ \eta _{1}\left ( 0\right ) & =-0.657\\ \eta _{1}^{\prime }\left ( 0\right ) & =-0.584 \end {align*}
and also \begin {align*} \eta _{2}^{\prime \prime }\left ( t\right ) +0.192\eta _{2}\left ( t\right ) & =0.534\sin \left ( 5t\right ) \\ \eta _{2}\left ( 0\right ) & =1.6\\ \eta _{2}^{\prime }\left ( 0\right ) & =5.37 \end {align*}
The solutions \(\eta _{1}\left ( t\right ) \) , \(\eta _{2}\left ( t\right ) \) are found using basic methods shown in other parts of these notes. The last step is to transform back to normal coordinates by applying step 7\begin {align*} \begin {Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {Bmatrix} & =\begin {bmatrix} 0.925 & 0.380\\ -0.219 & 0.534 \end {bmatrix}\begin {Bmatrix} \eta _{1}\left ( t\right ) \\ \eta _{2}\left ( t\right ) \end {Bmatrix} \\ & =\begin {Bmatrix} 0.925\,\eta _{1}+0.38\eta _{2}\\ 0.534\,\eta _{2}-0.219\,\eta _{1}\end {Bmatrix} \allowbreak \end {align*}
Hence\[ x_{1}\left ( t\right ) =0.925\eta _{1}\left ( t\right ) +0.38\eta _{2}\left ( t\right ) \]
and\[ x_{2}\left ( t\right ) =0.534\eta _{1}\left ( t\right ) -0.219\eta _{2}\left ( t\right ) \]
The above shows that the solution \(x_{1}\left ( t\right ) \) and \(x_{2}\left ( t\right ) \) has contributions from both nodal solutions.
Given a periodic function \(f\left ( t\right ) \) with period \(T\) then its Fourier series approximation \(\tilde {f}\left ( t\right ) \) using \(N\) terms is\begin {align*} \tilde {f}\left ( t\right ) & =\frac {1}{2}F_{0}+\operatorname {Re}\left ( {\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}t}\right ) \\ & =\frac {1}{2}F_{0}+\frac {1}{2}{\displaystyle \sum \limits _{n=1}^{N}} F_{n}e^{in\frac {2\pi }{T}t}+F_{n}^{\ast }e^{-in\frac {2\pi }{T}t}\\ & =\frac {1}{2}{\displaystyle \sum \limits _{n=-N}^{N}} F_{n}e^{in\frac {2\pi }{T}t} \end {align*}
Where\begin {align*} F_{n} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) e^{-in\frac {2\pi }{T}t}dt\\ F_{0} & =\frac {2}{T}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) dt \end {align*}
Another way to write the above is to use the classical representation using \(\cos \) and \(\sin \). The same coefficients (i.e. the same series) will result.\begin {align*} \tilde {f}\left ( t\right ) & =a_{0}+{\displaystyle \sum \limits _{n=1}^{N}} a_{n}\cos n\frac {2\pi }{T}t+{\displaystyle \sum \limits _{n=1}^{N}} b_{n}\sin n\frac {2\pi }{T}t\\ a_{0} & =\frac {1}{T}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) dt\\ a_{n} & =\frac {1}{T/2}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) \cos \left ( n\frac {2\pi }{T}t\right ) dt\\ b_{n} & =\frac {1}{T/2}{\displaystyle \int \limits _{0}^{T}} f\left ( t\right ) \sin \left ( n\frac {2\pi }{T}t\right ) dt \end {align*}
Just watch out in the above, that we divide by the full period when finding \(a_{0}\) and divide by half the period for all the other coefficients. In the end, when we find \(\tilde {f}\left ( t\right ) \) we can convert that to complex form. The complex form seems easier to use.
Let steady state \[ x_{ss}=\operatorname {Re}\left \{ \frac {\hat {F}}{k}D\left ( r,\zeta \right ) e^{i\varpi t}\right \} \]
Then\begin {align*} f_{tr}\left ( t\right ) & =f_{spring}+f_{damper}\\ & =kx+c\dot {x}\\ & =\operatorname {Re}\left \{ k\frac {\hat {F}}{k}D\left ( r,\zeta \right ) e^{i\varpi t}\right \} +\operatorname {Re}\left \{ ci\varpi \frac {\hat {F}}{k}D\left ( r,\zeta \right ) e^{i\varpi t}\right \} \\ & =\operatorname {Re}\left \{ \left ( \hat {F}+ci\varpi \frac {\hat {F}}{k}\right ) D\left ( r,\zeta \right ) e^{i\varpi t}\right \} \end {align*}
Hence \[ \left \vert f_{tr}\left ( t\right ) \right \vert _{\max }=\left \vert \hat {F}\right \vert \left \vert D\right \vert \sqrt {1+c^{2}\frac {\varpi ^{2}}{k^{2}}}=\left \vert \hat {F}\right \vert \left \vert D\right \vert \sqrt {1+\left ( 2\zeta r\right ) ^{2}} \]
So TR or force transmissibility is\[ TR=\frac {\left \vert f_{tr}\left ( t\right ) \right \vert _{\max }}{\left \vert \hat {F}\right \vert }=\left \vert D\right \vert \sqrt {1+\left ( 2\zeta r\right ) ^{2}} \]
If \(r>\sqrt {2}\) then we want small \(\zeta \) to reduce force transmitted to base. For \(r<\sqrt {2}\), it is the other way round.
We need transfer function between \(y\) and \(z.\) Equation of motion\begin {align*} my^{\prime \prime } & =-c\left ( y^{\prime }-z^{\prime }\right ) -k\left ( y-z\right ) \\ my^{\prime \prime }+cy^{\prime }+ky & =cz^{\prime }+kz \end {align*}
Let \(z=\operatorname {Re}\left \{ Ze^{i\omega t}\right \} ,z^{\prime }=\operatorname {Re}\left \{ i\omega Ze^{i\omega t}\right \} \) and let \(y=\operatorname {Re}\left \{ Ye^{i\omega t}\right \} ,y^{\prime }=\operatorname {Re}\left \{ i\omega Ye^{i\omega t}\right \} ,y^{\prime \prime }=\operatorname {Re}\left \{ -\omega ^{2}Ye^{i\omega t}\right \} \), hence the above becomes\begin {align*} m\operatorname {Re}\left \{ -\omega ^{2}Ye^{i\omega t}\right \} +c\operatorname {Re}\left \{ i\omega Ye^{i\omega t}\right \} +k\operatorname {Re}\left \{ Ye^{i\omega t}\right \} & =c\operatorname {Re}\left \{ i\omega Ze^{i\omega t}\right \} +k\operatorname {Re}\left \{ Ze^{i\omega t}\right \} \\ Y & =\frac {ci\omega +k}{-\omega ^{2}m+ci\omega +k}Z\\ & =\frac {i2\zeta \omega _{n}m\omega +k}{-\omega ^{2}m+i2\zeta \omega _{n}m\omega +k}Z\\ & =\frac {i2\zeta \omega _{n}\omega +\omega _{n}^{2}}{-\omega ^{2}+i2\zeta \omega _{n}\omega +\omega _{n}^{2}}Z\\ & =\frac {i2\zeta r+1}{\left ( 1-r^{2}\right ) +i2\zeta r}Z \end {align*}
Hence \(\left \vert D\left ( r,\zeta \right ) \right \vert =\frac {\sqrt {1+\left ( 2\zeta r\right ) ^{2}}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}}\) and \(\arg \left ( D\right ) =\tan ^{-1}\left ( 2\zeta r\right ) -\tan ^{-1}\left ( \frac {2\zeta r}{1-r^{2}}\right ) \) where \(r=\frac {\omega }{\omega _{n}}\).
Hence for good vibration isolation we need \(\frac {\left \vert Y\right \vert _{\max }}{\left \vert Z\right \vert }\) to be small. i.e. \(\left \vert D\right \vert \sqrt {1+\left ( 2\zeta r\right ) ^{2}}\) to be small. This is the same TR as for force isolation above.
For small \(\left \vert D\right \vert \), we need small \(\zeta \) and small \(k\) (the small \(k\) is to make \(r>\sqrt {2}\)) see plot
In Matlab, the above can be plotted using
close all; zeta = linspace(0.1, 0.7, 10); r = linspace(0, 3, 10); D0 = @(r,z) (sqrt(1+(2*z*r).^2)./sqrt((1-r.^2).^2+(2*z*r).^2)); figure; hold on; for i = 1:length(zeta) plot(r,D0(r,zeta(i))); end grid on;
We need transfer function between \(u\) and \(z_{a}\) where now \(z_{a}\) is the amplitude of the ground acceleration. This device is used to measure base acceleration by relating it linearly to relative displacement of \(m\) to base.
Equation of motion. We use relative distance now.\begin {align*} m\left ( u^{\prime \prime }+z^{\prime \prime }\right ) +cu^{\prime }+ku & =0\\ mu^{\prime \prime }+cu^{\prime }+ku & =-mz^{\prime \prime } \end {align*}
Let \(z^{\prime \prime }=\operatorname {Re}\left \{ Z_{a}e^{i\omega t}\right \} .\) Notice we here jumped right away to the \(z^{\prime \prime }\) itself and wrote it as \(\operatorname {Re}\left \{ Z_{a}e^{i\omega t}\right \} \) and we did not go through the steps as above starting from base motion. This is because we want the transfer function between relative motion \(u\) and acceleration of base.
