Chapter 1
Vibration

1.1 Modal analysis for two degrees of freedom system

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

Figure 1.1: 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).

1.1.1 Step one. Finding the equations of motion in normal coordinates space

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

Figure 1.2: general 2 degrees of freedom system

Hence, from the above the equations of motion are$$\begin{array}{rl}{m}_1{x}_1^{\prime \prime }+k_1{x}_1-k_2(x_2-x_1)& =f_1(t)\\ m_2{x}_2^{\prime \prime }+k_2(x_2-x_1)& =f_2(t)\end{array}$$

or$$\begin{array}{rl}{m}_1{x}_1^{\prime \prime }+x_1(k_1+k_2)-k_2{x}_2& =f_1(t)\\ m_2{x}_2^{\prime \prime }+k_2{x}_2-k_2{x}_1& =f_2(t)\end{array}$$

In Matrix form $$\left[\begin{array}{cc}{m}_1& 0\\ 0& m_2\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime \prime }\\ x_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{k}_1+k_2& -k_2\\ -k_2& k_2\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}=\left\{\begin{array}{c}{f}_1(t)\\ f_2(t)\end{array}\right\}$$

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

$$\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime \prime }\\ x_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}=\left\{\begin{array}{c}{f}_1(t)\\ f_2(t)\end{array}\right\}$$

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.

1.1.2 Step 2. Solving the eigenvalue problem, finding the natural frequencies

The first step in modal analysis is to solve the eigenvalue problem $det([K]-{\omega }^2[M])=0$ 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. $$\begin{array}{rl}det(\left[K\right]-{\omega }^2\left[M\right])& =0\\ det(\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]-{\omega }^2\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right])& =0\\ det\left[\begin{array}{cc}{k}_{11}-{\omega }^2{m}_{11}& k_{12}-{\omega }^2{m}_{12}\\ k_{21}-{\omega }^2{m}_{21}& k_{22}-{\omega }^2{m}_{22}\end{array}\right]& =0\\ (k_{11}-{\omega }^2{m}_{11})(k_{22}-{\omega }^2{m}_{22})-(k_{12}-{\omega }^2{m}_{12})(k_{21}-{\omega }^2{m}_{21})& =0\\ {\omega }^4(m_{11}{m}_{22}-m_{12}{m}_{21})+{\omega }^2(-k_{11}{m}_{22}+k_{12}{m}_{21}+k_{21}{m}_{12}-k_{22}{m}_{11})+k_{11}{k}_{22}-k_{12}{k}_{21}& =0\end{array}$$

The above is a polynomial in ${\omega }^4$. Let ${\omega }^2=\lambda$ it becomes$${\lambda }^2(m_{11}{m}_{22}-m_{12}{m}_{21})+\lambda (-k_{11}{m}_{22}+k_{12}{m}_{21}+k_{21}{m}_{12}-k_{22}{m}_{11})+k_{11}{k}_{22}-k_{12}{k}_{21}=0$$

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 ${\mathit{\phi }}_1,{\mathit{\phi }}_2$ also called the eigenvectors associated with ${\omega }_1$ and ${\omega }_2$

1.1.3 Step 3. Finding 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 vectors ${\phi }_1,{\phi }_2$$$\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]-{\omega }_1^2\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$$

and$$\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]-{\omega }_2^2\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$$

For ${\omega }_1$, let ${\phi }_{11}=1$ and solve for$$\begin{array}{rl}\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]-{\omega }_1^2\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}1\\ {\phi }_{21}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\\ \left[\begin{array}{cc}{k}_{11}-{\omega }_1^2{m}_{11}& k_{12}-{\omega }_1^2{m}_{12}\\ k_{21}-{\omega }_1^2{m}_{21}& k_{22}-{\omega }_1^2{m}_{22}\end{array}\right]\left\{\begin{array}{c}1\\ {\phi }_{21}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\end{array}$$

Which gives one equation now to solve for ${\phi }_{21}$ (the first row equation is only used)$$(k_{11}-{\omega }_1^2{m}_{11})+{\phi }_{21}(k_{12}-{\omega }_1^2{m}_{12})=0$$

Hence $${\phi }_{21}=\frac{-(k_{11}-{\omega }_1^2{m}_{11})}{(k_{12}-{\omega }_1^2{m}_{12})}$$

Therefore the first shape vector is$${\mathit{\phi }}_1=\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}=\left\{\begin{array}{c}1\\ \frac{-(k_{11}-{\omega }_1^2{m}_{11})}{(k_{12}-{\omega }_1^2{m}_{12})}\end{array}\right\}$$

Similarly the second shape function is obtained. For ${\omega }_2$, let ${\phi }_{12}=1$ and solve for$$\begin{array}{rl}\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]-{\omega }_2^2\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}1\\ {\phi }_{22}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\\ \left[\begin{array}{cc}{k}_{11}-{\omega }_2^2{m}_{11}& k_{12}-{\omega }_2^2{m}_{12}\\ k_{21}-{\omega }_2^2{m}_{21}& k_{22}-{\omega }_2^2{m}_{22}\end{array}\right]\left\{\begin{array}{c}1\\ {\phi }_{22}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\end{array}$$

Which gives one equation now to solve for ${\phi }_{22}$ (the first row equation is only used)$$(k_{11}-{\omega }_2^2{m}_{11})+{\phi }_{22}(k_{12}-{\omega }_2^2{m}_{12})=0$$

Hence $${\phi }_{22}=\frac{-(k_{11}-{\omega }_2^2{m}_{11})}{(k_{12}-{\omega }_2^2{m}_{12})}$$

Therefore the second shape vector is$${\mathit{\phi }}_2=\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}=\left\{\begin{array}{c}1\\ \frac{-(k_{11}-{\omega }_2^2{m}_{11})}{(k_{12}-{\omega }_2^2{m}_{12})}\end{array}\right\}$$

Now that the two non-normalized shape vectors are found, the next step is to perform mass normalization

1.1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors)

Let $${\mu }_1={\mathit{\phi }}_1^T\left[M\right]{\mathit{\phi }}_1$$

This results in a scalar value ${\mu }_1$, which is later used to normalize ${\mathit{\phi }}_1$. Similarly$${\mu }_2={\mathit{\phi }}_2^T\left[M\right]{\mathit{\phi }}_2$$

For example, to find ${\mu }_1$$$\begin{array}{rl}{\mu }_1& ={\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}\\ & =\left\{\begin{array}{cc}{\phi }_{11}& {\phi }_{21}\end{array}\right\}\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}\\ & =\left\{\begin{array}{cc}{\phi }_{11}{m}_{11}+{\phi }_{21}{m}_{21}& {\phi }_{11}{m}_{12}+{\phi }_{21}{m}_{22}\end{array}\right\}\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}\\ & ={\phi }_{11}({\phi }_{11}{m}_{11}+{\phi }_{21}{m}_{21})+{\phi }_{21}({\phi }_{11}{m}_{12}+{\phi }_{21}{m}_{22})\end{array}$$

Similarly, ${\mu }_2$ is found$$\begin{array}{rl}{\mu }_2& ={\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}\\ & =\left\{\begin{array}{cc}{\phi }_{12}& {\phi }_{22}\end{array}\right\}\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}\\ & =\left\{\begin{array}{cc}{\phi }_{12}{m}_{11}+{\phi }_{22}{m}_{21}& {\phi }_{12}{m}_{12}+{\phi }_{22}{m}_{22}\end{array}\right\}\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}\\ & ={\phi }_{12}({\phi }_{12}{m}_{11}+{\phi }_{22}{m}_{21})+{\phi }_{22}({\phi }_{12}{m}_{12}+{\phi }_{22}{m}_{22})\end{array}$$

Now that ${\mu }_1,{\mu }_2$ are obtained, the mass normalized shape vectors are found. They are called ${\mathbf{\Phi }}_1,{\mathbf{\Phi }}_2$$${\mathbf{\Phi }}_1=\frac{{\mathit{\phi }}_1}{\sqrt{{\mu }_1}}=\frac{\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}}{\sqrt{{\mu }_1}}=\left\{\begin{array}{c}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}\end{array}\right\}$$

Similarly$${\mathbf{\Phi }}_2=\frac{{\mathit{\phi }}_2}{\sqrt{{\mu }_2}}=\frac{\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}}{\sqrt{{\mu }_2}}=\left\{\begin{array}{c}\frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right\}$$

1.1.5 Step 5, obtain the modal transformation matrix $\mathrm{\Phi }$

The modal transformation matrix is the $2\times 2$ matrix made of of ${\mathbf{\Phi }}_1,{\mathbf{\Phi }}_2$ in each of its columns $$\begin{array}{rl}\left[\mathrm{\Phi }\right]& =\left[{\mathbf{\Phi }}_1{\mathbf{\Phi }}_2\right]\\ & =\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]\end{array}$$

Now the $\left[\mathrm{\Phi }\right]$ is found, the transformation from the normal coordinates $\left\{x\right\}$ to modal coordinates, which is called $\left\{\eta \right\}$ is found

$$\begin{array}{rl}\left\{x\right\}& =\left[\mathrm{\Phi }\right]\left\{\eta \right\}\\ \left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}& =\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]\left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}\end{array}$$

The transformation from modal coordinates back to normal coordinates is$$\begin{array}{rl}\left\{\eta \right\}& ={\left[\mathrm{\Phi }\right]}^{-1}\left\{x\right\}\\ \left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}& ={\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]}^{-1}\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}\end{array}$$

However, ${\left[\mathrm{\Phi }\right]}^{-1}={\left[\mathrm{\Phi }\right]}^T\left[M\right]$ therefore$$\begin{array}{rl}\left\{\eta \right\}& ={\left[\mathrm{\Phi }\right]}^T\left[M\right]\left\{x\right\}\\ \left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}& ={\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}\end{array}$$

The next step is to apply this transformation to the original equations of motion in order to decouple them

1.1.6 Step 6. Applying modal transformation to decouple the original equations of motion

The EOM in normal coordinates is$$\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime \prime }\\ x_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}=\left\{\begin{array}{c}{f}_1\left(t\right)\\ f_2\left(t\right)\end{array}\right\}$$

Applying the above modal transformation $\left\{x\right\}=\left[\mathrm{\Phi }\right]\left\{\eta \right\}$ on the above results in $$\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left[\mathrm{\Phi }\right]\left\{\begin{array}{c}{\eta }_1^{\prime \prime }\\ {\eta }_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]\left[\mathrm{\Phi }\right]\left\{\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right\}=\left\{\begin{array}{c}{f}_1\left(t\right)\\ f_2\left(t\right)\end{array}\right\}$$

pre-multiplying by ${\left[\mathrm{\Phi }\right]}^T$ results in$${\left[\mathrm{\Phi }\right]}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left[\mathrm{\Phi }\right]\left\{\begin{array}{c}{\eta }_1^{\prime \prime }\\ {\eta }_2^{\prime \prime }\end{array}\right\}+{\left[\mathrm{\Phi }\right]}^T\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]\left[\mathrm{\Phi }\right]\left\{\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right\}={\left[\mathrm{\Phi }\right]}^T\left\{\begin{array}{c}{f}_1\left(t\right)\\ f_2\left(t\right)\end{array}\right\}$$

The result of ${\left[\mathrm{\Phi }\right]}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left[\mathrm{\Phi }\right]$ will always be $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$. 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 ${\left[\mathrm{\Phi }\right]}^T\left[\begin{array}{cc}{k}_{11}& k_{12}\\ k_{21}& k_{22}\end{array}\right]\left[\mathrm{\Phi }\right]$ will be $\left[\begin{array}{cc}{\omega }_1^2& 0\\ 0& {\omega }_2^2\end{array}\right]$.

