## Using Simulink to analyze two degrees of freedom system

Spring 2009   Compiled on May 26, 2022 at 1:34am

### Abstract

A two degrees of freedom system consisting of two masses connected by springs and subject to 3 diﬀerent type of input forces is analyzed and simulated using Simulink

### 1 Introduction and Theory

The system that is being analyzed is show in the following diagram

In the above, $$F\left ( t\right )$$ is to be taken as each of the following

1. Unit impulse force.
2. Unit step force.
3. $$\sin \omega t$$

It is required to ﬁnd $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ analytically and then to use Matlab’s Simulink software for the analysis.

The mathematical model of the system is ﬁrst developed and the equation of motions obtained using Lagrangian formulation then the analytical solution is found by solving the resulting coupled second order diﬀerential equations for $$m_{1}$$ and $$m_{2}$$. Next, a simulink model is developed to implement the diﬀerential equations and the output $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ from Simulink is shown and compared to the output from the analytical solution.

### 2 Analytical solution

The following is the free body diagram of the above system

Assuming positive is downwards and that $$x_{2}>x_{1}$$, force-balance equations for $$m_{1}$$ results in$m_{1}\ddot {x}_{1}=F\left ( t\right ) +k_{1}\left ( x_{2}-x_{1}\right )$ And force-balance equations for $$m_{2}$$ results in$m_{2}\ddot {x}_{2}=-k_{1}\left ( x_{2}-x_{1}\right ) -k_{2}x_{2}$ Hence the EQM for the system become\begin {align*} m_{1}\ddot {x}_{1}-k_{1}x_{2}+k_{1}x_{1} & =F\left ( t\right ) \\ m_{2}\ddot {x}_{2}+\left ( k_{1}+k_{2}\right ) x_{2}-k_{1}x_{1} & =0 \end {align*}

Or in matrix form$\begin {pmatrix} m_{1} & 0\\ 0 & m_{2}\end {pmatrix}\begin {pmatrix} \ddot {x}_{1}\\ \ddot {x}_{2}\end {pmatrix} +\begin {pmatrix} k_{1} & -k_{1}\\ -k_{1} & \left ( k_{1}+k_{2}\right ) \end {pmatrix}\begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} =\begin {pmatrix} F\left ( t\right ) \\ 0 \end {pmatrix}$ The above can be written in matrix form as$M\mathbf {\ddot {x}}+K\mathbf {x}=\mathbf {F}$ Where $$\mathbf {\ddot {x},x,F\,\ }$$are 2 by 1 vectors and $$M$$ and $$K$$ are the mass and stiﬀness matrices. The solution to the above is \begin {equation} \mathbf {x}=\mathbf {x}_{h}+\mathbf {x}_{p} \tag {1} \end {equation}