Now, \(u=\operatorname {Re}\left \{ Ue^{i\omega t}\right \} ,u^{\prime }=\operatorname {Re}\left \{ i\omega Ue^{i\omega t}\right \} ,u^{\prime \prime }=\operatorname {Re}\left \{ -\omega ^{2}Ue^{i\omega t}\right \} \), hence the above becomes\begin {align*} m\operatorname {Re}\left \{ -\omega ^{2}Ue^{i\omega t}\right \} +c\operatorname {Re}\left \{ i\omega Ue^{i\omega t}\right \} +k\operatorname {Re}\left \{ Ue^{i\omega t}\right \} & =-m\operatorname {Re}\left \{ Z_{a}e^{i\omega t}\right \} \\ U & =\frac {-m}{-\omega ^{2}m+i\omega c+k}Z_{a}\\ & =\frac {-1}{-\omega ^{2}+i\omega 2\zeta \omega _{n}+\omega _{n}^{2}}Z_{a}\\ & =\frac {-1}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) +i\omega 2\zeta \omega _{n}}Z_{a} \end {align*}
Hence \(\left \vert D\left ( r,\zeta \right ) \right \vert =\frac {-1}{\sqrt {\left ( \omega _{n}^{2}-\omega ^{2}\right ) ^{2}+\left ( 2\omega \zeta \omega _{n}\right ) ^{2}}}\) and \(\arg \left ( D\right ) =-180^{0}-\tan ^{-1}\left ( \frac {2\omega \zeta \omega _{n}}{\omega _{n}^{2}-\omega ^{2}}\right ) \)
When system is very stiff, which means \(\omega _{n}\) very large compared to \(\omega \) , then \(D\left ( r,\zeta \right ) \approx \) \(\frac {-1}{\omega _{n}^{2}}Z_{a}\), hence by measuring \(u\) we estimate \(Z_{a}\) the amplitude of the ground acceleration since \(\omega _{n}^{2}\) is known. For accuracy, need \(\omega _{n}>5\omega \) at least.
Now we need to measure the base motion (not base acceleration like above). But we still use the relative displacement. Now the transfer function is between \(u\) and \(z\) where now \(z\) is the base motion amplitude.
Equation of motion. We use relative distance now.\begin {align*} m\left ( u^{\prime \prime }+z^{\prime \prime }\right ) +cu^{\prime }+ku & =0\\ mu^{\prime \prime }+cu^{\prime }+ku & =-mz^{\prime \prime } \end {align*}
Let \(z=\operatorname {Re}\left \{ Ze^{i\omega t}\right \} ,z^{\prime }=\operatorname {Re}\left \{ i\omega Ze^{i\omega t}\right \} \) ,\(z^{\prime \prime }=\operatorname {Re}\left \{ -\omega ^{2}Ze^{i\omega t}\right \} ,\)and let \(u=\operatorname {Re}\left \{ Ue^{i\omega t}\right \} ,u^{\prime }=\operatorname {Re}\left \{ i\omega Ue^{i\omega t}\right \} ,u^{\prime \prime }=\operatorname {Re}\left \{ -\omega ^{2}Ue^{i\omega t}\right \} \), hence the above becomes
Now, \(u=\operatorname {Re}\left \{ Ue^{i\omega t}\right \} ,u^{\prime }=\operatorname {Re}\left \{ i\omega Ue^{i\omega t}\right \} ,u^{\prime \prime }=\operatorname {Re}\left \{ -\omega ^{2}Ue^{i\omega t}\right \} \), hence the above becomes\begin {align*} m\operatorname {Re}\left \{ -\omega ^{2}Ue^{i\omega t}\right \} +c\operatorname {Re}\left \{ i\omega Ue^{i\omega t}\right \} +k\operatorname {Re}\left \{ Ue^{i\omega t}\right \} & =-m\operatorname {Re}\left \{ -\omega ^{2}Ze^{i\omega t}\right \} \\ U & =\frac {m\omega ^{2}}{-\omega ^{2}m+i\omega c+k}Z\\ & =\frac {\omega ^{2}}{-\omega ^{2}+i\omega 2\zeta \omega _{n}+\omega _{n}^{2}}Z\\ & =\frac {r^{2}}{\left ( 1-r^{2}\right ) +i2\zeta r}Z \end {align*}
Hence \(\left \vert D\left ( r,\zeta \right ) \right \vert =\frac {r^{2}}{\sqrt {\left ( 1-r^{2}\right ) +i2\zeta r}}\) and \(\arg \left ( D\right ) =-\tan ^{-1}\left ( \frac {2\zeta r}{1-r^{2}}\right ) \)
Now if \(r\) is very large, which happens when \(\omega _{n}\ll \omega \), then \(\frac {1}{\left ( 1-r^{2}\right ) +i2\zeta r}\Rightarrow \frac {1}{-r^{2}}\) since \(r^{2}\) is the dominant factor. Therefore \(U=\frac {r^{2}}{\left ( 1-r^{2}\right ) +i2\zeta r}Z_{a}\) now becomes \(U\simeq -Z_{a}\) therefore measuring the relative displacement \(U\) gives linear estimate of the ground motion. However, this device requires that \(\omega _{n}\) be much smaller than \(\omega \), which means that \(m\) has to be massive. So this device is heavy compared to accelerometer.
For good isolation of mass from ground motion, rule of thumb: Make damping low, and stiffness low (soft spring).
Equation used \(f_{\text {tr}}(t)=f_{\text {spring}}+f_{\text {damper}}\)
Transfer function \(\frac {\left \vert f_{\text {tr}}(t)\right \vert _{\max }}{\left \vert \hat {F}\right \vert }=\left \vert D\right \vert \sqrt {1+\left ( 2\zeta r\right ) ^{2}}\)
Equation used. Use absolute mass position \[ my^{\prime \prime }+cy^{\prime }+ky=cz^{\prime }+kz \]
Transfer function \[ \frac {\left \vert Y\right \vert _{\max }}{\left \vert Z\right \vert }=\left \vert D\right \vert \sqrt {1+\left ( 2\zeta r\right ) ^{2}}\]
Equation used. Use relative mass position \[ mu^{\prime \prime }+cu^{\prime }+ku=-mz^{\prime \prime }\]
Transfer function \begin {align*} U & =\frac {-1}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) +i\omega 2\zeta \omega _{n}}Z_{a}\Rightarrow \left \vert D\left ( r,\zeta \right ) \right \vert \\ & =\frac {-1}{\sqrt {\left ( \omega _{n}^{2}-\omega ^{2}\right ) ^{2}+\left ( 2\omega \zeta \omega _{n}\right ) ^{2}}} \end {align*}
Equation used. Use relative mass position \[ mu^{\prime \prime }+cu^{\prime }+ku=-mz^{\prime \prime }\]
Transfer function \begin {align*} U & =\frac {r^{2}}{(1-r^{2})+i2\zeta r}Z\rightarrow \left \vert D(r,\zeta )\right \vert \\ & =\frac {r^{2}}{\sqrt {(1-r^{2})+i2\zeta r}} \end {align*}
These definitions are used throughout the derivations below.\begin {align*} \xi & =\frac {c}{c_{r}}=\frac {c}{2\sqrt {km}}=\frac {c}{2\omega _{n}m}\\ u_{st} & =\frac {F}{k}\text { static deflection}\\ \omega _{n} & =\sqrt {\frac {k}{m}}\\ \omega _{D} & =\omega _{n}\sqrt {1-\xi ^{2}}\text {note: not defined for }\xi >1\text { since becomes complex}\\ r & =\frac {\varpi }{\omega _{n}}\\ T_{d} & =\frac {2\pi }{\omega _{d}}\ \text {damped period of oscillation}\\ \tau & =\left \{ \frac {-1}{\lambda _{1}},\frac {-1}{\lambda 2}\right \} \text { time constants where }\lambda _{i}\text { are roots of characteristic equation}\\ \beta & =\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\xi \right ) ^{2}}}\text { magnification factor}\\ \text { }\beta _{\max }\text { when }r & =\sqrt {1-2\xi ^{2}}\\ \beta _{\max } & =\frac {1}{2\xi \sqrt {1-\xi ^{2}}}\\ \frac {y_{n}}{y_{n+1}} & =e^{\frac {\zeta \omega _{n}2\pi }{\omega _{D}}}\overset {\text {small damping}}{\Rightarrow }e^{\frac {\zeta 2\pi }{\sqrt {1-\zeta ^{2}}}}\Rightarrow e^{\zeta 2\pi }\\ \ln \left ( \frac {y_{n}}{y_{n+1}}\right ) & =\zeta 2\pi \\ \frac {1}{M}\ln \left ( \frac {y_{n}}{y_{n+M}}\right ) & =\zeta 2\pi \end {align*}
\[ mu^{\prime \prime }+ku=F\sin \varpi t \]
Since there is no damping in the system, then there is no steady state solution. In other words, the particular solution is not the same as the steady state solution in this case. We need to find the particular solution using method on undetermined coefficients.