Let the result of ${\left[\mathrm{\Phi }\right]}^T\left\{\begin{array}{c}{f}_1\left(t\right)\\ f_2\left(t\right)\end{array}\right\}$ be $\left\{\begin{array}{c}{\tilde{f}}_1\left(t\right)\\ {\tilde{f}}_2\left(t\right)\end{array}\right\},$Therefore, in modal coordinates the original EOM becomes$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left\{\begin{array}{c}{\eta }_1^{\prime \prime }\\ {\eta }_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{\omega }_1^2& 0\\ 0& {\omega }_2^2\end{array}\right]\left\{\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right\}=\left\{\begin{array}{c}{\tilde{f}}_1\left(t\right)\\ {\tilde{f}}_2\left(t\right)\end{array}\right\}$$

The EOM are now decouples and each can be solved as follows$$\begin{array}{rl}{\eta }_1^{\prime \prime }\left(t\right)+{\omega }_1^2{\eta }_1\left(t\right)& ={\tilde{f}}_1\left(t\right)\\ {\eta }_2^{\prime \prime }\left(t\right)+{\omega }_2^2{\eta }_2\left(t\right)& ={\tilde{f}}_2\left(t\right)\end{array}$$

To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules

$$\begin{array}{rl}\left\{\eta \left(0\right)\right\}& ={\left[\mathrm{\Phi }\right]}^T\left[M\right]\left\{x\left(0\right)\right\}\\ \left\{{\eta }^{\prime }\left(0\right)\right\}& ={\left[\mathrm{\Phi }\right]}^T\left[M\right]\left\{x^{\prime }\left(0\right)\right\}\end{array}$$

Or in full form$$\left\{\begin{array}{c}{\eta }_1\left(0\right)\\ {\eta }_2\left(0\right)\end{array}\right\}={\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1\left(0\right)\\ x_2\left(0\right)\end{array}\right\}$$

and$$\left\{\begin{array}{c}{\eta }_1^{\prime }\left(0\right)\\ {\eta }_2^{\prime }\left(0\right)\end{array}\right\}={\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]}^T\left[\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime }\left(0\right)\\ x_2^{\prime }\left(0\right)\end{array}\right\}$$

Each of these EOM are solved using any of the standard methods. This will result is solutions ${\eta }_1\left(t\right)$ and ${\eta }_2\left(t\right)$

1.1.7 Step 7. Converting modal solution to normal coordinates solution

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 $\left\{x\right\}=\left[\mathrm{\Phi }\right]\left\{\eta \right\}$ is now applied to convert the solution to normal coordinates$$\begin{array}{rl}\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}& =\left[\begin{array}{cc}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{12}}{\sqrt{{\mu }_2}}\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}& \frac{{\phi }_{22}}{\sqrt{{\mu }_2}}\end{array}\right]\left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}\\ & =\left\{\begin{array}{c}\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}{\eta }_1\left(t\right)+\frac{{\phi }_{12}}{\sqrt{{\mu }_2}}{\eta }_2\left(t\right)\\ \frac{{\phi }_{21}}{\sqrt{{\mu }_1}}{\eta }_1\left(t\right)+\frac{{\phi }_{22}}{\sqrt{{\mu }_2}}{\eta }_2\left(t\right)\end{array}\right\}\end{array}$$

Hence$$x_1\left(t\right)=\frac{{\phi }_{11}}{\sqrt{{\mu }_1}}{\eta }_1\left(t\right)+\frac{{\phi }_{12}}{\sqrt{{\mu }_2}}{\eta }_2\left(t\right)$$

and$$x_2\left(t\right)=\frac{{\phi }_{21}}{\sqrt{{\mu }_1}}{\eta }_1\left(t\right)+\frac{{\phi }_{22}}{\sqrt{{\mu }_2}}{\eta }_2\left(t\right)$$

Notice that the solution in normal coordinates is a linear combination of the modal solutions. The terms $\frac{{\phi }_{ij}}{\sqrt{\mu }}$ are just scaling factors that represent the contribution of each modal solution to the final solution. This completes modal analysis

1.1.8 Numerical solution using 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

$$\left\{\begin{array}{c}{x}_1\left(0\right)\\ x_2\left(0\right)\end{array}\right\}=\left\{\begin{array}{c}0\\ 1\end{array}\right\}$$

and

$$\left\{\begin{array}{c}{x}_1^{\prime }\left(0\right)\\ x_2^{\prime }\left(0\right)\end{array}\right\}=\left\{\begin{array}{c}1.5\\ 3\end{array}\right\}$$

In normal coordinates, the EOM are

$$\begin{array}{rl}\left[\begin{array}{cc}{m}_1& 0\\ 0& m_2\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime \prime }\\ x_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{k}_1+k_2& -k_2\\ -k_2& k_2\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}& =\left\{\begin{array}{c}{f}_1\left(t\right)\\ f_2\left(t\right)\end{array}\right\}\\ \left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime \prime }\\ x_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}3& -2\\ -2& 2\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}& =\left\{\begin{array}{c}0\\ \sin \left(5t\right)\end{array}\right\}\end{array}$$

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

$$\begin{array}{rl}det(\left[K\right]-{\omega }^2\left[M\right])& =0\\ det(\left[\begin{array}{cc}3& -2\\ -2& 2\end{array}\right]-{\omega }^2\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right])& =0\\ det\left[\begin{array}{cc}3-{\omega }^2& -2\\ -2& 2-3{\omega }^2\end{array}\right]& =0\\ (3-{\omega }^2)(2-3{\omega }^2)-(-2)(-2)& =0\\ 3{\omega }^4-11{\omega }^2+2& =0\end{array}$$

Let ${\omega }^2=\lambda$ hence$$3{\lambda }^2-11\lambda +2=0$$ The solution is ${\lambda }_1=3.475$ 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 ${\phi }_1,{\phi }_2$$$(\left[\begin{array}{cc}3& -2\\ -2& 2\end{array}\right]-{\omega }_1^2\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right])\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$$

For ${\omega }_1=1.864$$$\begin{array}{rl}(\left[\begin{array}{cc}3& -2\\ -2& 2\end{array}\right]-{1.864}^2\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right])\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\\ \left[\begin{array}{cc}-0.475& -2\\ -2& -8.424\end{array}\right]\left\{\begin{array}{c}1\\ {\phi }_{21}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\end{array}$$

This gives one equation to solve for ${\phi }_{21}$ (the first row equation is only used)$$-0.475-2{\phi }_{21}=0$$

Hence $${\phi }_{21}=\frac{0.475}{-2}=-0.237$$

The first eigen vector is$${\mathit{\phi }}_1=\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}=\left\{\begin{array}{c}1\\ -0.237\end{array}\right\}$$

Similarly for ${\omega }_2=0.438$$$\begin{array}{rl}(\left[\begin{array}{cc}3& -2\\ -2& 2\end{array}\right]-{0.438}^2\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right])\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\\ \left[\begin{array}{cc}2.808& -2\\ -2& 1.425\end{array}\right]\left\{\begin{array}{c}1\\ {\phi }_{22}\end{array}\right\}& =\left\{\begin{array}{c}0\\ 0\end{array}\right\}\end{array}$$

This gives one equation to solve for ${\phi }_{22}$ (the first row equation is only used)$$2.808-2{\phi }_{22}=0$$

Hence $${\phi }_{22}=\frac{-2.808}{-2}=1.404$$

The second eigen vector is$${\mathit{\phi }}_2=\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}=\left\{\begin{array}{c}1\\ 1.404\end{array}\right\}$$

Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors) $${\mu }_1={\mathit{\phi }}_1^T\left[M\right]{\mathit{\phi }}_1$$

Hence$$\begin{array}{rl}{\mu }_1& ={\left\{\begin{array}{c}1\\ -0.237\end{array}\right\}}^T\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]\left\{\begin{array}{c}1\\ -0.237\end{array}\right\}\\ & =1.169\end{array}$$

Similarly, ${\mu }_2$ is found$${\mu }_2={\mathit{\phi }}_2^T\left[M\right]{\mathit{\phi }}_2$$

Hence$$\begin{array}{rl}{\mu }_2& ={\left\{\begin{array}{c}1\\ 1.404\end{array}\right\}}^T\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]\left\{\begin{array}{c}1\\ 1.404\end{array}\right\}\\ & =6.914\end{array}$$

Now that ${\mu }_1,{\mu }_2$ are found, the mass normalized eigen vectors are found. They are called ${\mathbf{\Phi }}_1,{\mathbf{\Phi }}_2$$$\begin{array}{rl}{\mathbf{\Phi }}_1& =\frac{{\mathit{\phi }}_1}{\sqrt{{\mu }_1}}=\frac{\left\{\begin{array}{c}{\phi }_{11}\\ {\phi }_{21}\end{array}\right\}}{\sqrt{{\mu }_1}}=\frac{\left\{\begin{array}{c}1\\ -0.237\end{array}\right\}}{\sqrt{1.169}}\\ & =\left\{\begin{array}{c}0.925\\ -0.219\end{array}\right\}\end{array}$$

Similarly$$\begin{array}{rl}{\mathbf{\Phi }}_2& =\frac{{\mathit{\phi }}_2}{\sqrt{{\mu }_2}}=\frac{\left\{\begin{array}{c}{\phi }_{12}\\ {\phi }_{22}\end{array}\right\}}{\sqrt{{\mu }_2}}=\frac{\left\{\begin{array}{c}1\\ 1.404\end{array}\right\}}{\sqrt{6.914}}\\ & =\left\{\begin{array}{c}0.380\\ 0.534\end{array}\right\}\end{array}$$

Therefore, the modal transformation matrix is $$\begin{array}{rl}\left[\mathrm{\Phi }\right]& =\left[{\mathbf{\Phi }}_1{\mathbf{\Phi }}_2\right]\\ & =\left[\begin{array}{cc}0.925& 0.380\\ -0.219& 0.534\end{array}\right]\end{array}$$

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 $\left[\mathrm{\Phi }\right]$ is found, the transformation from the normal coordinates $\left\{x\right\}$ to modal coordinates, called $\left\{\eta \right\},$ is obtained$$\begin{array}{rl}\left\{x\right\}& =\left[\mathrm{\Phi }\right]\left\{\eta \right\}\\ \left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}& =\left[\begin{array}{cc}0.925& 0.380\\ -0.219& 0.534\end{array}\right]\left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}\end{array}$$

The transformation from modal coordinates back to normal coordinates is$$\begin{array}{rl}\left\{\eta \right\}& ={\left[\mathrm{\Phi }\right]}^{-1}\left\{x\right\}\\ \left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}& ={\left[\begin{array}{cc}0.925& 0.380\\ -0.219& 0.534\end{array}\right]}^{-1}\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}\end{array}$$

However, ${\left[\mathrm{\Phi }\right]}^{-1}={\left[\mathrm{\Phi }\right]}^T\left[M\right]$ therefore$$\begin{array}{rl}\left\{\eta \right\}& ={\left[\mathrm{\Phi }\right]}^T\left[M\right]\left\{x\right\}\\ \left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}& ={\left[\begin{array}{cc}0.925& 0.380\\ -0.219& 0.534\end{array}\right]}^T\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}\\ & =\left[\begin{array}{cc}0.925& -0.657\\ 0.38& 1.6\end{array}\right]\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}\end{array}$$

The next step is to apply this transformation to the original equations of motion in order to decouple them.

Applying step 6 results in$$\begin{array}{rl}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left\{\begin{array}{c}{\eta }_1^{\prime \prime }\\ {\eta }_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{\omega }_1^2& 0\\ 0& {\omega }_2^2\end{array}\right]\left\{\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right\}& ={\left[\mathrm{\Phi }\right]}^T\left\{\begin{array}{c}0\\ \sin \left(5t\right)\end{array}\right\}\\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left\{\begin{array}{c}{\eta }_1^{\prime \prime }\\ {\eta }_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}{1.864}^2& 0\\ 0& {0.438}^2\end{array}\right]\left\{\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right\}& ={\left[\begin{array}{cc}0.925& 0.380\\ -0.219& 0.534\end{array}\right]}^T\left\{\begin{array}{c}0\\ \sin \left(5t\right)\end{array}\right\}\\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left\{\begin{array}{c}{\eta }_1^{\prime \prime }\\ {\eta }_2^{\prime \prime }\end{array}\right\}+\left[\begin{array}{cc}3.47& 0\\ 0& 0.192\end{array}\right]\left\{\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right\}& =\left\{\begin{array}{c}-0.219\sin \left(5t\right)\\ 0.534\sin \left(5t\right)\end{array}\right\}\end{array}$$

The EOM are now decoupled and each EOM can be solved easily as follows$$\begin{array}{rl}{\eta }_1^{\prime \prime }\left(t\right)+3.47{\eta }_1\left(t\right)& =-0.219\sin \left(5t\right)\\ {\eta }_2^{\prime \prime }\left(t\right)+0.192{\eta }_2\left(t\right)& =0.534\sin \left(5t\right)\end{array}$$