#### 2.1 Finding the homogenous solution

We start by ﬁnding $$\mathbf {x}_{h}$$ from the following$\begin {pmatrix} m_{1} & 0\\ 0 & m_{2}\end {pmatrix}\begin {pmatrix} \ddot {x}_{1}\\ \ddot {x}_{2}\end {pmatrix} +\begin {pmatrix} k_{1} & -k_{1}\\ -k_{1} & \left ( k_{1}+k_{2}\right ) \end {pmatrix}\begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} =\begin {pmatrix} 0\\ 0 \end {pmatrix}$ Assuming that $$x_{1}\left ( t\right ) =A_{1}\cos \left ( \omega t+\phi \right )$$ and $$x_{2}\left ( t\right ) =A_{2}\cos \left ( \omega t+\phi \right )$$, hence $$\dot {x}_{1}\left ( t\right ) =-\omega A_{1}\sin \left ( \omega t+\phi \right )$$ and $$\dot {x}_{2}\left ( t\right ) =-\omega A_{2}\sin \left ( \omega t+\phi \right )$$ and $$\ddot {x}_{1}\left ( t\right ) =-\omega ^{2}A_{1}\cos \left ( \omega t+\phi \right )$$ and $$\dot {x}_{2}\left ( t\right ) =-\omega ^{2}A_{2}\cos \left ( \omega t+\phi \right )$$. Substituting the above values in the above system results in$\begin {pmatrix} m_{1} & 0\\ 0 & m_{2}\end {pmatrix}\begin {pmatrix} -\omega ^{2}A_{1}\cos \left ( \omega t+\phi \right ) \\ -\omega ^{2}A_{2}\cos \left ( \omega t+\phi \right ) \end {pmatrix} +\begin {pmatrix} k_{1} & -k_{1}\\ -k_{1} & \left ( k_{1}+k_{2}\right ) \end {pmatrix}\begin {pmatrix} A_{1}\cos \left ( \omega t+\phi \right ) \\ A_{2}\cos \left ( \omega t+\phi \right ) \end {pmatrix} =\begin {pmatrix} 0\\ 0 \end {pmatrix}$ Divide by $$\cos \left ( \omega t+\phi \right )$$ since not zero (else no solution exist) we obtain$\begin {pmatrix} m_{1} & 0\\ 0 & m_{2}\end {pmatrix}\begin {pmatrix} -\omega ^{2}A_{1}\\ -\omega ^{2}A_{2}\end {pmatrix} +\begin {pmatrix} k_{1} & -k_{1}\\ -k_{1} & \left ( k_{1}+k_{2}\right ) \end {pmatrix}\begin {pmatrix} A_{1}\\ A_{2}\end {pmatrix} =\begin {pmatrix} 0\\ 0 \end {pmatrix}$ Rewrite the above as\begin {align} \begin {pmatrix} -\omega ^{2}m_{1}A_{1}\\ -\omega ^{2}m_{2}A_{2}\end {pmatrix} +\begin {pmatrix} k_{1}A_{1}-k_{1}A_{2}\\ -k_{1}A_{1}+\left ( k_{1}+k_{2}\right ) A_{2}\end {pmatrix} & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \nonumber \\\begin {pmatrix} -\omega ^{2}m_{1}A_{1}+k_{1}A_{1}-k_{1}A_{2}\\ -\omega ^{2}m_{2}A_{2}-k_{1}A_{1}+\left ( k_{1}+k_{2}\right ) A_{2}\end {pmatrix} & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \nonumber \\\begin {pmatrix} \left ( -\omega ^{2}m_{1}+k_{1}\right ) A_{1}-k_{1}A_{2}\\ -\omega ^{2}m_{2}A_{2}+\left ( k_{1}+k_{2}\right ) A_{2}-k_{1}A_{1}\end {pmatrix} & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \nonumber \\\begin {pmatrix} -\omega ^{2}m_{1}+k_{1} & -k_{1}\\ -k_{1} & -\omega ^{2}m_{2}+\left ( k_{1}+k_{2}\right ) \end {pmatrix}\begin {pmatrix} A_{1}\\ A_{2}\end {pmatrix} & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \tag {2} \end {align}

From the last equation above, we see that to obtain a solution we must have $\begin {vmatrix} -\omega ^{2}m_{1}+k_{1} & -k_{1}\\ -k_{1} & -\omega ^{2}m_{2}+\left ( k_{1}+k_{2}\right ) \end {vmatrix} =0$ since if we had $$\begin {pmatrix} A_{1}\\ A_{2}\end {pmatrix} =0$$ then no solution will exist. Therefore, taking the determinant and setting it to zero results in

\begin {align*} \left ( -\omega ^{2}m_{1}+k_{1}\right ) \left ( -\omega ^{2}m_{2}+\left ( k_{1}+k_{2}\right ) \right ) -k_{1}^{2} & =0\\ \allowbreak k_{1}^{2}+k_{1}k_{2}-\omega ^{2}k_{1}m_{1}-\omega ^{2}k_{1}m_{2}-\omega ^{2}k_{2}m_{1}+\omega ^{4}m_{1}m_{2}-k_{1}^{2} & =0\\ \allowbreak \omega ^{4}m_{1}m_{2}+\omega ^{2}\left ( -k_{1}m_{1}-k_{2}m_{1}-k_{1}m_{2}\right ) +\left ( k_{1}^{2}+k_{1}k_{2}-k_{1}^{2}\right ) & =0 \end {align*}

Let $$\omega ^{4}=\lambda ^{2}$$, hence the above becomes$\lambda ^{2}m_{1}m_{2}+\lambda \left ( -k_{1}m_{1}-k_{2}m_{1}-k_{1}m_{2}\right ) +\left ( k_{1}^{2}+k_{1}k_{2}-k_{1}^{2}\right ) =0$ Solving for $$\lambda$$ gives\begin {equation} \begin {array} [c]{c}\lambda _{1}=\frac {1}{2m_{1}m_{2}}\left ( k_{1}m_{1}+k_{1}m_{2}+k_{2}m_{1}+\sqrt {k_{1}^{2}m_{1}^{2}+k_{1}^{2}m_{2}^{2}+k_{2}^{2}m_{1}^{2}+2k_{1}k_{2}m_{1}^{2}+2k_{1}^{2}m_{1}m_{2}-2k_{1}k_{2}m_{1}m_{2}}\right ) \\ \lambda _{2}=\frac {1}{2m_{1}m_{2}}\left ( k_{1}m_{1}+k_{1}m_{2}+k_{2}m_{1}-\sqrt {k_{1}^{2}m_{1}^{2}+k_{1}^{2}m_{2}^{2}+k_{2}^{2}m_{1}^{2}+2k_{1}k_{2}m_{1}^{2}+2k_{1}^{2}m_{1}m_{2}-2k_{1}k_{2}m_{1}m_{2}}\right ) \end {array} \tag {3} \end {equation} For each of the above solutions, we obtain a diﬀerent $$\begin {pmatrix} A_{1}\\ A_{2}\end {pmatrix}$$ from equation (2) as follows