Let \(u=\) \(u_{h}+u_{p}.\) By guessing that \(u_{p}=c_{1}\sin \varpi t\) then we find the solution to be
\[ u=A\cos \omega _{n}t+B\sin \omega _{n}t+\frac {F}{k}\frac {1}{1-r^{2}}\sin \varpi t \]
Applying initial conditions is always done on the full solution. Applying initial conditions gives\[ u\left ( 0\right ) =A \]\begin {align*} u^{\prime }\left ( t\right ) & =-A\omega \sin \omega _{n}t+B\omega \cos \omega _{n}t+\varpi \frac {F}{k}\frac {1}{1-r^{2}}\cos \varpi t\\ u^{\prime }\left ( 0\right ) & =B\omega _{n}+\varpi \frac {F}{k}\frac {1}{1-r^{2}}\\ B & =\frac {u^{\prime }\left ( 0\right ) }{\omega _{n}}-\frac {F}{k}\frac {r}{1-r^{2}} \end {align*}
Where \(r=\frac {\varpi }{\omega _{n}}\)
The complete solution is
\begin {equation} u\left ( t\right ) =u\left ( 0\right ) \cos \omega _{n}t+\left ( \frac {u^{\prime }\left ( 0\right ) }{\omega _{n}}-\frac {F}{k}\frac {r}{1-r^{2}}\right ) \sin \omega _{n}t+\frac {F}{k}\frac {1}{1-r^{2}}\sin \varpi t \tag {1} \end {equation}
Example: Given force \(f\left ( t\right ) =3\sin \left ( 5t\right ) \) then \(\varpi =5\) rad/sec, and \(\hat {F}=3\). Let \(m=1,k=1\), then \(\omega _{n}=1\) rad/sec. Hence \(r=5\), Let initial conditions be zero, then
\begin {align*} u & =\left ( -3\frac {5}{1-5^{2}}\right ) \sin t+3\frac {1}{1-5^{2}}\sin 5t\\ & =0.625\,\sin t-0.125\,\sin 5.0t \end {align*}
1.4.2.1.1 Resonance forced vibration When \(\varpi \approx \omega \) we obtain resonance since \(r\rightarrow 1\) in the solution given in Eq (1) above and as written the solution can not be used for analysis. To obtain a solution for resonance some calculus is needed. Eq (1) is written as
\begin {equation} u\left ( t\right ) =u\left ( 0\right ) \cos \omega t+\left ( \frac {u^{\prime }\left ( 0\right ) }{\omega }-\frac {F}{k}\frac {\omega \varpi }{\omega ^{2}-\varpi ^{2}}\right ) \sin \omega t+\frac {F}{k}\frac {\omega ^{2}}{\omega ^{2}-\varpi ^{2}}\sin \varpi t \tag {1A} \end {equation} When \(\varpi \approx \omega \) but less than \(\omega \), letting \begin {equation} \omega -\varpi =2\Delta \tag {2} \end {equation} where \(\Delta \) is very small positive quantity. And since \(\varpi \approx \omega \) let \begin {equation} \omega +\varpi \approx 2\varpi \tag {3} \end {equation} Multiplying Eq (2) and (3) gives\begin {equation} \omega ^{2}-\varpi ^{2}=4\Delta \varpi \tag {4} \end {equation} Eq (1A) can now be written in terms of Eqs (2,3) as\begin {align*} u\left ( t\right ) & =u\left ( 0\right ) \cos \omega t+\left ( \frac {u^{\prime }\left ( 0\right ) }{\omega }-\frac {F}{k}\frac {\omega \varpi }{4\Delta \varpi }\right ) \sin \omega t+\frac {F}{k}\frac {\omega ^{2}}{4\Delta \varpi }\sin \varpi t\\ & =u\left ( 0\right ) \cos \omega t+\left ( \frac {v_{0}}{\omega }-\frac {F}{k}\frac {\omega }{4\Delta }\right ) \sin \omega t+\frac {F}{k}\frac {\omega ^{2}}{4\Delta \varpi }\sin \varpi t \end {align*}
Since \(\varpi \approx \omega \) the above becomes\begin {align*} u\left ( t\right ) & =u\left ( 0\right ) \cos \omega t+\left ( \frac {u^{\prime }\left ( 0\right ) }{\omega }-\frac {F}{k}\frac {\omega }{4\Delta }\right ) \sin \omega t+\frac {F}{k}\frac {\omega }{4\Delta }\sin \varpi t\\ & =u\left ( 0\right ) \cos \omega t+\frac {u^{\prime }\left ( 0\right ) }{\omega }\sin \omega t+\frac {F}{k}\frac {\omega }{4\Delta }\left ( \sin \varpi t-\sin \omega t\right ) \end {align*}
Using \(\sin \varpi t-\sin \omega t=2\sin \left ( \frac {\varpi -\omega }{2}t\right ) \cos \left ( \frac {\varpi +\omega }{2}t\right ) \) the above becomes\[ u\left ( t\right ) =u\left ( 0\right ) \cos \omega t+\frac {u^{\prime }\left ( 0\right ) }{\omega }\sin \omega t+\frac {F}{k}\frac {\omega }{2\Delta }\left ( \sin \left ( \frac {\varpi -\omega }{2}t\right ) \cos \left ( \frac {\varpi +\omega }{2}t\right ) \right ) \] From Eqs (2,3) the above can be written as\[ u\left ( t\right ) =u\left ( 0\right ) \cos \omega t+\frac {u^{\prime }\left ( 0\right ) }{\omega }\sin \omega t+\frac {F}{k}\frac {\omega }{2\Delta }\left ( \sin \left ( -\Delta t\right ) \cos \left ( \varpi t\right ) \right ) \] Since \(\lim _{\Delta \rightarrow 0}\frac {\sin \left ( \Delta t\right ) }{\Delta }=t \) the above becomes\[ u\left ( t\right ) =u\left ( 0\right ) \cos \omega t+\frac {u^{\prime }\left ( 0\right ) }{\omega }\sin \omega t-\frac {F}{k}\frac {\omega t}{2}\cos \left ( \omega t\right ) \] This is the solution to use for resonance.
\[ mu^{\prime \prime }+cu^{\prime }+ku=F\sin \varpi t \]\[ u^{\prime \prime }+2\xi \omega u^{\prime }+\omega ^{2}u=\frac {F}{m}\sin \varpi t \] The solution is\[ u\left ( t\right ) =u_{h}+u_{p} \] where \[ u_{h}\left ( t\right ) =e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \] and \[ u_{p}\left ( t\right ) =\frac {F}{\sqrt {\left ( k-m\varpi \right ) ^{2}+\left ( c\varpi \right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \] where \[ \tan \theta =\frac {c\varpi }{k-m\varpi ^{2}}=\frac {2\xi r}{1-r^{2}} \]
Very important note here in the calculations of \(\tan \theta \) above, one should be careful on the sign of the denominator. When the forcing frequency \(\varpi >\omega \) the denominator will become negative (the case of \(\varpi =\omega \) is resonance and is handled separately). Therefore, one should use \(\arctan \) that takes care of which quadrant the angle is. For example, in Mathematica use
ArcTan[1 - r^2, 2 Zeta r]]
and in Matlab use
atan2(2 Zeta r,1 - r^2)
Otherwise, wrong solution will result when \(\varpi >\omega \) The full solution is\begin {equation} u\left ( t\right ) =e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \tag {1} \end {equation} Applying initial conditions gives\begin {align*} A & =u\left ( 0\right ) +\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \theta \\ B & =\frac {u^{\prime }\left ( 0\right ) }{\omega _{d}}+\frac {u\left ( 0\right ) \xi \omega }{\omega _{d}}+\frac {F}{k}\frac {1}{\omega _{d}\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\left ( \xi \omega \sin \theta -\varpi \cos \theta \right ) \end {align*}
Another form of these equations is given as follows
\[ u_{p}=\frac {p_{0}}{k}\frac {1}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}\left ( \left ( 1-r^{2}\right ) \sin \varpi t-2\zeta r\cos \varpi t\right ) \]
Hence the full solution is
\begin {equation} u\left ( t\right ) =e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\frac {F}{k}\frac {1}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}\left ( \left ( 1-r^{2}\right ) \sin \varpi t-2\zeta r\cos \varpi t\right ) \label {eq1} \end {equation}
Applying initial conditions now gives
\begin {align*} A & =u\left ( 0\right ) +\frac {2Fr\xi }{k}\frac {1}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}\\ B & =\frac {u^{\prime }\left ( 0\right ) }{\omega _{d}}+\frac {u\left ( 0\right ) \xi \omega _{n}}{\omega _{d}}-\frac {F\left ( 1-r^{2}\right ) }{k\omega _{d}}\frac {\varpi }{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}+\frac {2Fr\zeta }{k\omega _{d}}\frac {\omega _{n}}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}} \end {align*}
The above 2 sets of equations are equivalent. One uses the phase angle explicitly and the second ones do not. Also, the above assume the force is \(F\sin \varpi t\) and not \(F\cos \varpi t\). If the force is \(F\cos \varpi t\) then in Eq 1.7 above, the term reverse places as in
\[ u\left ( t\right ) =e^{-\xi \omega _{n}t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\frac {F}{k}\frac {1}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}\left ( \left ( 1-r^{2}\right ) \cos \varpi t-2\zeta r\sin \varpi t\right ) \]
Applying initial conditions now gives
\begin {align*} A & =u\left ( 0\right ) +\frac {F}{k}\frac {\left ( 1-r^{2}\right ) }{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}\\ B & =\frac {u^{\prime }\left ( 0\right ) }{\omega _{d}}+\frac {u\left ( 0\right ) \xi \omega _{n}}{\omega _{d}}+\frac {2Fr\zeta }{k\omega _{d}}\frac {\varpi }{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}-\frac {F\left ( 1-r^{2}\right ) }{k\omega _{d}}\frac {\omega _{n}}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}} \end {align*}
When a system is damped, the problem with the divide by zero when \(r=1\) does not occur here as was the case with undamped system, since when when \(\varpi \approx \omega \) or \(r=1\), the solution in Eq (1) becomes\begin {multline*} u\left ( t\right ) =e^{-\xi \omega t}\left ( \left ( u\left ( 0\right ) +\frac {F}{k}\frac {1}{2\xi }\sin \theta \right ) \cos \omega _{d}t+\left ( \frac {u^{\prime }\left ( 0\right ) }{\omega _{d}}+\frac {u\left ( 0\right ) \xi \omega }{\omega _{d}}+\frac {F}{k}\frac {1}{2\omega _{d}\xi }\left ( \xi \omega \sin \theta -\varpi \cos \theta \right ) \right ) \sin \omega _{d}t\right ) \\ +\frac {F}{k}\frac {1}{2}\sin \left ( \varpi t-\theta \right ) \end {multline*} and the problem with the denominator going to zero does not show up here. The amplitude when steady state response is maximum can be found as follows. The amplitude of steady state motion is \(\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\). This is maximum when the magnification factor \(\beta =\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\) is maximum or when \(\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}\) or \(\sqrt {\left ( 1-\left ( \frac {\varpi }{\omega }\right ) ^{2}\right ) ^{2}+\left ( 2\xi \frac {\varpi }{\omega }\right ) ^{2}}\) is minimum. Taking derivative w.r.t. \(\varpi \) and equating the result to zero and solving for \(\varpi \) gives\[ \varpi =\omega \sqrt {1-2\xi ^{2}} \] We are looking for positive \(\varpi \), hence when \(\varpi =\omega \sqrt {1-2\xi ^{2}}\) the under-damped response is maximum.