To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules$$\begin{array}{rl}\left\{\begin{array}{c}{\eta }_1\left(0\right)\\ {\eta }_2\left(0\right)\end{array}\right\}& =\left[\begin{array}{cc}0.925& -0.657\\ 0.38& 1.6\end{array}\right]\left\{\begin{array}{c}{x}_1\left(0\right)\\ x_2\left(0\right)\end{array}\right\}\\ \left\{\begin{array}{c}{\eta }_1\left(0\right)\\ {\eta }_2\left(0\right)\end{array}\right\}& =\left[\begin{array}{cc}0.925& -0.657\\ 0.38& 1.6\end{array}\right]\left\{\begin{array}{c}0\\ 1\end{array}\right\}\\ & =\left\{\begin{array}{c}-0.657\\ 1.6\end{array}\right\}\end{array}$$

and$$\begin{array}{rl}\left\{\begin{array}{c}{\eta }_1^{\prime }\left(0\right)\\ {\eta }_2^{\prime }\left(0\right)\end{array}\right\}& =\left[\begin{array}{cc}0.925& -0.657\\ 0.38& 1.6\end{array}\right]\left\{\begin{array}{c}{x}_1^{\prime }\left(0\right)\\ x_2^{\prime }\left(0\right)\end{array}\right\}\\ & =\left[\begin{array}{cc}0.925& -0.657\\ 0.38& 1.6\end{array}\right]\left\{\begin{array}{c}1.5\\ 3\end{array}\right\}\\ & =\left\{\begin{array}{c}-0.584\\ 5.37\end{array}\right\}\end{array}$$

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 $$\begin{array}{rl}{\eta }_1^{\prime \prime }\left(t\right)+3.47{\eta }_1\left(t\right)& =-0.219\sin \left(5t\right)\\ {\eta }_1\left(0\right)& =-0.657\\ {\eta }_1^{\prime }\left(0\right)& =-0.584\end{array}$$

and also $$\begin{array}{rl}{\eta }_2^{\prime \prime }\left(t\right)+0.192{\eta }_2\left(t\right)& =0.534\sin \left(5t\right)\\ {\eta }_2\left(0\right)& =1.6\\ {\eta }_2^{\prime }\left(0\right)& =5.37\end{array}$$

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$$\begin{array}{rl}\left\{\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right\}& =\left[\begin{array}{cc}0.925& 0.380\\ -0.219& 0.534\end{array}\right]\left\{\begin{array}{c}{\eta }_1\left(t\right)\\ {\eta }_2\left(t\right)\end{array}\right\}\\ & =\left\{\begin{array}{c}0.925{\eta }_1+0.38{\eta }_2\\ 0.534{\eta }_2-0.219{\eta }_1\end{array}\right\}\end{array}$$

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.

1.2 Fourier series representation of a periodic function

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$$\begin{array}{rl}\tilde{f}\left(t\right)& =\frac{1}{2}{F}_0+\mathrm{Re}\left({\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}t}\right)\\ & =\frac{1}{2}{F}_0+\frac{1}{2}{\displaystyle \sum _{n=1}^N}{F}_n{e}^{in\frac{2\pi }{T}t}+F_n^{\ast }{e}^{-in\frac{2\pi }{T}t}\\ & =\frac{1}{2}{\displaystyle \sum _{n=-N}^N}{F}_n{e}^{in\frac{2\pi }{T}t}\end{array}$$

Where$$\begin{array}{rl}{F}_n& =\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}f\left(t\right)e^{-in\frac{2\pi }{T}t}dt\\ F_0& =\frac{2}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}f\left(t\right)dt\end{array}$$

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.$$\begin{array}{rl}\tilde{f}\left(t\right)& =a_0+{\displaystyle \sum _{n=1}^N}{a}_n\cos n\frac{2\pi }{T}t+{\displaystyle \sum _{n=1}^N}{b}_n\sin n\frac{2\pi }{T}t\\ a_0& =\frac{1}{T}{\displaystyle \underset{0}{\overset{T}{\int }}}f\left(t\right)dt\\ a_n& =\frac{1}{T/2}{\displaystyle \underset{0}{\overset{T}{\int }}}f\left(t\right)\cos \left(n\frac{2\pi }{T}t\right)dt\\ b_n& =\frac{1}{T/2}{\displaystyle \underset{0}{\overset{T}{\int }}}f\left(t\right)\sin \left(n\frac{2\pi }{T}t\right)dt\end{array}$$

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.

1.3 Generating Transfer functions for different vibration systems

Figure 1.3: vibration model

1.3.1 Force transmissibility

Let steady state $$x_{ss}=\mathrm{Re}\left\{\frac{\hat{F}}{k}D(r,\zeta )e^{i\varpi t}\right\}$$

Then$$\begin{array}{rl}{f}_{tr}\left(t\right)& =f_{spring}+f_{damper}\\ & =kx+c\dot{x}\\ & =\mathrm{Re}\left\{k\frac{\hat{F}}{k}D(r,\zeta )e^{i\varpi t}\right\}+\mathrm{Re}\left\{ci\varpi \frac{\hat{F}}{k}D(r,\zeta )e^{i\varpi t}\right\}\\ & =\mathrm{Re}\left\{(\hat{F}+ci\varpi \frac{\hat{F}}{k})D(r,\zeta )e^{i\varpi t}\right\}\end{array}$$

Hence $${\left|f_{tr}\left(t\right)\right|}_{max}=\left|\hat{F}\right|\left|D\right|\sqrt{1+c^2\frac{{\varpi }^2}{k^2}}=\left|\hat{F}\right|\left|D\right|\sqrt{1+{\left(2\zeta r\right)}^2}$$

So TR or force transmissibility is$$TR=\frac{{\left|f_{tr}\left(t\right)\right|}_{max}}{\left|\hat{F}\right|}=\left|D\right|\sqrt{1+{\left(2\zeta r\right)}^2}$$

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.

1.3.2 vibration isolation

We need transfer function between $y$ and $z.$ Equation of motion$$\begin{array}{rl}m{y}^{\prime \prime }& =-c(y^{\prime }-z^{\prime })-k(y-z)\\ m{y}^{\prime \prime }+c{y}^{\prime }+ky& =c{z}^{\prime }+kz\end{array}$$

Let $z=\mathrm{Re}\left\{Z{e}^{i\omega t}\right\},z^{\prime }=\mathrm{Re}\left\{i\omega Z{e}^{i\omega t}\right\}$ and let $y=\mathrm{Re}\left\{Y{e}^{i\omega t}\right\},y^{\prime }=\mathrm{Re}\left\{i\omega Y{e}^{i\omega t}\right\},y^{\prime \prime }=\mathrm{Re}\{-{\omega }^{2}Y{e}^{i\omega t}\}$, hence the above becomes$$\begin{array}{rl}m\mathrm{Re}\{-{\omega }^{2}Y{e}^{i\omega t}\}+c\mathrm{Re}\left\{i\omega Y{e}^{i\omega t}\right\}+k\mathrm{Re}\left\{Y{e}^{i\omega t}\right\}& =c\mathrm{Re}\left\{i\omega Z{e}^{i\omega t}\right\}+k\mathrm{Re}\left\{Z{e}^{i\omega t}\right\}\\ Y& =\frac{ci\omega +k}{-{\omega }^{2}m+ci\omega +k}Z\\ & =\frac{i2\zeta {\omega }_{n}m\omega +k}{-{\omega }^{2}m+i2\zeta {\omega }_{n}m\omega +k}Z\\ & =\frac{i2\zeta {\omega }_n\omega +{\omega }_n^2}{-{\omega }^2+i2\zeta {\omega }_n\omega +{\omega }_n^2}Z\\ & =\frac{i2\zeta r+1}{(1-r^2)+i2\zeta r}Z\end{array}$$

Hence $\left|D(r,\zeta )\right|=\frac{\sqrt{1+{\left(2\zeta r\right)}^2}}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}$ 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 $\frac{{\left|Y\right|}_{max}}{\left|Z\right|}$ to be small. i.e. $\left|D\right|\sqrt{1+{\left(2\zeta r\right)}^2}$ to be small. This is the same TR as for force isolation above.

For small $\left|D\right|$, we need small $\zeta$ and small $k$ (the small $k$ is to make $r>\sqrt{2}$) see plot

Figure 1.4: force transibility

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;
 

1.3.3 Accelerometer

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.$$\begin{array}{rl}m(u^{\prime \prime }+z^{\prime \prime })+c{u}^{\prime }+ku& =0\\ m{u}^{\prime \prime }+c{u}^{\prime }+ku& =-m{z}^{\prime \prime }\end{array}$$

Let $z^{\prime \prime }=\mathrm{Re}\left\{Z_a{e}^{i\omega t}\right\}.$ Notice we here jumped right away to the $z^{\prime \prime }$ itself and wrote it as $\mathrm{Re}\left\{Z_a{e}^{i\omega t}\right\}$ 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, $u=\mathrm{Re}\left\{U{e}^{i\omega t}\right\},u^{\prime }=\mathrm{Re}\left\{i\omega U{e}^{i\omega t}\right\},u^{\prime \prime }=\mathrm{Re}\{-{\omega }^{2}U{e}^{i\omega t}\}$, hence the above becomes$$\begin{array}{rl}m\mathrm{Re}\{-{\omega }^{2}U{e}^{i\omega t}\}+c\mathrm{Re}\left\{i\omega U{e}^{i\omega t}\right\}+k\mathrm{Re}\left\{U{e}^{i\omega t}\right\}& =-m\mathrm{Re}\left\{Z_a{e}^{i\omega t}\right\}\\ U& =\frac{-m}{-{\omega }^{2}m+i\omega c+k}{Z}_a\\ & =\frac{-1}{-{\omega }^2+i\omega 2\zeta {\omega }_n+{\omega }_n^2}{Z}_a\\ & =\frac{-1}{({\omega }_n^2-{\omega }^2)+i\omega 2\zeta {\omega }_n}{Z}_a\end{array}$$

Hence $\left|D(r,\zeta )\right|=\frac{-1}{\sqrt{{({\omega }_n^2-{\omega }^2)}^2+{\left(2\omega \zeta {\omega }_n\right)}^2}}$ 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 $D(r,\zeta )\approx$ $\frac{-1}{{\omega }_n^2}{Z}_a$, 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.

1.3.4 Seismometer

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.$$\begin{array}{rl}m(u^{\prime \prime }+z^{\prime \prime })+c{u}^{\prime }+ku& =0\\ m{u}^{\prime \prime }+c{u}^{\prime }+ku& =-m{z}^{\prime \prime }\end{array}$$

Let $z=\mathrm{Re}\left\{Z{e}^{i\omega t}\right\},z^{\prime }=\mathrm{Re}\left\{i\omega Z{e}^{i\omega t}\right\}$ ,$z^{\prime \prime }=\mathrm{Re}\{-{\omega }^{2}Z{e}^{i\omega t}\},$and let $u=\mathrm{Re}\left\{U{e}^{i\omega t}\right\},u^{\prime }=\mathrm{Re}\left\{i\omega U{e}^{i\omega t}\right\},u^{\prime \prime }=\mathrm{Re}\{-{\omega }^{2}U{e}^{i\omega t}\}$, hence the above becomes

Now, $u=\mathrm{Re}\left\{U{e}^{i\omega t}\right\},u^{\prime }=\mathrm{Re}\left\{i\omega U{e}^{i\omega t}\right\},u^{\prime \prime }=\mathrm{Re}\{-{\omega }^{2}U{e}^{i\omega t}\}$, hence the above becomes$$\begin{array}{rl}m\mathrm{Re}\{-{\omega }^{2}U{e}^{i\omega t}\}+c\mathrm{Re}\left\{i\omega U{e}^{i\omega t}\right\}+k\mathrm{Re}\left\{U{e}^{i\omega t}\right\}& =-m\mathrm{Re}\{-{\omega }^{2}Z{e}^{i\omega t}\}\\ U& =\frac{m{\omega }^2}{-{\omega }^{2}m+i\omega c+k}Z\\ & =\frac{{\omega }^2}{-{\omega }^2+i\omega 2\zeta {\omega }_n+{\omega }_n^2}Z\\ & =\frac{r^2}{(1-r^2)+i2\zeta r}Z\end{array}$$

Hence $\left|D(r,\zeta )\right|=\frac{r^2}{\sqrt{(1-r^2)+i2\zeta r}}$ 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 $\frac{1}{(1-r^2)+i2\zeta r}\Rightarrow \frac{1}{-r^2}$ since $r^2$ is the dominant factor. Therefore $U=\frac{r^2}{(1-r^2)+i2\zeta r}{Z}_a$ 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.

1.3.5 Summary of vibration transfer functions

For good isolation of mass from ground motion, rule of thumb: Make damping low, and stiffness low (soft spring).