For $$\lambda _{1}$$, (2) becomes\begin {align*} \begin {pmatrix} -\lambda _{1}m_{1}+k_{1} & -k_{1}\\ -k_{1} & -\lambda _{1}m_{2}+\left ( k_{1}+k_{2}\right ) \end {pmatrix}\begin {pmatrix} A_{1}\\ A_{2}\end {pmatrix} & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \\\begin {pmatrix} \left ( -\lambda _{1}m_{1}+k_{1}\right ) A_{1}-k_{1}A_{2}\\ -k_{1}A_{1}-\lambda _{1}m_{2}A_{2}+\left ( k_{1}+k_{2}\right ) A_{2}\end {pmatrix} & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \end {align*}

From the ﬁrst equation above, we have\begin {align*} \left ( -\lambda _{1}m_{1}+k_{1}\right ) A_{1}-k_{1}A_{2} & =0\\ \frac {-\lambda _{1}m_{1}+k_{1}}{k_{1}} & =\left ( \frac {A_{2}}{A_{1}}\right ) ^{\left ( 1\right ) } \end {align*}

Similarly for $$\lambda _{2}$$, $\frac {-\lambda _{2}m_{1}+k_{1}}{k_{1}}=\left ( \frac {A_{2}}{A_{1}}\right ) ^{\left ( 2\right ) }$ Let\begin {align} r_{1} & =\frac {-\lambda _{1}m_{1}+k_{1}}{k_{1}}=\left ( \frac {A_{2}}{A_{1}}\right ) ^{\left ( 1\right ) }\tag {4}\\ r_{2} & =\frac {-\lambda _{2}m_{1}+k_{1}}{k_{1}}=\left ( \frac {A_{2}}{A_{1}}\right ) ^{\left ( 2\right ) }\nonumber \end {align}

Hence now $$\mathbf {x}_{h}$$ can be written as$\begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} _{h}=\begin {pmatrix} A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \\ A_{2}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +A_{2}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \end {pmatrix}$ But $$A_{2}^{\left ( 1\right ) }=r_{1}A_{1}^{\left ( 1\right ) }$$ and $$A_{2}^{\left ( 2\right ) }=r_{2}A_{1}^{\left ( 2\right ) }$$, hence the above becomes\begin {equation} \begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} _{h}=\begin {pmatrix} A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \\ r_{1}A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +r_{2}A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \end {pmatrix} \tag {5} \end {equation} Now, given numerical values for $$k_{1},k_{2},m_{1},m_{2}$$ we can ﬁnd $$\omega _{1},\omega _{2}$$ from (3) above, and next ﬁnd $$r_{1},r_{2}$$ from (4).  Hence (5) contains 4 unknowns, $$A_{1}^{\left ( 1\right ) },A_{1}^{\left ( 2\right ) },\phi _{1},\phi _{2}$$ which now can be found from initial conditions (after we ﬁnd the particular solution) which we will now proceed to do.

#### 2.2 Finding particular solutions

There are 3 diﬀerent $$F\left ( t\right )$$ which we are asked to consider

1. Unit impulse force.
2. Unit step force.
3. $$\sin \omega t$$

For each of the above, we ﬁnd $$\mathbf {x}_{p}$$ and then add it to $$\mathbf {x}_{h}$$ found above in (5) to obtain (1).

##### 2.2.1 Finding the particular solution for unit impulse input

Using the standard response for a unit impulse which for a single degree of freedom system is $$x\left ( t\right ) =\frac {1}{m\omega _{n}}\sin \omega _{n}t$$, then we write $$\mathbf {x}_{p}$$ as$\mathbf {x}_{p}=\begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} _{p}=\begin {pmatrix} \frac {1}{m\omega _{1}}\sin \omega _{1}t+\frac {1}{m\omega _{2}}\sin \omega _{2}t\\ 0 \end {pmatrix}$ Hence, the general solution becomes\begin {equation} \begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} =\begin {pmatrix} A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \\ r_{1}A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +r_{2}A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \end {pmatrix} +\begin {pmatrix} \frac {1}{m\omega _{1}}\sin \omega _{1}t+\frac {1}{m\omega _{2}}\sin \omega _{2}t\\ 0 \end {pmatrix} \tag {6} \end {equation}