The solution is\[ u\left ( t\right ) =u_{h}+u_{p} \] Where \(u_{h}=\left ( A+Bt\right ) e^{-\omega t}\) and \(u_{p}=\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \) where \(\tan \theta =\frac {2r}{1-r^{2}}\)(making sure to use correct \(\arctan \) definition). Hence\[ u\left ( t\right ) =\left ( A+Bt\right ) e^{-\omega t}+\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \] where \(A,B\) are found from initial conditions\begin {align*} A & =u\left ( 0\right ) +\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \theta \\ B & =u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega +\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\left ( \omega \sin \theta -\varpi \cos \theta \right ) \end {align*}
The solution is\[ u\left ( t\right ) =u_{h}+u_{p} \] where\[ u_{h}\left ( t\right ) =Ae^{p_{1}t}+Be^{p_{2}t} \] and \[ u_{p}\left ( t\right ) =\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \] hence\[ u=Ae^{p_{1}t}+Be^{p_{2}t}+\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \] where \(\tan \theta =\frac {2\xi r}{1-r^{2}}\) and\begin {align*} p_{1} & =-\frac {c}{2m}+\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\omega \xi +\omega _{n}\sqrt {\xi ^{2}-1}\\ p_{2} & =-\frac {c}{2m}-\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\omega \xi -\omega _{n}\sqrt {\xi ^{2}-1} \end {align*}
Hence the solution is\begin {align*} u\left ( t\right ) & =Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t}+Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t}+\frac {F}{k}\beta \sin \left ( \varpi t-\theta \right ) \\ A & =\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega \xi +u\left ( 0\right ) \omega \sqrt {\xi ^{2}-1}+\frac {F}{k}\beta \left ( \left ( \xi +\sqrt {\xi ^{2}-1}\right ) \omega \sin \theta -\varpi \cos \theta \right ) }{2\omega \sqrt {\xi ^{2}-1}}\\ B & =-\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega \xi -u\left ( 0\right ) \omega \sqrt {\xi ^{2}-1}+\frac {F}{k}\beta \left ( \left ( \xi -\sqrt {\xi ^{2}-1}\right ) \omega \sin \theta -\varpi \cos \theta \right ) }{2\omega \sqrt {\xi ^{2}-1}} \end {align*}
\begin {align*} my^{\prime \prime }+cy^{\prime }+ky & =\operatorname {Re}\left ( \hat {F}e^{i\varpi t}\right ) \\ x & =\operatorname {Re}\left \{ \hat {X}e^{i\varpi t}\right \} \\ \hat {X} & =\frac {\hat {F}}{k}D\left ( r,\zeta \right ) \\ D\left ( r,\zeta \right ) & =\frac {1}{\left ( 1-r^{2}\right ) +2i\zeta r}\\ x & =\operatorname {Re}\left \{ \frac {\hat {F}}{k}\left \vert D\left ( r,\zeta \right ) \right \vert e^{i\left ( \varpi t-\theta \right ) }\right \} \\ \theta & =\tan ^{-1}\frac {2\zeta r}{1-r^{2}} \end {align*}
Let load be harmonic and represented in general as \(\operatorname {Re}\left ( \hat {F}e^{i\varpi t}\right ) \) where \(\hat {F}\) is the complex amplitude of the force.
Hence system is represented by \begin {align*} my^{\prime \prime }+cy^{\prime }+ky & =\operatorname {Re}\left ( \hat {F}e^{i\varpi t}\right ) \\ y^{\prime \prime }+2\zeta \omega _{n}y^{\prime }+\omega _{n}^{2}y & =\operatorname {Re}\left ( \frac {\hat {F}}{m}e^{i\varpi t}\right ) \end {align*}
Let \(y=\operatorname {Re}\left ( \hat {Y}e^{i\varpi t}\right ) \) Hence \(y^{\prime }=\operatorname {Re}\left ( i\varpi \hat {Y}e^{i\varpi t}\right ) ,y^{\prime \prime }=\operatorname {Re}\left ( -\varpi ^{2}\hat {Y}e^{i\varpi t}\right ) \), therefore the differential equation becomes
\begin {align*} \operatorname {Re}\left ( -\varpi ^{2}\hat {Y}e^{i\varpi t}\right ) +2\zeta \omega _{n}\operatorname {Re}\left ( i\varpi \hat {Y}e^{i\varpi t}\right ) +\omega _{n}^{2}\operatorname {Re}\left ( \hat {Y}e^{i\varpi t}\right ) & =\operatorname {Re}\left ( \frac {\hat {F}}{m}e^{i\varpi t}\right ) \\ \left ( -\varpi ^{2}+2\zeta \omega _{n}i\varpi +\omega _{n}^{2}\right ) \hat {Y} & =\frac {\hat {F}}{m}\\ \hat {Y} & =\frac {\frac {\hat {F}}{m}}{\left ( -\varpi ^{2}+2\zeta \omega _{n}i\varpi +\omega _{n}^{2}\right ) } \end {align*}
Dividing numerator and denominator \(\omega _{n}^{2}\) gives\[ \hat {Y}=\frac {\hat {F}}{k}\frac {1}{\left ( 1-r^{2}\right ) +i2\zeta r} \]
Where \(r=\frac {\varpi }{\omega _{n}},\) hence the response is\[ y=\operatorname {Re}\left ( \frac {\hat {F}}{k}\frac {1}{\left ( 1-r^{2}\right ) +i2\zeta r}e^{i\varpi t}\right ) \]
Therefore, the phase of the response is\[ \arg \left ( y\right ) =\arg \left ( \hat {F}\right ) -\tan ^{-1}\left ( \frac {2\zeta r}{\left ( 1-r^{2}\right ) }\right ) +\varpi t \]
Hence at \(t=0\) the phase of the response will be\[ \arg \left ( y\right ) =\arg \left ( \hat {F}\right ) -\tan ^{-1}\left ( \frac {2\zeta r}{\left ( 1-r^{2}\right ) }\right ) \]
So when \(\hat {F}\) is real, the phase of the response is simply \(-\tan ^{-1}\left ( \frac {2\zeta r}{\left ( 1-r^{2}\right ) }\right ) \)
Undamped case
When \(\zeta =0\) the above becomes\begin {align*} y & =\operatorname {Re}\left ( \frac {\hat {F}}{k}\frac {1}{\left ( 1-r^{2}\right ) }e^{i\varpi t}\right ) \\ & =\frac {\left \vert \hat {F}\right \vert }{k}\frac {1}{\left ( 1-r^{2}\right ) }\cos \left ( \varpi t+\arg \left ( \hat {F}\right ) \right ) \end {align*}
For real force this becomes\[ y=\frac {F}{k}\frac {1}{\left ( 1-r^{2}\right ) }\cos \left ( \varpi t\right ) \]
The magnitude \(\left \vert \hat {Y}\right \vert =\frac {F}{k}\frac {1}{\left ( 1-r^{2}\right ) }\) and phase zero.
damped cases
\(\zeta >0\)\[ y=\operatorname {Re}\left ( \frac {\hat {F}}{k}\frac {1}{\left ( 1-r^{2}\right ) +i2\zeta r}e^{i\varpi t}\right ) \]
\begin {align*} \left \vert \hat {Y}\right \vert & =\frac {\left \vert \hat {F}\right \vert }{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}}\\ \arg \left ( \hat {Y}\right ) & =\phi =\arg \left ( \hat {F}\right ) -\tan ^{-1}\left ( \frac {2\zeta r}{1-r^{2}}\right ) +\varpi t \end {align*}
Hence for real force and at \(t=0\) the phase of displacement is \[ -\tan ^{-1}\left ( \frac {2\zeta r}{1-r^{2}}\right ) \] lag behind the load.
When \(r<1\) then \(\phi \) goes from \(0\) to \(-90^{0}\) Therefore phase of displacement is \(0\) to \(-90^{0}\) behind force. The minus sign at the front was added since the complex number is in the denominator. Hence the response will always be lagging in phase relative for load.
For \(r>1\)
Now \(1-r^{2}\) is negative, hence the phase will be from \(-90^{\circ }\) to \(-180^{\circ }\)
When \(r=1\)
\[ y=\operatorname {Re}\left ( \frac {\hat {F}}{k}\frac {1}{i2\zeta }e^{i\varpi t}\right ) \]
\begin {align*} \left \vert \hat {Y}\right \vert & =\frac {\left \vert \hat {F}\right \vert }{k}\frac {1}{2\zeta }\\ \arg \left ( \hat {Y}\right ) & =-90^{\circ } \end {align*}
Now phase is \(-90^{\circ }\)
Examples. System has \(\zeta =0.1\) and \(m=1,k=1\) subjected for force \(3\cos \left ( 0.5t\right ) \) find the steady state solution.