Isolate base from force. transmitted by machine

Equation used $f_{\text{tr}}(t)=f_{\text{spring}}+f_{\text{damper}}$

Transfer function $\frac{{|f_{\text{tr}}(t)|}_{max}}{\left|\hat{F}\right|}=\left|D\right|\sqrt{1+{\left(2\zeta r\right)}^2}$

Isolate machine from motion of base

Equation used. Use absolute mass position $$m{y}^{\prime \prime }+c{y}^{\prime }+ky=c{z}^{\prime }+kz$$

Transfer function $$\frac{{\left|Y\right|}_{max}}{\left|Z\right|}=\left|D\right|\sqrt{1+{\left(2\zeta r\right)}^2}$$

Accelerometer: Measure base acc. using relative displacement

Equation used. Use relative mass position $$m{u}^{\prime \prime }+c{u}^{\prime }+ku=-m{z}^{\prime \prime }$$

Transfer function $$\begin{array}{rl}U& =\frac{-1}{({\omega }_n^2-{\omega }^2)+i\omega 2\zeta {\omega }_n}{Z}_a\Rightarrow \left|D(r,\zeta )\right|\\ & =\frac{-1}{\sqrt{{({\omega }_n^2-{\omega }^2)}^2+{\left(2\omega \zeta {\omega }_n\right)}^2}}\end{array}$$

Seismometer: Measure base motion using relative displacement

Equation used. Use relative mass position $$m{u}^{\prime \prime }+c{u}^{\prime }+ku=-m{z}^{\prime \prime }$$

Transfer function $$\begin{array}{rl}U& =\frac{r^2}{(1-r^2)+i2\zeta r}Z\to |D(r,\zeta )|\\ & =\frac{r^2}{\sqrt{(1-r^2)+i2\zeta r}}\end{array}$$

1.4 Solution of Vibration equation of motion for different loading

1.4.1 common definitions

These definitions are used throughout the derivations below.$$\begin{array}{rl}\xi & =\frac{c}{c_r}=\frac{c}{2\sqrt{km}}=\frac{c}{2{\omega }_{n}m}\\ u_{st}& =\frac{F}{k}\text{ static deflection}\\ {\omega }_n& =\sqrt{\frac{k}{m}}\\ {\omega }_D& ={\omega }_n\sqrt{1-{\xi }^2}\text{note: not defined for }\xi >1\text{ since becomes complex}\\ r& =\frac{\varpi }{{\omega }_n}\\ T_d& =\frac{2\pi }{{\omega }_d}\text{damped period of oscillation}\\ \tau & =\{\frac{-1}{{\lambda }_1},\frac{-1}{\lambda 2}\}\text{ time constants where }{\lambda }_i\text{ are roots of characteristic equation}\\ \beta & =\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\xi \right)}^2}}\text{ magnification factor}\\ {\beta }_{max}\text{ when }r& =\sqrt{1-2{\xi }^2}\\ {\beta }_{max}& =\frac{1}{2\xi \sqrt{1-{\xi }^2}}\\ \frac{y_n}{y_{n+1}}& =e^{\frac{\zeta {\omega }_{n}2\pi }{{\omega }_D}}\stackrel{\text{small damping}}{\Rightarrow }{e}^{\frac{\zeta 2\pi }{\sqrt{1-{\zeta }^2}}}\Rightarrow e^{\zeta 2\pi }\\ \ln \left(\frac{y_n}{y_{n+1}}\right)& =\zeta 2\pi \\ \frac{1}{M}\ln \left(\frac{y_n}{y_{n+M}}\right)& =\zeta 2\pi \end{array}$$

1.4.2 Harmonic loading $m{u}^{\prime \prime }+c{u}^{\prime }+ku=F\sin \varpi t$

1.4.2.1 Undamped Harmonic loading

Figure 1.5: single degree no damping forced

$$m{u}^{\prime \prime }+ku=F\sin \varpi t$$

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$$u\left(0\right)=A$$$$\begin{array}{rl}{u}^{\prime }\left(t\right)& =-A\omega \sin {\omega }_{n}t+B\omega \cos {\omega }_{n}t+\varpi \frac{F}{k}\frac{1}{1-r^2}\cos \varpi t\\ u^{\prime }\left(0\right)& =B{\omega }_n+\varpi \frac{F}{k}\frac{1}{1-r^2}\\ B& =\frac{u^{\prime }\left(0\right)}{{\omega }_n}-\frac{F}{k}\frac{r}{1-r^2}\end{array}$$

Where $r=\frac{\varpi }{{\omega }_n}$

The complete solution is

$$\begin{array}{}\text{(1)}& u\left(t\right)=u\left(0\right)\cos {\omega }_{n}t+(\frac{u^{\prime }\left(0\right)}{{\omega }_n}-\frac{F}{k}\frac{r}{1-r^2})\sin {\omega }_{n}t+\frac{F}{k}\frac{1}{1-r^2}\sin \varpi t\end{array}$$

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

$$\begin{array}{rl}u& =(-3\frac{5}{1-5^2})\sin t+3\frac{1}{1-5^2}\sin 5t\\ & =0.625\sin t-0.125\sin 5.0t\end{array}$$

1.4.2.1.1 Resonance forced vibration When $\varpi \approx \omega$ we obtain resonance since $r\to 1$ 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

$$\begin{array}{}\text{(1A)}& u\left(t\right)=u\left(0\right)\cos \omega t+(\frac{u^{\prime }\left(0\right)}{\omega }-\frac{F}{k}\frac{\omega \varpi }{{\omega }^2-{\varpi }^2})\sin \omega t+\frac{F}{k}\frac{{\omega }^2}{{\omega }^2-{\varpi }^2}\sin \varpi t\end{array}$$ When $\varpi \approx \omega$ but less than $\omega$, letting $$\begin{array}{}\text{(2)}& \omega -\varpi =2\mathrm{\Delta }\end{array}$$ where $\mathrm{\Delta }$ is very small positive quantity. And since $\varpi \approx \omega$ let $$\begin{array}{}\text{(3)}& \omega +\varpi \approx 2\varpi \end{array}$$ Multiplying Eq (2) and (3) gives$$\begin{array}{}\text{(4)}& {\omega }^2-{\varpi }^2=4\mathrm{\Delta }\varpi \end{array}$$ Eq (1A) can now be written in terms of Eqs (2,3) as$$\begin{array}{rl}u\left(t\right)& =u\left(0\right)\cos \omega t+(\frac{u^{\prime }\left(0\right)}{\omega }-\frac{F}{k}\frac{\omega \varpi }{4\mathrm{\Delta }\varpi })\sin \omega t+\frac{F}{k}\frac{{\omega }^2}{4\mathrm{\Delta }\varpi }\sin \varpi t\\ & =u\left(0\right)\cos \omega t+(\frac{v_0}{\omega }-\frac{F}{k}\frac{\omega }{4\mathrm{\Delta }})\sin \omega t+\frac{F}{k}\frac{{\omega }^2}{4\mathrm{\Delta }\varpi }\sin \varpi t\end{array}$$

Since $\varpi \approx \omega$ the above becomes$$\begin{array}{rl}u\left(t\right)& =u\left(0\right)\cos \omega t+(\frac{u^{\prime }\left(0\right)}{\omega }-\frac{F}{k}\frac{\omega }{4\mathrm{\Delta }})\sin \omega t+\frac{F}{k}\frac{\omega }{4\mathrm{\Delta }}\sin \varpi t\\ & =u\left(0\right)\cos \omega t+\frac{u^{\prime }\left(0\right)}{\omega }\sin \omega t+\frac{F}{k}\frac{\omega }{4\mathrm{\Delta }}(\sin \varpi t-\sin \omega t)\end{array}$$

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$$u\left(t\right)=u\left(0\right)\cos \omega t+\frac{u^{\prime }\left(0\right)}{\omega }\sin \omega t+\frac{F}{k}\frac{\omega }{2\mathrm{\Delta }}(\sin \left(\frac{\varpi -\omega }{2}t\right)\cos \left(\frac{\varpi +\omega }{2}t\right))$$ From Eqs (2,3) the above can be written as$$u\left(t\right)=u\left(0\right)\cos \omega t+\frac{u^{\prime }\left(0\right)}{\omega }\sin \omega t+\frac{F}{k}\frac{\omega }{2\mathrm{\Delta }}(\sin (-\mathrm{\Delta }t)\cos \left(\varpi t\right))$$ Since $\underset{\mathrm{\Delta }\to 0}{lim}\frac{\sin \left(\mathrm{\Delta }t\right)}{\mathrm{\Delta }}=t$ 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.

1.4.2.2 Underdamped harmonic loading $c<c_r,\xi <1$

Figure 1.6: single degree damping forced

$$m{u}^{\prime \prime }+c{u}^{\prime }+ku=F\sin \varpi t$$$$u^{\prime \prime }+2\xi \omega u^{\prime }+{\omega }^{2}u=\frac{F}{m}\sin \varpi t$$ The solution is$$u\left(t\right)=u_h+u_p$$ where $$u_h\left(t\right)=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$$ and $$u_p\left(t\right)=\frac{F}{\sqrt{{(k-m\varpi )}^2+{\left(c\varpi \right)}^2}}\sin (\varpi t-\theta )$$ 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$$\begin{array}{}\text{(1)}& u\left(t\right)=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t-\theta )\end{array}$$ Applying initial conditions gives$$\begin{array}{rl}A& =u\left(0\right)+\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin \theta \\ B& =\frac{u^{\prime }\left(0\right)}{{\omega }_d}+\frac{u\left(0\right)\xi \omega }{{\omega }_d}+\frac{F}{k}\frac{1}{{\omega }_d\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}(\xi \omega \sin \theta -\varpi \cos \theta )\end{array}$$

Another form of these equations is given as follows

$$u_p=\frac{p_0}{k}\frac{1}{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}((1-r^2)\sin \varpi t-2\zeta r\cos \varpi t)$$

Hence the full solution is

$$\begin{array}{}\text{(1)}& u\left(t\right)=e^{-\xi {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{F}{k}\frac{1}{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}((1-r^2)\sin \varpi t-2\zeta r\cos \varpi t)\end{array}$$

Applying initial conditions now gives

$$\begin{array}{rl}A& =u\left(0\right)+\frac{2Fr\xi }{k}\frac{1}{{(1-r^2)}^2+{\left(2\xi r\right)}^2}\\ B& =\frac{u^{\prime }\left(0\right)}{{\omega }_d}+\frac{u\left(0\right)\xi {\omega }_n}{{\omega }_d}-\frac{F(1-r^2)}{k{\omega }_d}\frac{\varpi }{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}+\frac{2Fr\zeta }{k{\omega }_d}\frac{{\omega }_n}{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}\end{array}$$

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

$$u\left(t\right)=e^{-\xi {\omega }_{n}t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{F}{k}\frac{1}{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}((1-r^2)\cos \varpi t-2\zeta r\sin \varpi t)$$

Applying initial conditions now gives

$$\begin{array}{rl}A& =u\left(0\right)+\frac{F}{k}\frac{(1-r^2)}{{(1-r^2)}^2+{\left(2\xi r\right)}^2}\\ B& =\frac{u^{\prime }\left(0\right)}{{\omega }_d}+\frac{u\left(0\right)\xi {\omega }_n}{{\omega }_d}+\frac{2Fr\zeta }{k{\omega }_d}\frac{\varpi }{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}-\frac{F(1-r^2)}{k{\omega }_d}\frac{{\omega }_n}{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}\end{array}$$

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$$\begin{array}{c}u\left(t\right)=e^{-\xi \omega t}((u\left(0\right)+\frac{F}{k}\frac{1}{2\xi }\sin \theta )\cos {\omega }_{d}t+(\frac{u^{\prime }\left(0\right)}{{\omega }_d}+\frac{u\left(0\right)\xi \omega }{{\omega }_d}+\frac{F}{k}\frac{1}{2{\omega }_d\xi }(\xi \omega \sin \theta -\varpi \cos \theta ))\sin {\omega }_{d}t)\hfill \\ \hfill +\frac{F}{k}\frac{1}{2}\sin (\varpi t-\theta )\end{array}$$ 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 $\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}$. This is maximum when the magnification factor $\beta =\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}$ is maximum or when $\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}$ or $\sqrt{{(1-{\left(\frac{\varpi }{\omega }\right)}^2)}^2+{\left(2\xi \frac{\varpi }{\omega }\right)}^2}$ 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.

1.4.2.3 critically damping harmonic loading $\xi =\frac{c}{c_r}=1$

The solution is$$u\left(t\right)=u_h+u_p$$ Where $u_h=(A+Bt)e^{-\omega t}$ and $u_p=\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t-\theta )$ where $\tan \theta =\frac{2r}{1-r^2}$(making sure to use correct $\arctan$ definition). Hence$$u\left(t\right)=(A+Bt)e^{-\omega t}+\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t-\theta )$$ where $A,B$ are found from initial conditions$$\begin{array}{rl}A& =u\left(0\right)+\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin \theta \\ B& =u^{\prime }\left(0\right)+u\left(0\right)\omega +\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}(\omega \sin \theta -\varpi \cos \theta )\end{array}$$

1.4.2.4 overdamped harmonic loading $\xi =\frac{c}{c_r}>1$

The solution is$$u\left(t\right)=u_h+u_p$$ where$$u_h\left(t\right)=A{e}^{p_{1}t}+B{e}^{p_{2}t}$$ and $$u_p\left(t\right)=\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t-\theta )$$ hence$$u=A{e}^{p_{1}t}+B{e}^{p_{2}t}+\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t-\theta )$$ where $\tan \theta =\frac{2\xi r}{1-r^2}$ and$$\begin{array}{rl}{p}_1& =-\frac{c}{2m}+\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\omega \xi +{\omega }_n\sqrt{{\xi }^2-1}\\ p_2& =-\frac{c}{2m}-\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\omega \xi -{\omega }_n\sqrt{{\xi }^2-1}\end{array}$$

Hence the solution is$$\begin{array}{rl}u\left(t\right)& =A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t}+B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t}+\frac{F}{k}\beta \sin (\varpi t-\theta )\\ A& =\frac{u^{\prime }\left(0\right)+u\left(0\right)\omega \xi +u\left(0\right)\omega \sqrt{{\xi }^2-1}+\frac{F}{k}\beta ((\xi +\sqrt{{\xi }^2-1})\omega \sin \theta -\varpi \cos \theta )}{2\omega \sqrt{{\xi }^2-1}}\\ B& =-\frac{u^{\prime }\left(0\right)+u\left(0\right)\omega \xi -u\left(0\right)\omega \sqrt{{\xi }^2-1}+\frac{F}{k}\beta ((\xi -\sqrt{{\xi }^2-1})\omega \sin \theta -\varpi \cos \theta )}{2\omega \sqrt{{\xi }^2-1}}\end{array}$$

1.4.2.5 Solution using frequency approach to harmonic loading

$$\begin{array}{rl}m{y}^{\prime \prime }+c{y}^{\prime }+ky& =\mathrm{Re}\left(\hat{F}{e}^{i\varpi t}\right)\\ x& =\mathrm{Re}\left\{\hat{X}{e}^{i\varpi t}\right\}\\ \hat{X}& =\frac{\hat{F}}{k}D(r,\zeta )\\ D(r,\zeta )& =\frac{1}{(1-r^2)+2i\zeta r}\\ x& =\mathrm{Re}\left\{\frac{\hat{F}}{k}\left|D(r,\zeta )\right|e^{i(\varpi t-\theta )}\right\}\\ \theta & ={\tan }^{-1}\frac{2\zeta r}{1-r^2}\end{array}$$

Let load be harmonic and represented in general as $\mathrm{Re}\left(\hat{F}{e}^{i\varpi t}\right)$ where $\hat{F}$ is the complex amplitude of the force.

Hence system is represented by $$\begin{array}{rl}m{y}^{\prime \prime }+c{y}^{\prime }+ky& =\mathrm{Re}\left(\hat{F}{e}^{i\varpi t}\right)\\ y^{\prime \prime }+2\zeta {\omega }_n{y}^{\prime }+{\omega }_n^{2}y& =\mathrm{Re}\left(\frac{\hat{F}}{m}{e}^{i\varpi t}\right)\end{array}$$

Let $y=\mathrm{Re}\left(\hat{Y}{e}^{i\varpi t}\right)$ Hence $y^{\prime }=\mathrm{Re}\left(i\varpi \hat{Y}{e}^{i\varpi t}\right),y^{\prime \prime }=\mathrm{Re}(-{\varpi }^2\hat{Y}{e}^{i\varpi t})$, therefore the differential equation becomes

$$\begin{array}{rl}\mathrm{Re}(-{\varpi }^2\hat{Y}{e}^{i\varpi t})+2\zeta {\omega }_n\mathrm{Re}\left(i\varpi \hat{Y}{e}^{i\varpi t}\right)+{\omega }_n^2\mathrm{Re}\left(\hat{Y}{e}^{i\varpi t}\right)& =\mathrm{Re}\left(\frac{\hat{F}}{m}{e}^{i\varpi t}\right)\\ (-{\varpi }^2+2\zeta {\omega }_{n}i\varpi +{\omega }_n^2)\hat{Y}& =\frac{\hat{F}}{m}\\ \hat{Y}& =\frac{\frac{\hat{F}}{m}}{(-{\varpi }^2+2\zeta {\omega }_{n}i\varpi +{\omega }_n^2)}\end{array}$$

Dividing numerator and denominator ${\omega }_n^2$ gives$$\hat{Y}=\frac{\hat{F}}{k}\frac{1}{(1-r^2)+i2\zeta r}$$

Where $r=\frac{\varpi }{{\omega }_n},$ hence the response is$$y=\mathrm{Re}\left(\frac{\hat{F}}{k}\frac{1}{(1-r^2)+i2\zeta r}{e}^{i\varpi t}\right)$$

Therefore, the phase of the response is$$\arg \left(y\right)=\arg \left(\hat{F}\right)-{\tan }^{-1}\left(\frac{2\zeta r}{(1-r^2)}\right)+\varpi t$$

Hence at $t=0$ the phase of the response will be$$\arg \left(y\right)=\arg \left(\hat{F}\right)-{\tan }^{-1}\left(\frac{2\zeta r}{(1-r^2)}\right)$$

So when $\hat{F}$ is real, the phase of the response is simply $-{\tan }^{-1}\left(\frac{2\zeta r}{(1-r^2)}\right)$

Undamped case

When $\zeta =0$ the above becomes$$\begin{array}{rl}y& =\mathrm{Re}\left(\frac{\hat{F}}{k}\frac{1}{(1-r^2)}{e}^{i\varpi t}\right)\\ & =\frac{\left|\hat{F}\right|}{k}\frac{1}{(1-r^2)}\cos (\varpi t+\arg \left(\hat{F}\right))\end{array}$$

For real force this becomes$$y=\frac{F}{k}\frac{1}{(1-r^2)}\cos \left(\varpi t\right)$$

The magnitude $\left|\hat{Y}\right|=\frac{F}{k}\frac{1}{(1-r^2)}$ and phase zero.

damped cases

$\zeta >0$$$y=\mathrm{Re}\left(\frac{\hat{F}}{k}\frac{1}{(1-r^2)+i2\zeta r}{e}^{i\varpi t}\right)$$

$$\begin{array}{rl}\left|\hat{Y}\right|& =\frac{\left|\hat{F}\right|}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}\\ \arg \left(\hat{Y}\right)& =\varphi =\arg \left(\hat{F}\right)-{\tan }^{-1}\left(\frac{2\zeta r}{1-r^2}\right)+\varpi t\end{array}$$

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 $\varphi$ 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 $r=1$

$$y=\mathrm{Re}\left(\frac{\hat{F}}{k}\frac{1}{i2\zeta }{e}^{i\varpi t}\right)$$

$$\begin{array}{rl}\left|\hat{Y}\right|& =\frac{\left|\hat{F}\right|}{k}\frac{1}{2\zeta }\\ \arg \left(\hat{Y}\right)& =-{90}^{\circ }\end{array}$$

Now phase is $-{90}^{\circ }$

Figure 1.7: steady state

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 $y\left(t\right)=\mathrm{Re}\left(\hat{Y}{e}^{i\varpi t}\right)$, ${\omega }_n=\sqrt{\frac{k}{m}}=1$ rad/sec, hence $r=0.5$ under the response is

$$\begin{array}{rl}y\left(t\right)& =\mathrm{Re}\left(\left|\hat{Y}\right|e^{i\varpi t}\right)\\ & =\left|\hat{Y}\right|\cos \left(\varpi t\right)\\ & =\frac{F}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}\cos (.5t-{\tan }^{-1}\left(\frac{2\left(0.1\right)0.5}{1-{0.5}^2}\right))\\ & =3\frac{1}{\sqrt{{(1-{0.5}^2)}^2+{\left(2\left(0.1\right)0.5\right)}^2}}\cos (.5t-{7.59}^{\circ })\\ & =3.9649\cos (.5t-{7.59}^{\circ })\end{array}$$

Figure 1.8: steady state plot

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.

1.4.2.6 Summary table

$\zeta =0$	roots $\{\begin{array}{l}-i\omega \\ +i\omega \end{array}$ $u\left(t\right)$ $\{\begin{array}{l}\varpi =\omega \to u\left(0\right)\cos \varpi t+\frac{u^{\prime }\left(0\right)}{\varpi }\sin \varpi t-\frac{F}{K}\frac{\varpi t}{2}\cos \left(\varpi t\right)\\ \varpi \ne \omega \to u\left(0\right)\cos \omega t+(\frac{u^{\prime }\left(0\right)}{\omega }-\frac{F}{K}\frac{r}{1-r^2})\sin \omega t+\frac{F}{K}\frac{1}{1-r^2}\sin \varpi t\end{array}$
$\zeta <1$	roots $\{\begin{array}{l}-\xi \omega +i{\omega }_n\sqrt{1-{\xi }^2}\\ -\xi \omega -i{\omega }_n\sqrt{1-{\xi }^2}\end{array}$ $u\left(t\right)=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{F}{K}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t-\theta )$ $A=u_0+\frac{F}{K}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin \theta$ $B=\frac{v_0}{{\omega }_d}+\frac{u_0\xi \omega }{{\omega }_d}+\frac{F}{K}\frac{1}{{\omega }_d\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}(\xi \omega \sin \theta -\varpi \cos \theta )$
$\zeta =1$	roots $\{\begin{array}{l}-\omega \\ -\omega \end{array}$ $u\left(t\right)=(A+Bt)e^{-\omega t}+\frac{F}{K}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t-\theta )$ $A=u_0+\frac{F}{K}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin \theta$ $B=v_0+u_0\omega +\frac{F/k}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}(\omega \sin \theta -\varpi \cos \theta )$
$\zeta >1$	roots $\{\begin{array}{l}-{\omega }_n\xi +{\omega }_n\sqrt{{\xi }^2-1}\\ -{\omega }_n\xi -{\omega }_n\sqrt{{\xi }^2-1}\end{array}$ $u\left(t\right)=A{e}^{(-\xi +\sqrt{{\xi }^2-1}){\omega }_{n}t}+B{e}^{(-\xi -\sqrt{{\xi }^2-1}){\omega }_{n}t}+\frac{F}{K}\beta \sin (\varpi t-\theta )$ $A=\frac{v_0+u_0\omega \xi +u_0\omega \sqrt{{\xi }^2-1}+\frac{F}{K}\beta ((\xi +\sqrt{{\xi }^2-1})\omega \sin \theta -\varpi \cos \theta )}{2\omega \sqrt{{\xi }^2-1}}$ $B=-\frac{v_0+u_0\omega \xi -u_0\omega \sqrt{{\xi }^2-1}+\frac{F}{K}\beta ((\xi -\sqrt{{\xi }^2-1})\omega \sin \theta -\varpi \cos \theta )}{2\omega \sqrt{{\xi }^2-1}}$

1.4.3 constant loading $m{u}^{\prime \prime }+c{u}^{\prime }+ku=F$

1.4.3.1 Undamped Constant loading case $\zeta =0$

$$m{u}^{\prime \prime }+ku=F$$$$u^{\prime \prime }+{\omega }^{2}u=F$$$$u\left(t\right)=u_h+u_p$$

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$$\begin{array}{rl}A& =u\left(0\right)-\frac{F}{k}\\ B& =\frac{u^{\prime }\left(0\right)}{\omega }\end{array}$$

And complete solution is$$u\left(t\right)=\frac{F}{k}+(u\left(0\right)-\frac{F}{k})\cos \omega t+\frac{u^{\prime }\left(0\right)}{\omega }\sin \omega t$$

1.4.3.2 underdamped constant loading $\zeta <1$

The general solution is $$u\left(t\right)=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{F}{k}$$ From initial conditions$$\begin{array}{rl}A& =u\left(0\right)-\frac{F}{k}\\ B& =\frac{u^{\prime }\left(0\right)+u\left(0\right)\xi \omega -\frac{F}{k}\xi \omega }{{\omega }_d}\end{array}$$

Hence the solution is$$u\left(t\right)=e^{-\xi \omega t}((u\left(0\right)-\frac{F}{k})\cos {\omega }_{d}t+\left(\frac{u^{\prime }\left(0\right)+u\left(0\right)\xi \omega -\frac{F}{k}\xi \omega }{{\omega }_d}\right)\sin {\omega }_{d}t)+\frac{F}{k}$$

1.4.3.3 Critical damping constant loading $\zeta =1$

The general solution is $$u(t)=(A+Bt)e^{-\omega t}+\frac{F}{k}$$ Where from initial conditions $$\begin{array}{rl}A& =u(0)-\frac{F}{k}\\ B& =u^{\prime }(0)+u(0)\omega -\frac{F}{k}\omega \end{array}$$