Answer \(y\left ( t\right ) =\operatorname {Re}\left ( \hat {Y}e^{i\varpi t}\right ) \), \(\omega _{n}=\sqrt {\frac {k}{m}}=1\) rad/sec, hence \(r=0.5\) under the response is
\begin {align*} y\left ( t\right ) & =\operatorname {Re}\left ( \left \vert \hat {Y}\right \vert e^{i\varpi t}\right ) \\ & =\left \vert \hat {Y}\right \vert \cos \left ( \varpi t\right ) \\ & =\frac {F}{k}\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\zeta r\right ) ^{2}}}\cos \left ( .5t-\tan ^{-1}\left ( \frac {2\left ( 0.1\right ) 0.5}{1-0.5^{2}}\right ) \right ) \\ & =3\frac {1}{\sqrt {\left ( 1-0.5^{2}\right ) ^{2}+\left ( 2\left ( 0.1\right ) 0.5\right ) ^{2}}}\cos \left ( .5t-7.59^{\circ }\right ) \\ & =3.9649\cos \left ( .5t-7.59^{\circ }\right ) \end {align*}
The equation of motion can also be written as \(u^{\prime \prime }+2\zeta \omega u^{\prime }+\omega ^{2}u=\frac {F}{m}\sin \varpi t\).
The following table gives the solutions for initial conditions are \(u\left ( 0\right ) \) and \(u^{\prime }\left ( 0\right ) \) under all damping conditions. The roots shown are the roots of the quadratic characteristic equation \(\lambda ^{2}+2\zeta \omega \lambda +\omega ^{2}\lambda =0\). Special handling is needed to obtain the solution of the differential equation for the case of \(\zeta =0\) and \(\varpi =\omega \) as described in the detailed section below.
| \(\zeta =0\) |
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| \(\zeta <1\) |
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| \(\zeta =1\) |
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| \(\zeta >1\) |
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\[ mu^{\prime \prime }+ku=F \]\[ u^{\prime \prime }+\omega ^{2}u=F \]\[ u\left ( t\right ) =u_{h}+u_{p} \]
Where \(u_{h}=A\cos \omega t+B\sin \omega t\) and \(u_{p}=\frac {F}{k}\), the solution is\[ u\left ( t\right ) =A\cos \omega t+B\sin \omega t+\frac {F}{k} \]
Applying initial conditions gives\begin {align*} A & =u\left ( 0\right ) -\frac {F}{k}\\ B & =\frac {u^{\prime }\left ( 0\right ) }{\omega } \end {align*}
And complete solution is\[ u\left ( t\right ) =\frac {F}{k}+\left ( u\left ( 0\right ) -\frac {F}{k}\right ) \cos \omega t+\frac {u^{\prime }\left ( 0\right ) }{\omega }\sin \omega t \]
The general solution is \[ u\left ( t\right ) =e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\frac {F}{k} \] From initial conditions\begin {align*} A & =u\left ( 0\right ) -\frac {F}{k}\\ B & =\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \xi \omega -\frac {F}{k}\xi \omega }{\omega _{d}} \end {align*}
Hence the solution is\[ u\left ( t\right ) =e^{-\xi \omega t}\left ( \left ( u\left ( 0\right ) -\frac {F}{k}\right ) \cos \omega _{d}t+\left ( \frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \xi \omega -\frac {F}{k}\xi \omega }{\omega _{d}}\right ) \sin \omega _{d}t\right ) +\frac {F}{k} \]
The general solution is \[ u(t)=(A + B t) e^{-\omega t} + \frac {F}{k} \] Where from initial conditions \begin {align*} A & =u(0) -\frac {F}{k}\\ B & = u^{\prime }(0) + u(0) \omega -\frac {F}{k} \omega \end {align*}
The solution is\[ u\left ( t\right ) =Ae^{p_{1}t}+Be^{p_{2}t}+\frac {F}{k} \] Where now\begin {align*} B & =\frac {\frac {F}{k}p_{1}-u_{0}p_{1}+u^{\prime }\left ( 0\right ) }{\left ( p_{2}-p_{1}\right ) }\\ A & =u\left ( 0\right ) -\frac {F}{k}-B \end {align*}
Hence the solution is\[ u\left ( t\right ) =Ae^{p_{1}t}+Be^{p_{2}t}+\frac {F}{k} \] Where \begin {align*} p_{1} & =-\frac {c}{2m}+\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\omega \xi +\omega _{n}\sqrt {\xi ^{2}-1}\\ p_{2} & =-\frac {c}{2m}-\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\omega \xi -\omega _{n}\sqrt {\xi ^{2}-1} \end {align*}
| \(\zeta =0\) |
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| \(\zeta <1\) |
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| \(\zeta =1\) |
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| \(\zeta >1\) |
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\[ mu^{\prime \prime }+ku=0 \]\[ u^{\prime \prime }+\omega ^{2}u=0 \] The solution is \[ u\left ( t\right ) =u\left ( 0\right ) \cos \omega t+\frac {u^{\prime }\left ( 0\right ) }{\omega }\sin \omega t \]
\[ mu^{\prime \prime }+cu^{\prime }+ku=0 \]\[ u^{\prime \prime }+2\xi \omega u^{\prime }+\omega ^{2}u=0 \]
The solution is\[ u=e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \] Applying initial conditions gives \(A=u\left ( 0\right ) \) and \(B=\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \xi \omega }{\omega _{d}}\). Therefore the solution becomes\[ u\left ( t\right ) =e^{-\xi \omega t}\left ( u\left ( 0\right ) \cos \omega _{d}t+\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \xi \omega }{\omega _{d}}\sin \omega _{d}t\right ) \]
The solution is\begin {align*} u\left ( t\right ) & =\left ( A+Bt\right ) e^{-\left ( \frac {c_{r}}{2m}\right ) t}\\ & =\left ( A+Bt\right ) e^{-\omega t} \end {align*}
where \(A,B\) are found from initial conditions \(A=u\left ( 0\right ) \),\(B=u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega \), hence\[ u\left ( t\right ) =\left ( u\left ( 0\right ) +\left ( u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega \right ) t\right ) e^{-\omega t} \]
The solution is\[ u\left ( t\right ) =Ae^{\lambda _{1}t}+Be^{\lambda _{2}t} \] where \(A,B\) are found from initial conditions.\begin {align*} A & =\frac {u^{\prime }(0)-u\left ( 0\right ) \lambda _{2}}{2\omega \sqrt {\xi ^{2}-1}}\\ B & =\frac {-u^{\prime }(0)+u\left ( 0\right ) \lambda _{1}}{2\omega \sqrt {\xi ^{2}-1}} \end {align*}
where \(\lambda _{1}\) and \(\lambda _{2}\) are the roots of the characteristic equation\begin {align*} \lambda _{1} & =-\frac {c}{2m}+\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\xi \omega +\omega \sqrt {\xi ^{2}-1}\\ \lambda _{2} & =-\frac {c}{2m}-\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\xi \omega -\omega \sqrt {\xi ^{2}-1} \end {align*}
| \(\begin {array} [c]{l}\zeta =0\\ u^{\prime \prime }+\omega ^{2}u=0 \end {array} \) |
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The roots of the characteristic equation for \(u^{\prime \prime }+2\xi \omega u^{\prime }+\omega ^{2}u=0\) are given in this table
| roots | time constant \(\tau \) | |
| \(\xi <1\) | \(\left \{ -\xi \omega +j\omega _{n}\sqrt {1-\xi ^{2}},-\xi \omega -i\omega _{n}\sqrt {1-\xi ^{2}}\right \} \) | \(\frac {1}{\xi \omega }\) |
| \(\xi =1\) | \(\left \{ -\omega ,-\omega \right \} \) | \(\frac {1}{\omega }\) |
| \(\xi >1\) | \(\left \{ -\omega _{n}\xi +\omega _{n}\sqrt {\xi ^{2}-1},-\omega _{n}\xi -\omega _{n}\sqrt {\xi ^{2}-1}\right \} \) | \(\frac {1}{\omega _{n}\xi -\omega _{n}\sqrt {\xi ^{2}-1}},\frac {1}{\omega _{n}\xi +\omega _{n}\sqrt {\xi ^{2}-1}}\)(which to use? the bigger?) |
1.4.5.1.1 Undamped system with impulse \[ m\ddot {u}+ku=F_{0}\delta (t) \]
with initial conditions \(u\left ( 0\right ) =0\) and \(u^{\prime }\left ( 0\right ) =0.\)Assuming the impulse acts for a very short time period from \(0\) to \(t_{1} \) seconds, where \(t_{1}\) is small amount. Integrating the above differential equation gives\[ \int _{0}^{t_{1}}m\ddot {u}dt+\int _{0}^{t_{1}}kudt=\int _{0}^{t_{1}}F_{0}\delta (t) \] Since \(t_{1}\) is very small, it can be assumed that \(u\) changes is negligible, hence the above reduces to\begin {align*} \int _{0}^{t_{1}}m\ddot {u}dt & =\int _{0}^{t_{1}}F_{0}\delta (t)\\ \int _{0}^{t_{1}}m\left ( \frac {d\dot {u}}{dt}\right ) dt & =\int _{0}^{t_{1}}F_{0}\delta (t)\\ \int _{\dot {u}\left ( 0\right ) }^{\dot {u}\left ( t_{1}\right ) }d\dot {u} & =\frac {F_{0}}{m}\int _{0}^{t_{1}}\delta (t)\\ \dot {u}\left ( t_{1}\right ) -\dot {u}\left ( 0\right ) & =\frac {F_{0}}{m}\int _{0}^{t_{1}}\delta (t)\\ \dot {u}\left ( t_{1}\right ) & =\frac {F_{0}}{m}\int _{0}^{t_{1}}\delta (t) \end {align*}
since we assumed \(u^{\prime }\left ( 0\right ) =0\) and since \(\int _{0}^{t_{1}}\delta (t)=1\) then the above reduces to\[ \dot {u}\left ( t_{1}\right ) =\frac {F_{0}}{m} \] Therefore, the effect of the impulse is the same as if the system was a free system but with initial velocity given by \(\frac {F_{0}}{m}\) and zero initial position. Hence the system is now solved as follows\[ m\ddot {u}+ku=0 \] With \(u\left ( 0\right ) =0\) and \(u^{\prime }\left ( 0\right ) =\frac {F_{0}}{m}\). The solution is\[ u_{impulse}\left ( t\right ) =\frac {F_{0}}{m\omega }\sin \omega t \] If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is \(u\left ( t\right ) =u(0)\cos \omega t+\frac {u^{\prime }(0)}{\omega }\sin \omega t\), therefore, the full solution is\[ u\left ( t\right ) =\overset {\text {due to IC only}}{\overbrace {u(0)\cos \omega t+\frac {u^{\prime }(0)}{\omega }\sin \omega t}}+\overset {\text {due to impulse}}{\overbrace {\frac {F_{0}}{m\omega }\sin \omega t}} \]
1.4.5.1.2 under-damped with impulse \(c<c_{r},\xi <1\) \[ m\ddot {u}+c\dot {u}+ku=\delta (t) \] \[ \ddot {u}+2\xi \omega \dot {u}+\omega ^{2}u=\delta (t) \] with initial conditions \(u\left ( 0\right ) =0\) and \(u^{\prime }\left ( 0\right ) =0.\)Integrating gives\[ \int _{0}^{t_{1}}m\ddot {u}dt+\int _{0}^{t_{1}}c\dot {u}dt+\int _{0}^{t_{1}}kudt=\int _{0}^{t_{1}}F_{0}\delta (t) \] Since \(t_{1}\) is very small, it can be assumed that \(u\) changes is negligible as well as the change in velocity, hence the above reduces to the same result as in the case of undamped. Therefore, the system is solved as free system, but with initial velocity \(u^{\prime }\left ( 0\right ) =\) \(F_{0}/m \) and zero initial position.