1.4.3.4 Over-damped constant loading $\zeta >0$

The solution is$$u\left(t\right)=A{e}^{p_{1}t}+B{e}^{p_{2}t}+\frac{F}{k}$$ Where now$$\begin{array}{rl}B& =\frac{\frac{F}{k}{p}_1-u_0{p}_1+u^{\prime }\left(0\right)}{(p_2-p_1)}\\ A& =u\left(0\right)-\frac{F}{k}-B\end{array}$$

Hence the solution is$$u\left(t\right)=A{e}^{p_{1}t}+B{e}^{p_{2}t}+\frac{F}{k}$$ Where $$\begin{array}{rl}{p}_1& =-\frac{c}{2m}+\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\omega \xi +{\omega }_n\sqrt{{\xi }^2-1}\\ p_2& =-\frac{c}{2m}-\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\omega \xi -{\omega }_n\sqrt{{\xi }^2-1}\end{array}$$

1.4.3.5 Summary table for constant loading solutions

$\zeta =0$	roots$\{\begin{array}{l}-i\omega \\ +i\omega \end{array}$ $u\left(t\right)=(u_0-\frac{F}{k})\cos \omega t+\frac{v_0}{\omega }\sin \omega t+\frac{F}{k}$
$\zeta <1$	roots$\{\begin{array}{l}-\xi \omega +i{\omega }_n\sqrt{1-{\xi }^2}\\ -\xi \omega -i{\omega }_n\sqrt{1-{\xi }^2}\end{array}$ $u\left(t\right)=e^{-\xi \omega t}((u_0-\frac{F}{k})\cos {\omega }_{d}t+\left(\frac{v_0+u_0\xi \omega -\frac{F}{k}\xi \omega }{{\omega }_d}\right)\sin {\omega }_{d}t)+\frac{F}{k}$
$\zeta =1$	roots $\{\begin{array}{l}-\omega \\ -\omega \end{array}$ $u\left(t\right)=((u_0-\frac{F}{k})+(v_0+u_0\omega -\frac{F}{k}\omega )t)e^{-\omega t}+\frac{F}{k}$
$\zeta >1$	roots $\{\begin{array}{l}-{\omega }_n\xi +{\omega }_n\sqrt{{\xi }^2-1}\\ -{\omega }_n\xi -{\omega }_n\sqrt{{\xi }^2-1}\end{array}$ $B=\frac{\frac{F}{k}{p}_1-u_0{p}_1+v_0}{(p_2-p_1)}$ $A=u_0-\frac{F}{k}-B$ $p_1=-\frac{c}{2m}+\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-{\omega }_n\xi +{\omega }_n\sqrt{{\xi }^2-1}$ $p_2=-\frac{c}{2m}-\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-{\omega }_n\xi -{\omega }_n\sqrt{{\xi }^2-1}$ $u\left(t\right)=A{e}^{p_{1}t}+B{e}^{p_{2}t}+\frac{F}{k}$

1.4.4 No loading (free vibration) $m{u}^{\prime \prime }+c{u}^{\prime }+ku=0$

1.4.4.1 Undamped free vibration

Figure 1.9: single degree mass no damping

$$m{u}^{\prime \prime }+ku=0$$$$u^{\prime \prime }+{\omega }^{2}u=0$$ The solution is $$u\left(t\right)=u\left(0\right)\cos \omega t+\frac{u^{\prime }\left(0\right)}{\omega }\sin \omega t$$

1.4.4.2 under-damped free vibration $c<c_r,\xi <1$

Figure 1.10: single degree mass damping

$$m{u}^{\prime \prime }+c{u}^{\prime }+ku=0$$$$u^{\prime \prime }+2\xi \omega u^{\prime }+{\omega }^{2}u=0$$

The solution is$$u=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$$ 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$$u\left(t\right)=e^{-\xi \omega t}(u\left(0\right)\cos {\omega }_{d}t+\frac{u^{\prime }\left(0\right)+u\left(0\right)\xi \omega }{{\omega }_d}\sin {\omega }_{d}t)$$

1.4.4.3 critically damped free vibration $\xi =\frac{c}{c_r}=1$

The solution is$$\begin{array}{rl}u\left(t\right)& =(A+Bt)e^{-\left(\frac{c_r}{2m}\right)t}\\ & =(A+Bt)e^{-\omega t}\end{array}$$

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$$u\left(t\right)=(u\left(0\right)+(u^{\prime }\left(0\right)+u\left(0\right)\omega )t)e^{-\omega t}$$

1.4.4.4 over-damped free vibration $\xi =\frac{c}{c_r}>1$

The solution is$$u\left(t\right)=A{e}^{{\lambda }_{1}t}+B{e}^{{\lambda }_{2}t}$$ where $A,B$ are found from initial conditions.$$\begin{array}{rl}A& =\frac{u^{\prime }(0)-u\left(0\right){\lambda }_2}{2\omega \sqrt{{\xi }^2-1}}\\ B& =\frac{-u^{\prime }(0)+u\left(0\right){\lambda }_1}{2\omega \sqrt{{\xi }^2-1}}\end{array}$$

where ${\lambda }_1$ and ${\lambda }_2$ are the roots of the characteristic equation$$\begin{array}{rl}{\lambda }_1& =-\frac{c}{2m}+\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\xi \omega +\omega \sqrt{{\xi }^2-1}\\ {\lambda }_2& =-\frac{c}{2m}-\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\xi \omega -\omega \sqrt{{\xi }^2-1}\end{array}$$

1.4.4.5 Summary table for free vibration solutions

$\begin{array}{l}\zeta =0\\ u^{\prime \prime }+{\omega }^{2}u=0\end{array}$	roots$\{\begin{array}{l}-i\omega \\ +i\omega \end{array}$ $u\left(t\right)=u(0)\cos \omega t+\frac{u^{\prime }(0)}{\omega }\sin \omega t$ $\{\begin{array}{l}u\left(t\right)=A\cos (\omega t-\varphi )\\ A=\sqrt{u^2(0)+{\left(\frac{u^{\prime }(0)}{\omega }\right)}^2}\\ \varphi ={\tan }^{-1}\left(\frac{u^{\prime }(0)/\omega }{u(0)}\right)\end{array}$
$\zeta <1$	roots$\{\begin{array}{l}-\xi \omega +i\omega \sqrt{1-{\xi }^2}\\ -\xi \omega -i\omega \sqrt{1-{\xi }^2}\end{array}$ $u\left(t\right)=e^{-\xi \omega t}(u(0)\cos {\omega }_{d}t+\frac{u^{\prime }(0)+u(0)\xi \omega }{{\omega }_d}\sin {\omega }_{d}t)$
$\zeta =1$	roots$\{\begin{array}{l}-\omega \\ -\omega \end{array}$ $u\left(t\right)=(u(0)(1+\omega t)+u^{\prime }(0)t)e^{-\omega t}$
$\zeta >1$	roots$\{\begin{array}{l}{\lambda }_1=-\omega \xi +\omega \sqrt{{\xi }^2-1}\\ {\lambda }_2=-\omega \xi -\omega \sqrt{{\xi }^2-1}\end{array}$ $\{\begin{array}{l}u\left(t\right)=A{e}^{{\lambda }_1\omega t}+B{e}^{{\lambda }_2\omega t}\\ A=\frac{u^{\prime }(0)-u\left(0\right){\lambda }_2}{2\omega \sqrt{{\xi }^2-1}}\\ B=\frac{-u^{\prime }(0)+u\left(0\right){\lambda }_1}{2\omega \sqrt{{\xi }^2-1}}\end{array}$

1.4.4.6 Roots of characteristic equation

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 $\tau$
$\xi <1$	$\{-\xi \omega +j{\omega }_n\sqrt{1-{\xi }^2},-\xi \omega -i{\omega }_n\sqrt{1-{\xi }^2}\}$	$\frac{1}{\xi \omega }$
$\xi =1$	$\{-\omega ,-\omega \}$	$\frac{1}{\omega }$
$\xi >1$	$\{-{\omega }_n\xi +{\omega }_n\sqrt{{\xi }^2-1},-{\omega }_n\xi -{\omega }_n\sqrt{{\xi }^2-1}\}$	$\frac{1}{{\omega }_n\xi -{\omega }_n\sqrt{{\xi }^2-1}},\frac{1}{{\omega }_n\xi +{\omega }_n\sqrt{{\xi }^2-1}}$(which to use? the bigger?)

1.4.5 impulse $F_0\delta \left(t\right)$ loading

1.4.5.1 impulse input

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$$\begin{array}{rl}{\int }_0^{t_1}m\ddot{u}dt& ={\int }_0^{t_1}{F}_0\delta (t)\\ {\int }_0^{t_1}m\left(\frac{d\dot{u}}{dt}\right)dt& ={\int }_0^{t_1}{F}_0\delta (t)\\ {\int }_{\dot{u}\left(0\right)}^{\dot{u}\left(t_1\right)}d\dot{u}& =\frac{F_0}{m}{\int }_0^{t_1}\delta (t)\\ \dot{u}\left(t_1\right)-\dot{u}\left(0\right)& =\frac{F_0}{m}{\int }_0^{t_1}\delta (t)\\ \dot{u}\left(t_1\right)& =\frac{F_0}{m}{\int }_0^{t_1}\delta (t)\end{array}$$

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$$u\left(t\right)=\stackrel{\text{due to IC only}}{\overbrace{u(0)\cos \omega t+\frac{u^{\prime }(0)}{\omega }\sin \omega t}}+\stackrel{\text{due to impulse}}{\overbrace{\frac{F_0}{m\omega }\sin \omega t}}$$

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$$u_{impulse}=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$$

applying initial conditions gives $A=0$ and $B=\frac{\left(\frac{F_0}{m}\right)}{{\omega }_d}$, hence$$u_{impulse}\left(t\right)=e^{-\xi \omega t}(\frac{F_0}{m{\omega }_d}\sin {\omega }_{d}t)$$

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 $u\left(t\right)=e^{-\xi \omega t}(u(0)\cos {\omega }_{d}t+\frac{u^{\prime }(0)+u(0)\xi \omega }{{\omega }_d}\sin {\omega }_{d}t)$, therefore, the full solution is$$u\left(t\right)=\stackrel{\text{due to IC only}}{\overbrace{e^{-\xi \omega t}(u(0)\cos {\omega }_{d}t+\frac{u^{\prime }(0)+u(0)\xi \omega }{{\omega }_d}\sin {\omega }_{d}t)}}+\stackrel{\text{due to impulse}}{\overbrace{e^{-\xi \omega t}(\frac{F_0}{m{\omega }_d}\sin {\omega }_{d}t)}}$$

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$$\begin{array}{rl}u\left(t\right)& =(A+Bt)e^{-\left(\frac{c_r}{2m}\right)t}\\ & =(A+Bt)e^{-\omega t}\end{array}$$

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 $u\left(t\right)=(u_0(1+\omega t)+u^{\prime }\left(0\right)t)e^{-\omega t}$, therefore, the full solution is$$u\left(t\right)=\stackrel{\text{due to IC only}}{\overbrace{(u\left(0\right)(1+\omega t)+u^{\prime }\left(0\right)t)e^{-\omega t}}}+\stackrel{\text{due to impulse}}{\overbrace{\frac{F_{0}t}{m}{e}^{-\omega t}}}$$

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$$u_{impulse}\left(t\right)=A{e}^{{\lambda }_1\omega t}+B{e}^{{\lambda }_2\omega t}$$ where $A,B$ are found from initial conditions and $$\begin{array}{rl}{\lambda }_1& =-\omega \xi +\omega \sqrt{{\xi }^2-1}\\ {\lambda }_2& =-\omega \xi -\omega \sqrt{{\xi }^2-1}\\ A& =\frac{u^{\prime }(0)-u\left(0\right){\lambda }_2}{2\omega \sqrt{{\xi }^2-1}}\\ B& =\frac{-u^{\prime }(0)+u\left(0\right){\lambda }_1}{2\omega \sqrt{{\xi }^2-1}}\end{array}$$

Hence the solution is$$u_{impulse}\left(t\right)=A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t}+B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t}$$

where$$\begin{array}{rl}A& =\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}\\ B& =\frac{-\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}\end{array}$$

Hence

$$u_{impulse}\left(t\right)=\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t}-\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t}$$

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 $u\left(t\right)=A{e}^{p_{1}t}+B{e}^{p_{2}t}$, therefore, the full solution is$$u\left(t\right)=A{e}^{{\lambda }_1\omega t}+B{e}^{{\lambda }_2\omega t}+\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}{e}^{{\lambda }_1\omega t}-\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}{e}^{{\lambda }_2\omega t}$$

$$\begin{array}{rl}A& =\frac{u^{\prime }\left(0\right)-u\left(0\right){\lambda }_2}{2\omega \sqrt{{\xi }^2-1}}\\ B& =\frac{-u^{\prime }\left(0\right)+u\left(0\right){\lambda }_1}{2\omega \sqrt{{\xi }^2-1}}\end{array}$$