Initial conditions are \(u\left ( 0\right ) =0\) and \(u^{\prime }\left ( 0\right ) =0\) then the solution is\[ u_{impulse}=e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \]
applying initial conditions gives \(A=0\) and \(B=\frac {\left ( \frac {F_{0}}{m}\right ) }{\omega _{d}}\), hence\[ u_{impulse}\left ( t\right ) =e^{-\xi \omega t}\left ( \frac {F_{0}}{m\omega _{d}}\sin \omega _{d}t\right ) \]
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is \(u\left ( t\right ) =e^{-\xi \omega t}\left ( u(0)\cos \omega _{d}t+\frac {u^{\prime }(0)+u(0)\xi \omega }{\omega _{d}}\sin \omega _{d}t\right ) \), therefore, the full solution is\[ u\left ( t\right ) =\overset {\text {due to IC only}}{\overbrace {e^{-\xi \omega t}\left ( u(0)\cos \omega _{d}t+\frac {u^{\prime }(0)+u(0)\xi \omega }{\omega _{d}}\sin \omega _{d}t\right ) }}+\overset {\text {due to impulse}}{\overbrace {e^{-\xi \omega t}\left ( \frac {F_{0}}{m\omega _{d}}\sin \omega _{d}t\right ) }} \]
1.4.5.1.3 critically damped with impulse input \(\xi =\frac {c}{c_{r}}=1\) with initial conditions \(u\left ( 0\right ) =0\) and \(u^{\prime }\left ( 0\right ) =0\) then the solution is\begin {align*} u\left ( t\right ) & =\left ( A+Bt\right ) e^{-\left ( \frac {c_{r}}{2m}\right ) t}\\ & =\left ( A+Bt\right ) e^{-\omega t} \end {align*}
where \(A,B\) are found from initial conditions \(A=u\left ( 0\right ) =0\) and \(B=u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega =\frac {F_{0}}{m}\), hence the solution is\[ u_{impulse}\left ( t\right ) =\frac {F_{0}t}{m}e^{-\omega t} \]
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is \(u\left ( t\right ) =\left ( u_{0}(1+\omega t)+u^{\prime }\left ( 0\right ) t\right ) e^{-\omega t}\), therefore, the full solution is\[ u\left ( t\right ) =\overset {\text {due to IC only}}{\overbrace {\left ( u\left ( 0\right ) (1+\omega t)+u^{\prime }\left ( 0\right ) t\right ) e^{-\omega t}}}+\overset {\text {due to impulse}}{\overbrace {\frac {F_{0}t}{m}e^{-\omega t}}} \]
1.4.5.1.4 over-damped with impulse input \(\xi =\frac {c}{c_{r}}>1\) With initial conditions are \(u\left ( 0\right ) =0\) and \(u^{\prime }\left ( 0\right ) =0\) the solution is\[ u_{impulse}\left ( t\right ) =Ae^{\lambda _{1}\omega t}+Be^{\lambda _{2}\omega t} \] where \(A,B\) are found from initial conditions and \begin {align*} \lambda _{1} & =-\omega \xi +\omega \sqrt {\xi ^{2}-1}\\ \lambda _{2} & =-\omega \xi -\omega \sqrt {\xi ^{2}-1}\\ A & =\frac {u^{\prime }(0)-u\left ( 0\right ) \lambda _{2}}{2\omega \sqrt {\xi ^{2}-1}}\\ B & =\frac {-u^{\prime }(0)+u\left ( 0\right ) \lambda _{1}}{2\omega \sqrt {\xi ^{2}-1}} \end {align*}
Hence the solution is\[ u_{impulse}\left ( t\right ) =Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t}+Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t} \]
where\begin {align*} A & =\frac {\frac {F_{0}}{m}}{2\omega \sqrt {\xi ^{2}-1}}\\ B & =\frac {-\frac {F_{0}}{m}}{2\omega \sqrt {\xi ^{2}-1}} \end {align*}
Hence
\[ u_{impulse}\left ( t\right ) =\frac {\frac {F_{0}}{m}}{2\omega \sqrt {\xi ^{2}-1}}e^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t}-\frac {\frac {F_{0}}{m}}{2\omega \sqrt {\xi ^{2}-1}}e^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t} \]
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is \(u\left ( t\right ) =Ae^{p_{1}t}+Be^{p_{2}t}\), therefore, the full solution is\[ u\left ( t\right ) =Ae^{\lambda _{1}\omega t}+Be^{\lambda _{2}\omega t}+\frac {\frac {F_{0}}{m}}{2\omega \sqrt {\xi ^{2}-1}}e^{\lambda _{1}\omega t}-\frac {\frac {F_{0}}{m}}{2\omega \sqrt {\xi ^{2}-1}}e^{\lambda _{2}\omega t} \]
\begin {align*} A & =\frac {u^{\prime }\left ( 0\right ) -u\left ( 0\right ) \lambda _{2}}{2\omega \sqrt {\xi ^{2}-1}}\\ B & =\frac {-u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \lambda _{1}}{2\omega \sqrt {\xi ^{2}-1}} \end {align*}
| \(\begin {array} [c]{l}\zeta =0\\ u^{\prime \prime }+\omega ^{2}u=0 \end {array} \) |
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The impulse response can be implemented in Mathematica as
parms = {m -> 10, c -> 1.2, k -> 4.3, a -> 1}; tf = TransferFunctionModel[a/(m s^2 + c s + k) /. parms, s] sol = OutputResponse[tf, DiracDelta[t], t]; Plot[sol, {t, 0, 60}, PlotRange -> All, Frame -> True, FrameLabel -> {{z[t], None}, {Row[{t, " (sec)"}], eq}}, GridLines -> Automatic]
Now assume the input is as follows
given by \(F\left ( t\right ) =F_{0}\sin \left ( \varpi t\right ) \) where \(\varpi =\frac {2\pi }{2t_{1}}=\frac {\pi }{t_{1}}\)
1.4.5.2.1 undamped system with sin impulse \[ m\ddot {u}+ku=\left \{ \begin {array} [c]{ccc}F_{0}\sin \left ( \varpi t\right ) & & 0\leq t\leq t_{1}\\ 0 & & t>t_{1}\end {array} \right . \]
with \(u\left ( 0\right ) =u_{0}\) and \(\dot {u}\left ( 0\right ) =v_{0}.\) For \(0\leq t\leq t_{1}\) the solution is\[ u\left ( t\right ) =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {r}{1-r^{2}}\right ) \sin \omega t+u_{st}\frac {1}{1-r^{2}}\sin \left ( \frac {\pi }{t_{1}}t\right ) \]
where \(r=\frac {\varpi }{\omega }=\frac {\pi /t_{1}}{\omega }=\frac {T}{2t_{1}}\) where \(T\) is the natural period of the system. \(u_{st}=\frac {F_{0}}{k}\), hence the above becomes\begin {equation} u\left ( t\right ) =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-\frac {F_{0}}{k}\frac {\left ( \frac {\pi /t_{1}}{\omega }\right ) }{1-\left ( \frac {\pi /t_{1}}{\omega }\right ) ^{2}}\right ) \sin \omega t+\frac {F_{0}}{k}\frac {1}{1-\left ( \frac {\pi /t_{1}}{\omega }\right ) ^{2}}\sin \left ( \pi \frac {t}{t_{1}}\right ) \tag {1} \end {equation}
When \(u_{0}=0\) and \(v_{0}=0\) then\begin {align*} u\left ( t\right ) & =-\frac {F_{0}}{k}\frac {\left ( \frac {\pi /t_{1}}{\omega }\right ) }{1-\left ( \frac {\pi /t_{1}}{\omega }\right ) ^{2}}\sin \omega t+\frac {F_{0}}{k}\frac {1}{1-\left ( \frac {\pi /t_{1}}{\omega }\right ) ^{2}}\sin \left ( \pi \frac {t}{t_{1}}\right ) \\ u\left ( t\right ) & =\frac {F_{0}}{k}\frac {1}{1-\left ( \frac {\pi /t_{1}}{\omega }\right ) ^{2}}\left ( \sin \left ( \pi \frac {t}{t_{1}}\right ) -\frac {\pi /t_{1}}{\omega }\sin \omega t\right ) \end {align*}
The above Eq (1) gives solution during the time \(0\leq t\leq t_{1}\)
Now after \(t=t_{1}\) the force will disappear, the differential equation becomes\[ m\ddot {u}+ku=0\ \ \ \ \ \ t>t_{1} \]
but with the initial conditions evaluate at \(t=t_{1}.\) From (1)\begin {align} u\left ( t_{1}\right ) & =u_{0}\cos \omega t_{1}+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {r}{1-r^{2}}\right ) \sin \omega t_{1}+u_{st}\frac {1}{1-r^{2}}\sin \varpi t_{1}\nonumber \\ & =u_{0}\cos \omega t_{1}+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {r}{1-r^{2}}\right ) u_{st}\frac {r}{1-r^{2}}\sin \omega t_{1} \tag {2} \end {align}
since \(\sin \varpi t_{1}=0\). taking derivative of Eq (1)\[ \dot {u}\left ( t\right ) =-\omega u_{0}\sin \omega t+\omega \left ( \frac {v_{0}}{\omega }-u_{st}\frac {r}{1-r^{2}}\right ) \cos \omega t+\varpi \frac {1}{1-r^{2}}\cos \varpi t \]
and at \(t=t_{1}\) the above becomes\begin {align} \dot {u}\left ( t_{1}\right ) & =-\omega u_{0}\sin \omega t_{1}+\omega \left ( \frac {v_{0}}{\omega }-u_{st}\frac {r}{1-r^{2}}\right ) \cos \omega t_{1}+\varpi \frac {1}{1-r^{2}}\cos \varpi t_{1}\nonumber \\ & =-\omega u_{0}\sin \omega t_{1}+\omega \left ( \frac {v_{0}}{\omega }-u_{st}\frac {r}{1-r^{2}}\right ) \cos \omega t_{1}-\varpi \frac {1}{1-r^{2}} \tag {3} \end {align}