1.4.5.1.5 Summary table

$\begin{array}{l}\zeta =0\\ u^{\prime \prime }+{\omega }^{2}u=0\end{array}$	roots$\{\begin{array}{l}-i\omega \\ +i\omega \end{array}$ $u\left(t\right)=\stackrel{\text{transient}}{\overbrace{u(0)\cos \omega t+\frac{u^{\prime }(0)}{\omega }\sin \omega t}}+\stackrel{\text{steady state}}{\overbrace{\frac{F_0}{m\omega }\sin \omega t}}$
$\zeta <1$	roots$\{\begin{array}{l}-\xi \omega +i\omega \sqrt{1-{\xi }^2}\\ -\xi \omega -i\omega \sqrt{1-{\xi }^2}\end{array}$ $u\left(t\right)=\stackrel{\text{transient}}{\overbrace{e^{-\xi \omega t}(u(0)\cos {\omega }_{d}t+\frac{u^{\prime }(0)+u(0)\xi \omega }{{\omega }_d}\sin {\omega }_{d}t)}}+\stackrel{\text{steady state}}{\overbrace{e^{-\xi \omega t}(\frac{F_0}{m{\omega }_d}\sin {\omega }_{d}t)}}$
$\zeta =1$	roots$\{\begin{array}{l}-\omega \\ -\omega \end{array}$ $u\left(t\right)=(u\left(0\right)(1+\omega t)+u^{\prime }\left(0\right)t)e^{-\omega t}+\frac{F_{0}t}{m}{e}^{-\omega t}$
$\zeta >1$	roots$\{\begin{array}{l}{\lambda }_1=-\omega \xi +\omega \sqrt{{\xi }^2-1}\\ {\lambda }_2=-\omega \xi -\omega \sqrt{{\xi }^2-1}\end{array}$ $\{\begin{array}{l}u\left(t\right)=A{e}^{{\lambda }_1\omega t}+B{e}^{{\lambda }_2\omega t}+\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}{e}^{{\lambda }_1\omega t}-\frac{\frac{F_0}{m}}{2\omega \sqrt{{\xi }^2-1}}{e}^{{\lambda }_2\omega t}\\ A=\frac{u^{\prime }\left(0\right)-u\left(0\right){\lambda }_2}{2\omega \sqrt{{\xi }^2-1}}\\ B=\frac{-u^{\prime }\left(0\right)+u\left(0\right){\lambda }_1}{2\omega \sqrt{{\xi }^2-1}}\end{array}$

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]
 

Figure 1.11: impulse Response Diagram

1.4.5.2 Impulse sin function

Now assume the input is as follows

Figure 1.12: input function

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 $$m\ddot{u}+ku=\{\begin{array}{ccc}{F}_0\sin \left(\varpi t\right)& & 0\le t\le t_1\\ 0& & t>t_1\end{array}$$

with $u\left(0\right)=u_0$ and $\dot{u}\left(0\right)=v_0.$ For $0\le t\le t_1$ the solution is$$u\left(t\right)=u_0\cos \omega t+(\frac{v_0}{\omega }-u_{st}\frac{r}{1-r^2})\sin \omega t+u_{st}\frac{1}{1-r^2}\sin \left(\frac{\pi }{t_1}t\right)$$

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$$\begin{array}{}\text{(1)}& u\left(t\right)=u_0\cos \omega t+(\frac{v_0}{\omega }-\frac{F_0}{k}\frac{\left(\frac{\pi /t_1}{\omega }\right)}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2})\sin \omega t+\frac{F_0}{k}\frac{1}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2}\sin \left(\pi \frac{t}{t_1}\right)\end{array}$$

When $u_0=0$ and $v_0=0$ then$$\begin{array}{rl}u\left(t\right)& =-\frac{F_0}{k}\frac{\left(\frac{\pi /t_1}{\omega }\right)}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2}\sin \omega t+\frac{F_0}{k}\frac{1}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2}\sin \left(\pi \frac{t}{t_1}\right)\\ u\left(t\right)& =\frac{F_0}{k}\frac{1}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2}(\sin \left(\pi \frac{t}{t_1}\right)-\frac{\pi /t_1}{\omega }\sin \omega t)\end{array}$$

The above Eq (1) gives solution during the time $0\le t\le t_1$

Now after $t=t_1$ the force will disappear, the differential equation becomes$$m\ddot{u}+ku=0t>t_1$$

but with the initial conditions evaluate at $t=t_1.$ From (1)$$\begin{array}{rl}u\left(t_1\right)& =u_0\cos \omega t_1+(\frac{v_0}{\omega }-u_{st}\frac{r}{1-r^2})\sin \omega t_1+u_{st}\frac{1}{1-r^2}\sin \varpi t_1\\ \text{(2)}& & =u_0\cos \omega t_1+(\frac{v_0}{\omega }-u_{st}\frac{r}{1-r^2})u_{st}\frac{r}{1-r^2}\sin \omega t_1\end{array}$$

since $\sin \varpi t_1=0$. taking derivative of Eq (1)$$\dot{u}\left(t\right)=-\omega u_0\sin \omega t+\omega (\frac{v_0}{\omega }-u_{st}\frac{r}{1-r^2})\cos \omega t+\varpi \frac{1}{1-r^2}\cos \varpi t$$

and at $t=t_1$ the above becomes$$\begin{array}{rl}\dot{u}\left(t_1\right)& =-\omega u_0\sin \omega t_1+\omega (\frac{v_0}{\omega }-u_{st}\frac{r}{1-r^2})\cos \omega t_1+\varpi \frac{1}{1-r^2}\cos \varpi t_1\\ \text{(3)}& & =-\omega u_0\sin \omega t_1+\omega (\frac{v_0}{\omega }-u_{st}\frac{r}{1-r^2})\cos \omega t_1-\varpi \frac{1}{1-r^2}\end{array}$$

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 $t\le t_1$ we obtain resonance since $r\to 1$ 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$$\begin{array}{rl}u\left(t\right)& =u_0\cos \omega t+(\frac{v_0}{\omega }-u_{st}\frac{\frac{\varpi }{\omega }}{1-{\left(\frac{\varpi }{\omega }\right)}^2})\sin \omega t+u_{st}\frac{1}{1-{\left(\frac{\varpi }{\omega }\right)}^2}\sin \varpi t\\ \text{(1A)}& & =u_0\cos \omega t+(\frac{v_0}{\omega }-u_{st}\frac{\omega \varpi }{{\omega }^2-{\varpi }^2})\sin \omega t+u_{st}\frac{{\omega }^2}{{\omega }^2-{\varpi }^2}\sin \varpi t\end{array}$$

Now looking at case when $\varpi \approx \omega$ but less than $\omega$, hence let $$\begin{array}{}\text{(2)}& \omega -\varpi =2\mathrm{\Delta }\end{array}$$ where $\mathrm{\Delta }$ is very small positive quantity. and we also have $$\begin{array}{}\text{(3)}& \omega +\varpi \approx 2\varpi \end{array}$$

Multiplying Eq (2) and (3) with each others gives$$\begin{array}{}\text{(4)}& {\omega }^2-{\varpi }^2=4\mathrm{\Delta }\varpi \end{array}$$

Going back to Eq (1A) and rewriting it as$$\begin{array}{rl}u\left(t\right)& =u_0\cos \omega t+(\frac{v_0}{\omega }-u_{st}\frac{\omega \varpi }{4\mathrm{\Delta }\varpi })\sin \omega t+u_{st}\frac{{\omega }^2}{4\mathrm{\Delta }\varpi }\sin \varpi t\\ & =u_0\cos \omega t+(\frac{v_0}{\omega }-u_{st}\frac{\omega }{4\mathrm{\Delta }})\sin \omega t+u_{st}\frac{{\omega }^2}{4\mathrm{\Delta }\varpi }\sin \varpi t\end{array}$$

Since $\varpi \approx \omega$ the above becomes$$\begin{array}{rl}u\left(t\right)& =u_0\cos \omega t+(\frac{v_0}{\omega }-u_{st}\frac{\omega }{4\mathrm{\Delta }})\sin \omega t+u_{st}\frac{\omega }{4\mathrm{\Delta }}\sin \varpi t\\ & =u_0\cos \omega t+\frac{v_0}{\omega }\sin \omega t+u_{st}\frac{\omega }{4\mathrm{\Delta }}(\sin \varpi t-\sin \omega t)\end{array}$$

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$$u\left(t\right)=u_0\cos \omega t+\frac{v_0}{\omega }\sin \omega t+u_{st}\frac{\omega }{2\mathrm{\Delta }}(\sin \left(\frac{\varpi -\omega }{2}t\right)\cos \left(\frac{\varpi +\omega }{2}t\right))$$

From Eq(2) $\varpi -\omega =-2\mathrm{\Delta }$ and $\omega +\varpi \approx 2\varpi$ hence the above becomes$$u\left(t\right)=u_0\cos \omega t+\frac{v_0}{\omega }\sin \omega t+u_{st}\frac{\omega }{2\mathrm{\Delta }}(\sin (-\mathrm{\Delta }t)\cos \left(\varpi t\right))$$

or since $\varpi \approx \omega$$$u\left(t\right)=u_0\cos \omega t+\frac{v_0}{\omega }\sin \omega t-u_{st}\frac{\omega }{2\mathrm{\Delta }}(\sin \left(\mathrm{\Delta }t\right)\cos \left(\omega t\right))$$

Now $\underset{\mathrm{\Delta }\to 0}{lim}\frac{\sin \left(\mathrm{\Delta }t\right)}{\mathrm{\Delta }}=t$ 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$$\begin{array}{rl}\text{(1)}& u\left(t\right)& =u_0\cos \varpi t+\frac{v_0}{\varpi }\sin \varpi t-u_{st}\frac{\varpi t}{2}\cos \left(\varpi t\right)& =u_0\cos \left(\frac{\pi }{t_1}t\right)+\frac{v_0}{\varpi }\sin \left(\frac{\pi }{t_1}t\right)-u_{st}\left(\frac{\pi }{2t_1}t\right)\cos \left(\frac{\pi }{t_1}t\right)\end{array}$$

since $\varpi \approx \omega$ in this case. This is the solution to use for resonance and for $t\le t_1$

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$$$\begin{array}{rl}\dot{u}\left(t_1\right)& =-\varpi u_0\sin \varpi t_1+v_0\cos \varpi t_1+u_{st}\frac{{\varpi }^2{t}_1}{2}\sin \left(\varpi t_1\right)-u_{st}\frac{\varpi }{2}\cos \left(\varpi t_1\right)\\ & =-v_0+u_{st}\frac{\varpi }{2}\end{array}$$

Now the solution for $t>t_1$ is$$\begin{array}{rl}u\left(t\right)& =u\left(t_1\right)\cos \omega t+\frac{\dot{u}\left(t_1\right)}{\omega }\sin \omega t\\ & =(-u\left(0\right)+u_{st}\frac{\pi }{2})\cos \omega t+\frac{-u^{\prime }\left(0\right)+u_{st}\frac{\pi }{2t_1}}{\omega }\sin \omega t\end{array}$$

1.4.5.2.2 under-damped with sin impulse $c<c_r,\xi <1$ $$m\ddot{u}+c\dot{u}+ku=\{\begin{array}{ccc}{F}_0\sin \left(\varpi \right)& & 0\le t\le t_1\\ 0& & t>t_1\end{array}$$

or$$\ddot{u}+2\xi \omega \dot{u}+{\omega }^{2}u=\{\begin{array}{ccc}{F}_0\sin \left(\varpi \right)& & 0\le t\le t_1\\ 0& & t>t_1\end{array}$$

$$m\ddot{u}+c\dot{u}+ku=F\sin \varpi t$$

$$\ddot{u}+2\xi \omega \dot{u}+{\omega }^{2}u=\frac{F}{m}\sin \varpi t$$

For $t\le t_1$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$$\begin{array}{}\text{(1)}& u\left(t\right)=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t-\theta )\end{array}$$

Applying initial conditions gives$$\begin{array}{rl}A& =u_0+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin \theta \\ B& =\frac{v_0}{{\omega }_d}+\frac{u_0\xi \omega }{{\omega }_d}+\frac{u_{st}}{{\omega }_d\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}(\xi \omega \sin \theta -\varpi \cos \theta )\end{array}$$

For $t>t_1$. From (1)$$\begin{array}{}\text{(2)}& u\left(t_1\right)=e^{-\xi \omega t_1}(A\cos {\omega }_d{t}_1+B\sin {\omega }_d{t}_1)+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t_1-\theta )\end{array}$$