since \(\cos \varpi t_{1}=-1.\) Now (2) and (3) are used as initial conditions to solve \(m\ddot {u}+ku=0\,\). The solution for \(t>t_{1}\) is\[ u\left ( t\right ) =u\left ( t_{1}\right ) \cos \omega t+\frac {\dot {u}\left ( t_{1}\right ) }{\omega }\sin \omega t \]
Resonance with undamped sin impulse When \(\varpi \approx \omega \) and \(t\leq t_{1}\) we obtain resonance since \(r\rightarrow 1\) in the solution shown up and as written the solution can’t be used for analysis in this case. To obtain a solution for resonance some calculus is needed. Eq (1) is written as\begin {align} u\left ( t\right ) & =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {\frac {\varpi }{\omega }}{1-\left ( \frac {\varpi }{\omega }\right ) ^{2}}\right ) \sin \omega t+u_{st}\frac {1}{1-\left ( \frac {\varpi }{\omega }\right ) ^{2}}\sin \varpi t\nonumber \\ & =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {\omega \varpi }{\omega ^{2}-\varpi ^{2}}\right ) \sin \omega t+u_{st}\frac {\omega ^{2}}{\omega ^{2}-\varpi ^{2}}\sin \varpi t \tag {1A} \end {align}
Now looking at case when \(\varpi \approx \omega \) but less than \(\omega \), hence let \begin {equation} \omega -\varpi =2\Delta \tag {2} \end {equation} where \(\Delta \) is very small positive quantity. and we also have \begin {equation} \omega +\varpi \approx 2\varpi \tag {3} \end {equation}
Multiplying Eq (2) and (3) with each others gives\begin {equation} \omega ^{2}-\varpi ^{2}=4\Delta \varpi \tag {4} \end {equation}
Going back to Eq (1A) and rewriting it as\begin {align*} u\left ( t\right ) & =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {\omega \varpi }{4\Delta \varpi }\right ) \sin \omega t+u_{st}\frac {\omega ^{2}}{4\Delta \varpi }\sin \varpi t\\ & =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {\omega }{4\Delta }\right ) \sin \omega t+u_{st}\frac {\omega ^{2}}{4\Delta \varpi }\sin \varpi t \end {align*}
Since \(\varpi \approx \omega \) the above becomes\begin {align*} u\left ( t\right ) & =u_{0}\cos \omega t+\left ( \frac {v_{0}}{\omega }-u_{st}\frac {\omega }{4\Delta }\right ) \sin \omega t+u_{st}\frac {\omega }{4\Delta }\sin \varpi t\\ & =u_{0}\cos \omega t+\frac {v_{0}}{\omega }\sin \omega t+u_{st}\frac {\omega }{4\Delta }\left ( \sin \varpi t-\sin \omega t\right ) \end {align*}
now using \(\sin \varpi t-\sin \omega t=2\sin \left ( \frac {\varpi -\omega }{2}t\right ) \cos \left ( \frac {\varpi +\omega }{2}t\right ) \) the above becomes\[ u\left ( t\right ) =u_{0}\cos \omega t+\frac {v_{0}}{\omega }\sin \omega t+u_{st}\frac {\omega }{2\Delta }\left ( \sin \left ( \frac {\varpi -\omega }{2}t\right ) \cos \left ( \frac {\varpi +\omega }{2}t\right ) \right ) \]
From Eq(2) \(\varpi -\omega =-2\Delta \) and \(\omega +\varpi \approx 2\varpi \) hence the above becomes\[ u\left ( t\right ) =u_{0}\cos \omega t+\frac {v_{0}}{\omega }\sin \omega t+u_{st}\frac {\omega }{2\Delta }\left ( \sin \left ( -\Delta t\right ) \cos \left ( \varpi t\right ) \right ) \]
or since \(\varpi \approx \omega \)\[ u\left ( t\right ) =u_{0}\cos \omega t+\frac {v_{0}}{\omega }\sin \omega t-u_{st}\frac {\omega }{2\Delta }\left ( \sin \left ( \Delta t\right ) \cos \left ( \omega t\right ) \right ) \]
Now \(\lim _{\Delta \rightarrow 0}\frac {\sin \left ( \Delta t\right ) }{\Delta }=t\) hence the above becomes\[ u\left ( t\right ) =u_{0}\cos \omega t+\frac {v_{0}}{\omega }\sin \omega t-u_{st}\frac {\omega t}{2}\cos \left ( \omega t\right ) \]
This can also be written as\begin {align} u\left ( t\right ) & =u_{0}\cos \varpi t+\frac {v_{0}}{\varpi }\sin \varpi t-u_{st}\frac {\varpi t}{2}\cos \left ( \varpi t\right ) \tag {1}\\ & =u_{0}\cos \left ( \frac {\pi }{t_{1}}t\right ) +\frac {v_{0}}{\varpi }\sin \left ( \frac {\pi }{t_{1}}t\right ) -u_{st}\left ( \frac {\pi }{2t_{1}}t\right ) \cos \left ( \frac {\pi }{t_{1}}t\right ) \nonumber \end {align}
since \(\varpi \approx \omega \) in this case. This is the solution to use for resonance and for \(t\leq t_{1}\)
Hence for \(t>t_{1}\), the above equations is used to determine initial conditions at \(t=t_{1}\)\[ u\left ( t_{1}\right ) =u_{0}\cos \varpi t_{1}+\frac {v_{0}}{\varpi }\sin \varpi t_{1}-u_{st}\frac {\varpi t_{1}}{2}\cos \left ( \varpi t_{1}\right ) \]
but \(\cos \varpi t_{1}=\cos \frac {\pi }{t_{1}}t_{1}=-1\) and \(\sin \varpi t_{1}=0\) and \(\frac {\varpi t_{1}}{2}=\frac {\pi }{2}\), hence the above becomes\[ u\left ( t_{1}\right ) =-u_{0}+u_{st}\frac {\pi }{2} \]
Taking derivative of Eq (1) gives\[ \dot {u}\left ( t\right ) =-\varpi u_{0}\sin \varpi t+v_{0}\cos \varpi t+u_{st}\frac {\varpi ^{2}t}{2}\sin \left ( \varpi t\right ) -u_{st}\frac {\varpi }{2}\cos \left ( \varpi t\right ) \]
and at \(t=t_{1}\)\begin {align*} \dot {u}\left ( t_{1}\right ) & =-\varpi u_{0}\sin \varpi t_{1}+v_{0}\cos \varpi t_{1}+u_{st}\frac {\varpi ^{2}t_{1}}{2}\sin \left ( \varpi t_{1}\right ) -u_{st}\frac {\varpi }{2}\cos \left ( \varpi t_{1}\right ) \\ & =-v_{0}+u_{st}\frac {\varpi }{2} \end {align*}
Now the solution for \(t>t_{1}\) is\begin {align*} u\left ( t\right ) & =u\left ( t_{1}\right ) \cos \omega t+\frac {\dot {u}\left ( t_{1}\right ) }{\omega }\sin \omega t\\ & =\left ( -u\left ( 0\right ) +u_{st}\frac {\pi }{2}\right ) \cos \omega t+\frac {-u^{\prime }\left ( 0\right ) +u_{st}\frac {\pi }{2t_{1}}}{\omega }\sin \omega t \end {align*}
1.4.5.2.2 under-damped with sin impulse \(c<c_{r},\xi <1\) \[ m\ddot {u}+c\dot {u}+ku=\left \{ \begin {array} [c]{ccc}F_{0}\sin \left ( \varpi \right ) & & 0\leq t\leq t_{1}\\ 0 & & t>t_{1}\end {array} \right . \]
or\[ \ddot {u}+2\xi \omega \dot {u}+\omega ^{2}u=\left \{ \begin {array} [c]{ccc}F_{0}\sin \left ( \varpi \right ) & & 0\leq t\leq t_{1}\\ 0 & & t>t_{1}\end {array} \right . \]
\[ m\ddot {u}+c\dot {u}+ku=F\sin \varpi t \]
\[ \ddot {u}+2\xi \omega \dot {u}+\omega ^{2}u=\frac {F}{m}\sin \varpi t \]
For \(t\leq t_{1}\)Initial conditions are \(u\left ( 0\right ) =u_{0}\) and \(\dot {u}\left ( 0\right ) =v_{0}\) and \(u_{st}=\frac {F}{k}\) then the solution from above is\begin {equation} u\left ( t\right ) =e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \tag {1} \end {equation}
Applying initial conditions gives\begin {align*} A & =u_{0}+\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \theta \\ B & =\frac {v_{0}}{\omega _{d}}+\frac {u_{0}\xi \omega }{\omega _{d}}+\frac {u_{st}}{\omega _{d}\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\left ( \xi \omega \sin \theta -\varpi \cos \theta \right ) \end {align*}
For \(t>t_{1}\). From (1)\begin {equation} u\left ( t_{1}\right ) =e^{-\xi \omega t_{1}}\left ( A\cos \omega _{d}t_{1}+B\sin \omega _{d}t_{1}\right ) +\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \left ( \varpi t_{1}-\theta \right ) \tag {2} \end {equation}
Taking derivative of (1) gives\begin {multline*} \dot {u}\left ( t\right ) =-\xi \omega e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) +e^{-\xi \omega t}\left ( -A\omega _{d}\sin \omega _{d}t+\omega _{d}B\cos \omega _{d}t\right ) \\ +\varpi \frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\cos \left ( \varpi t-\theta \right ) \end {multline*}
at \(t=t_{1}\)\begin {multline} \dot {u}\left ( t_{1}\right ) =-\xi \omega e^{-\xi \omega t_{1}}\left ( A\cos \omega _{d}t_{1}+B\sin \omega _{d}t_{1}\right ) +e^{-\xi \omega t_{1}}\left ( -A\omega _{d}\sin \omega _{d}t_{1}+\omega _{d}B\cos \omega _{d}t_{1}\right ) \tag {3}\\ +\varpi \frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\cos \left ( \varpi t_{1}-\theta \right ) \nonumber \end {multline}
Now for \(t>t_{1}\) the equation becomes\[ m\ddot {u}+c\dot {u}+ku=0 \]
which has the solution \[ u=e^{-\xi \omega t}\left ( A\cos \omega _{d}t+B\sin \omega _{d}t\right ) \]
where \(A=u\left ( t_{1}\right ) \) and \(B=\frac {\dot {u}\left ( t_{1}\right ) +u\left ( t_{1}\right ) \xi \omega }{\omega _{d}}\)
1.4.5.2.3 critically damped with sin impulse \(\xi =\frac {c}{c_{r}}=1\) For \(t\leq t_{1}\)Initial conditions are \(u\left ( 0\right ) =u_{0}\) and \(\dot {u}\left ( 0\right ) =v_{0}\) then the solution is from above\begin {equation} u\left ( t\right ) =\left ( A+Bt\right ) e^{-\omega t}+\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \tag {1} \end {equation}
Where \(\tan \theta =\frac {c\varpi }{k-m\varpi ^{2}}=\frac {2\xi r}{1-r^{2}}.\) \(A,B \) are found from initial conditions\begin {align*} A & =u_{0}+\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \theta \\ B & =v_{0}+u_{0}\omega +\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\left ( \omega \sin \theta -\varpi \cos \theta \right ) \end {align*}
For \(t>t_{1}\) the solution is\begin {equation} u\left ( t\right ) =\left ( u\left ( t_{1}\right ) +\left ( \dot {u}\left ( t_{1}\right ) +u\left ( t_{1}\right ) \omega \right ) t\right ) e^{-\omega t} \tag {2} \end {equation}
To find \(u\left ( t_{1}\right ) ,\) from Eq(1)\[ u\left ( t_{1}\right ) =\left ( A+Bt\right ) e^{-\omega t_{1}}+\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \left ( \varpi t_{1}-\theta \right ) \]
taking derivative of (1) gives\begin {equation} \dot {u}\left ( t\right ) =-\omega \left ( A+Bt\right ) e^{-\omega t}+Be^{-\omega t}+\varpi \frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \tag {3} \end {equation}
at \(t=t_{1}\)\begin {equation} \dot {u}\left ( t_{1}\right ) =-\omega \left ( A+Bt_{1}\right ) e^{-\omega t_{1}}+Be^{-\omega t_{1}}+\varpi \frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2r\right ) ^{2}}}\sin \left ( \varpi t_{1}-\theta \right ) \tag {4} \end {equation}
Hence Eq (2) can now be evaluated using Eq(3,4)
1.4.5.2.4 over-damped with sin impulse \(\xi =\frac {c}{c_{r}}>1\) For \(t\leq t_{1}\)Initial conditions are \(u\left ( 0\right ) =u_{0}\) and \(\dot {u}\left ( 0\right ) =v_{0}\) then the solution is\[ u=Ae^{p_{1}t}+Be^{p_{2}t}+\frac {u_{st}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\sin \left ( \varpi t-\theta \right ) \]
where \(\tan \theta =\frac {2\xi r}{1-r^{2}}\)(make sure you use correct quadrant, see not above on \(\arctan \)) and\begin {align*} p_{1} & =-\frac {c}{2m}+\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}\\ & =-\omega \xi +\omega \sqrt {\xi ^{2}-1} \end {align*}
and\begin {align*} p_{2} & =-\frac {c}{2m}-\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}\\ & =-\omega \xi -\omega \sqrt {\xi ^{2}-1} \end {align*}
leading to the solution where \(\tan \theta =\frac {2\xi r}{1-r^{2}}\) and\begin {align*} p_{1} & =-\frac {c}{2m}+\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\omega \xi +\omega _{n}\sqrt {\xi ^{2}-1}\\ p_{2} & =-\frac {c}{2m}-\sqrt {\left ( \frac {c}{2m}\right ) ^{2}-\frac {k}{m}}=-\omega \xi -\omega _{n}\sqrt {\xi ^{2}-1} \end {align*}
is\begin {align*} u\left ( t\right ) & =Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t}+Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t}+\frac {F}{k}\beta \sin \left ( \varpi t-\theta \right ) \\ A & =\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega \xi +u\left ( 0\right ) \omega \sqrt {\xi ^{2}-1}+\frac {F}{k}\beta \left ( \left ( \xi +\sqrt {\xi ^{2}-1}\right ) \omega \sin \theta -\varpi \cos \theta \right ) }{2\omega \sqrt {\xi ^{2}-1}}\\ B & =-\frac {u^{\prime }\left ( 0\right ) +u\left ( 0\right ) \omega \xi -u\left ( 0\right ) \omega \sqrt {\xi ^{2}-1}+\frac {F}{k}\beta \left ( \left ( \xi -\sqrt {\xi ^{2}-1}\right ) \omega \sin \theta -\varpi \cos \theta \right ) }{2\omega \sqrt {\xi ^{2}-1}}\\ \beta & =\frac {1}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}} \end {align*}
For \(t>t_{1}\). From Eq(1) and at \(t=t_{1}\)\begin {equation} u\left ( t_{1}\right ) =Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t_{1}}+Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t_{1}}+D\sin \left ( \varpi t_{1}-\theta \right ) \tag {2} \end {equation}
Taking derivative of Eq (1)\[ \dot {u}\left ( t\right ) =\omega Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t}+\omega Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t}+\varpi D\cos \left ( \varpi t-\theta \right ) \]
At \(t=t_{1}\)\begin {equation} \dot {u}\left ( t_{1}\right ) =\omega Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega t_{1}}+\omega Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega t_{1}}+\varpi D\cos \left ( \varpi t_{1}-\theta \right ) \tag {3} \end {equation}
Equation of motion now is\[ \ddot {u}+2\xi \omega \dot {u}+\omega ^{2}u=0 \]
which has solution for over-damped given by\[ u\left ( t\right ) =Ae^{\left ( -\xi +\sqrt {\xi ^{2}-1}\right ) \omega _{n}t}+Be^{\left ( -\xi -\sqrt {\xi ^{2}-1}\right ) \omega _{n}t} \]
where\begin {align*} A & =-\frac {\dot {u}\left ( t_{1}\right ) +u\left ( t_{1}\right ) \omega _{n}\left ( \xi -\sqrt {\xi ^{2}-1}\right ) }{2\omega _{n}\sqrt {\xi ^{2}-1}}\\ B & =\frac {\dot {u}\left ( t_{1}\right ) +u\left ( t_{1}\right ) \xi \omega _{n}\left ( \xi +\sqrt {\xi ^{2}-1}\right ) }{2\omega _{n}\sqrt {\xi ^{2}-1}} \end {align*}
Input is given by \(F\left ( t\right ) =F_{0}\sin \left ( \varpi t\right ) \) where \(\varpi =\frac {2\pi }{2t_{1}}=\frac {\pi }{t_{1}}\)
t1 = 2; Plot[(UnitStep[t] - UnitStep[t - 2]) Sin[Pi/t1 t], {t, 0, 10}, PlotRange -> All, Ticks -> {{0, {2, "t1"}, 4}, Automatic}]
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This tree illustrates the different cases that needs to be considered for the solution of single degree of freedom system with harmonic loading.
There are \(12\) cases to consider. Resonance needs to be handled as special case when damping is absent due to the singularity in the standard solution when the forcing frequency is the same as the natural frequency. When damping is present, there is no resonance, however, there is what is called practical response which occur when the forcing frequency is almost the same as the natural frequency.
The following is another diagram made sometime ago which contains more useful information and is kept here for reference.
This table shows many cycles it takes for the peak to decay by half its original value as a function of the damping \(\zeta \). For example, we see that when \(\zeta =2.7\%\) then it takes \(4\) cycles for the peak (i.e. displacement) to reduce to half its value.
data = Table[{i, (1/i Log[2]/(2*Pi)*100)}, {i, 1, 20}]; TableForm[N@data, TableHeadings -> {None, {Column[{"number of cycles", "needed for peak", "to decay by half"}], "\[Zeta] (%)"}}]