Taking derivative of (1) gives$$\begin{array}{c}\dot{u}\left(t\right)=-\xi \omega e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+e^{-\xi \omega t}(-A{\omega }_d\sin {\omega }_{d}t+{\omega }_{d}B\cos {\omega }_{d}t)\hfill \\ \hfill +\varpi \frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\cos (\varpi t-\theta )\end{array}$$

at $t=t_1$$$\begin{array}{c}\dot{u}\left(t_1\right)=-\xi \omega e^{-\xi \omega t_1}(A\cos {\omega }_d{t}_1+B\sin {\omega }_d{t}_1)+e^{-\xi \omega t_1}(-A{\omega }_d\sin {\omega }_d{t}_1+{\omega }_{d}B\cos {\omega }_d{t}_1)\hfill \\ \hfill +\varpi \frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\cos (\varpi t_1-\theta )\end{array}$$

Now for $t>t_1$ the equation becomes$$m\ddot{u}+c\dot{u}+ku=0$$

which has the solution $$u=e^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$$

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 $t\le t_1$Initial conditions are $u\left(0\right)=u_0$ and $\dot{u}\left(0\right)=v_0$ then the solution is from above$$\begin{array}{}\text{(1)}& u\left(t\right)=(A+Bt)e^{-\omega t}+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t-\theta )\end{array}$$

Where $\tan \theta =\frac{c\varpi }{k-m{\varpi }^2}=\frac{2\xi r}{1-r^2}.$ $A,B$ are found from initial conditions$$\begin{array}{rl}A& =u_0+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin \theta \\ B& =v_0+u_0\omega +\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}(\omega \sin \theta -\varpi \cos \theta )\end{array}$$

For $t>t_1$ the solution is$$\begin{array}{}\text{(2)}& u\left(t\right)=(u\left(t_1\right)+(\dot{u}\left(t_1\right)+u\left(t_1\right)\omega )t)e^{-\omega t}\end{array}$$

To find $u\left(t_1\right),$ from Eq(1)$$u\left(t_1\right)=(A+Bt)e^{-\omega t_1}+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t_1-\theta )$$

taking derivative of (1) gives$$\begin{array}{}\text{(3)}& \dot{u}\left(t\right)=-\omega (A+Bt)e^{-\omega t}+B{e}^{-\omega t}+\varpi \frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t-\theta )\end{array}$$

at $t=t_1$$$\begin{array}{}\text{(4)}& \dot{u}\left(t_1\right)=-\omega (A+B{t}_1)e^{-\omega t_1}+B{e}^{-\omega t_1}+\varpi \frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\varpi t_1-\theta )\end{array}$$

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 $t\le t_1$Initial conditions are $u\left(0\right)=u_0$ and $\dot{u}\left(0\right)=v_0$ then the solution is$$u=A{e}^{p_{1}t}+B{e}^{p_{2}t}+\frac{u_{st}}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\varpi t-\theta )$$

where $\tan \theta =\frac{2\xi r}{1-r^2}$(make sure you use correct quadrant, see not above on $\arctan$) and$$\begin{array}{rl}{p}_1& =-\frac{c}{2m}+\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}\\ & =-\omega \xi +\omega \sqrt{{\xi }^2-1}\end{array}$$

and$$\begin{array}{rl}{p}_2& =-\frac{c}{2m}-\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}\\ & =-\omega \xi -\omega \sqrt{{\xi }^2-1}\end{array}$$

leading to the solution where $\tan \theta =\frac{2\xi r}{1-r^2}$ and$$\begin{array}{rl}{p}_1& =-\frac{c}{2m}+\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\omega \xi +{\omega }_n\sqrt{{\xi }^2-1}\\ p_2& =-\frac{c}{2m}-\sqrt{{\left(\frac{c}{2m}\right)}^2-\frac{k}{m}}=-\omega \xi -{\omega }_n\sqrt{{\xi }^2-1}\end{array}$$

is$$\begin{array}{rl}u\left(t\right)& =A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t}+B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t}+\frac{F}{k}\beta \sin (\varpi t-\theta )\\ A& =\frac{u^{\prime }\left(0\right)+u\left(0\right)\omega \xi +u\left(0\right)\omega \sqrt{{\xi }^2-1}+\frac{F}{k}\beta ((\xi +\sqrt{{\xi }^2-1})\omega \sin \theta -\varpi \cos \theta )}{2\omega \sqrt{{\xi }^2-1}}\\ B& =-\frac{u^{\prime }\left(0\right)+u\left(0\right)\omega \xi -u\left(0\right)\omega \sqrt{{\xi }^2-1}+\frac{F}{k}\beta ((\xi -\sqrt{{\xi }^2-1})\omega \sin \theta -\varpi \cos \theta )}{2\omega \sqrt{{\xi }^2-1}}\\ \beta & =\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\end{array}$$

For $t>t_1$. From Eq(1) and at $t=t_1$$$\begin{array}{}\text{(2)}& u\left(t_1\right)=A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t_1}+B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t_1}+D\sin (\varpi t_1-\theta )\end{array}$$

Taking derivative of Eq (1)$$\dot{u}\left(t\right)=\omega A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t}+\omega B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t}+\varpi D\cos (\varpi t-\theta )$$

At $t=t_1$$$\begin{array}{}\text{(3)}& \dot{u}\left(t_1\right)=\omega A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t_1}+\omega B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t_1}+\varpi D\cos (\varpi t_1-\theta )\end{array}$$

Equation of motion now is$$\ddot{u}+2\xi \omega \dot{u}+{\omega }^{2}u=0$$

which has solution for over-damped given by$$u\left(t\right)=A{e}^{(-\xi +\sqrt{{\xi }^2-1}){\omega }_{n}t}+B{e}^{(-\xi -\sqrt{{\xi }^2-1}){\omega }_{n}t}$$

where$$\begin{array}{rl}A& =-\frac{\dot{u}\left(t_1\right)+u\left(t_1\right){\omega }_n(\xi -\sqrt{{\xi }^2-1})}{2{\omega }_n\sqrt{{\xi }^2-1}}\\ B& =\frac{\dot{u}\left(t_1\right)+u\left(t_1\right)\xi {\omega }_n(\xi +\sqrt{{\xi }^2-1})}{2{\omega }_n\sqrt{{\xi }^2-1}}\end{array}$$

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}]
 

1.4.5.2.5 Summary table

$\zeta =0$	roots $\{\begin{array}{l}-i\omega \\ +i\omega \end{array}$ $u\left(t\right)\{\begin{array}{l}\varpi =\omega \to \{\begin{array}{ll}u\left(0\right)\cos \omega t+\frac{u^{\prime }\left(0\right)}{\omega }\sin \omega t-\frac{F_0}{k}\frac{\omega t}{2}\cos \left(\omega t\right)& 0\le t\le t_1\\ (-u\left(0\right)+\frac{F_0}{k}\frac{\pi }{2})\cos \omega t+\frac{-u^{\prime }\left(0\right)+\frac{F_0}{k}\frac{\pi }{2t_1}}{\omega }\sin \omega t& t>t_1\end{array}\\ \varpi \ne \omega \to \{\begin{array}{ll}u\left(0\right)\cos \omega t+(\frac{u^{\prime }\left(0\right)}{\varpi }-\frac{\left(F_0/k\right)\left(\frac{\pi /t_1}{\omega }\right)}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2})\sin \omega t+\frac{F_0/k}{1-{\left(\frac{\pi /t_1}{\omega }\right)}^2}\sin \left(\frac{\pi t}{t_1}\right)& 0\le t\le t_1\\ u\left(t_1\right)\cos \omega t+\frac{u^{\prime }\left(t_1\right)}{\omega }\sin \omega t& t>t_1\end{array}\end{array}$
$\zeta <1$	roots $\{\begin{array}{l}-\xi \omega +i{\omega }_n\sqrt{1-{\xi }^2}\\ -\xi \omega -i{\omega }_n\sqrt{1-{\xi }^2}\end{array}$ time constant $\tau =\frac{1}{\zeta {\omega }_n}$ $u\left(t\right)=\{\begin{array}{ll}{e}^{-\xi \omega t}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)+\frac{F_0}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin (\pi \frac{t}{t_1}-\theta )& 0\le t\le t_1\\ e^{-\xi \omega t}(u\left(t_1\right)\cos {\omega }_{d}t+\frac{u^{\prime }\left(t_1\right)+u\left(t_1\right)\xi \omega }{{\omega }_d}\sin {\omega }_{d}t)& t>t_1\end{array}$ $A=u\left(0\right)+\frac{F_0}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}\sin \theta$ $B=\frac{u^{\prime }\left(0\right)}{{\omega }_d}+\frac{u\left(0\right)\xi \omega }{{\omega }_d}+\frac{F_0}{k}\frac{1}{{\omega }_d\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}(\xi \omega \sin \theta -\varpi \cos \theta )$
$\zeta =1$	roots $\{\begin{array}{l}-\omega \\ -\omega \end{array}$ $u\left(t\right)=\{\begin{array}{ll}(A+Bt)e^{-\omega t}+\frac{F_0}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin (\pi \frac{t}{t_1}-\theta )& 0\le t\le t_1\\ u\left(t\right)=(u\left(t_1\right)+(u^{\prime }\left(t_1\right)+u\left(t_1\right)\omega )t)e^{-\omega t}& t>t_1\end{array}$ $A=u\left(0\right)+\frac{F_0}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}\sin \theta$ $B=u^{\prime }\left(0\right)+u\left(0\right)\omega +\frac{F_0}{k}\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2r\right)}^2}}(\omega \sin \theta -\frac{\pi }{t_1}\cos \theta )$
$\zeta >1$	roots $\{\begin{array}{l}-{\omega }_n\xi +{\omega }_n\sqrt{{\xi }^2-1}\\ -{\omega }_n\xi -{\omega }_n\sqrt{{\xi }^2-1}\end{array}$ $u\left(t\right)=\{\begin{array}{ll}A{e}^{(-\xi +\sqrt{{\xi }^2-1})\omega t}+B{e}^{(-\xi -\sqrt{{\xi }^2-1})\omega t}+\frac{F}{k}\beta \sin (\pi \frac{t}{t_1}-\theta )& 0\le t\le t_1\\ u\left(t\right)=A_1{e}^{(-\xi +\sqrt{{\xi }^2-1}){\omega }_{n}t}+B_1{e}^{(-\xi -\sqrt{{\xi }^2-1}){\omega }_{n}t}& t>t_1\end{array}$ $A=\frac{u^{\prime }\left(0\right)+u\left(0\right)\omega \xi +u\left(0\right)\omega \sqrt{{\xi }^2-1}+\frac{F}{k}\beta ((\xi +\sqrt{{\xi }^2-1})\omega \sin \theta -\frac{\pi }{t_1}\cos \theta )}{2\omega \sqrt{{\xi }^2-1}}$ $B=-\frac{u^{\prime }\left(0\right)+u\left(0\right)\omega \xi -u\left(0\right)\omega \sqrt{{\xi }^2-1}+\frac{F}{k}\beta ((\xi -\sqrt{{\xi }^2-1})\omega \sin \theta -\frac{\pi }{t_1}\cos \theta )}{2\omega \sqrt{{\xi }^2-1}}$ $\beta =\frac{1}{\sqrt{{(1-r^2)}^2+{\left(2\xi r\right)}^2}}$ $A_1=-\frac{\dot{u}\left(t_1\right)+u\left(t_1\right){\omega }_n(\xi -\sqrt{{\xi }^2-1})}{2{\omega }_n\sqrt{{\xi }^2-1}}$ $B_1=\frac{\dot{u}\left(t_1\right)+u\left(t_1\right)\xi {\omega }_n(\xi +\sqrt{{\xi }^2-1})}{2{\omega }_n\sqrt{{\xi }^2-1}}$

1.4.6 Tree view look at the different cases

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.

Figure 1.13: single degree system tree

The following is another diagram made sometime ago which contains more useful information and is kept here for reference.

Figure 1.14: one DOF system

1.4.7 Cycles for the peak to decay by half its original value

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 $\zeta =2.7\mathrm{\%}$ 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] (%)"}}]
 

Figure 1.15: peak table

1.4.8 references

1. Vibration analysis by Robert K. Vierck
2. Structural dynamics theory and computation, 5th edition by Mario Paz, William Leigh
3. Dynamic of structures, Ray W. Clough and Joseph Penzien
4. Theory of vibration,volume 1, by A.A.Shabana
5. Notes on Diffy Qs, Differential equations for engineers, by Jiri Lebl, online PDF book, chapter 2.6, oct 1,2012 http://www.jirka.org/diffyqs/

¹http://en.wikipedia.org/wiki/Fundamental\_equation\_of\_constrained\_motion