##### 2.2.2 Finding the particular solution for unit step input

Since unit step is $$1$$ for $$t>0$$, then, using convolution we write\begin {align*} x_{p}\left ( t\right ) & =\int _{0}^{t}f\left ( \tau \right ) h\left ( t-\tau \right ) d\tau \\ & =\int _{0}^{t}h\left ( t-\tau \right ) d\tau \\ & =\int _{0}^{t}\frac {1}{m\omega _{n}}\sin \omega _{n}\left ( t-\tau \right ) d\tau \\ & =\frac {1}{m\omega _{n}}\left [ \frac {-\cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) }{-\omega _{n}}\right ] _{0}^{t}\\ & =\frac {1}{m\omega _{n}^{2}}\left [ \cos \left ( \omega _{n}\left ( t-\tau \right ) \right ) \right ] _{0}^{t}\\ & =\frac {1}{m\omega _{n}^{2}}\left [ 1-\cos \left ( \omega _{n}t\right ) \right ] \end {align*}

Then, since now we have 2 natural frequencies, we can write $$\mathbf {x}_{p}$$ as$\mathbf {x}_{p}=\begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} _{p}=\begin {pmatrix} \frac {1}{m\omega _{1}^{2}}\left [ 1-\cos \left ( \omega _{1}t\right ) \right ] +\frac {1}{m\omega _{2}^{2}}\left [ 1-\cos \left ( \omega _{2}t\right ) \right ] \\ 0 \end {pmatrix}$ Hence, the general solution becomes\begin {equation} \begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} =\begin {pmatrix} A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \\ r_{1}A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +r_{2}A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \end {pmatrix} +\begin {pmatrix} \frac {1}{m\omega _{1}^{2}}\left [ 1-\cos \left ( \omega _{1}t\right ) \right ] +\frac {1}{m\omega _{2}^{2}}\left [ 1-\cos \left ( \omega _{2}t\right ) \right ] \\ 0 \end {pmatrix} \nonumber \end {equation}

##### 2.2.3 Finding the particular solution for $$\sin \omega t$$

In this case, we guess that $$x_{1p}=c_{1}\cos \omega t+c_{2}\sin \omega t$$, and since there is no forcing function being applied directly on $$m_{2}$$ then $$x_{2p}=0$$ hence$\mathbf {x}_{p}=\begin {pmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {pmatrix} _{p}=\begin {pmatrix} c_{1}\cos \omega t+c_{2}\sin \omega t\\ 0 \end {pmatrix}$ Then $$\dot {x}_{1p}\left ( t\right ) =-\omega c_{1}\sin \omega t+\omega c_{2}\cos \omega t$$ and $$\ddot {x}_{1p}\left ( t\right ) =-\omega ^{2}c_{1}\cos \omega t+\omega ^{2}c_{2}\sin \omega t$$ and now we substitute these into the original ODE for $$x_{1}$$ which is\begin {align*} m_{1}\ddot {x}_{1}-k_{1}x_{2}+k_{1}x_{1} & =F\left ( t\right ) \\ m_{1}\ddot {x}_{1p}-k_{1}x_{2p}+k_{1}x_{1p} & =\sin \omega t \end {align*}

We obtain the following\begin {align*} m_{1}\left ( -\omega ^{2}c_{1}\cos \omega t+\omega ^{2}c_{2}\sin \omega t\right ) -k_{1}\left ( 0\right ) +k_{1}\left ( c_{1}\cos \omega t+c_{2}\sin \omega t\right ) & =\sin \omega t\\ \cos \omega t\left ( -\omega ^{2}c_{1}m_{1}+k_{1}c_{1}\right ) +\sin \omega t\left ( m_{1}\omega ^{2}c_{2}+k_{1}c_{2}\right ) & =\sin \omega t \end {align*}

Hence by comparing coeﬃcients, we obtain\begin {align*} -\omega ^{2}c_{1}m_{1}+k_{1}c_{1} & =0\\ m_{1}\omega ^{2}c_{2}+k_{1}c_{2} & =1 \end {align*}

or\begin {align*} c_{1}\left ( -\omega ^{2}m_{1}+k_{1}\right ) & =0\\ c_{2}\left ( m_{1}\omega ^{2}+k_{1}\right ) & =1 \end {align*}

$$c_{1}$$ must be zero since $$k_{1}-\omega ^{2}m_{1}=0$$ only when $$\omega =\omega _{n}$$ and we assume that this is not the case here. Hence\begin {align*} c_{1} & =0\\ c_{2} & =\frac {1}{\left ( m_{1}\omega ^{2}+k_{1}\right ) } \end {align*}

Therefore $$\mathbf {x}_{p}$$ becomes$\mathbf {x}_{p}=\begin {pmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {pmatrix} _{p}=\begin {pmatrix} \frac {1}{\left ( m_{1}\omega ^{2}+k_{1}\right ) }\sin \omega t\\ 0 \end {pmatrix}$ And the general solution becomes\begin {equation} \begin {pmatrix} x_{1}\\ x_{2}\end {pmatrix} =\begin {pmatrix} A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \\ r_{1}A_{1}^{\left ( 1\right ) }\cos \left ( \omega _{1}t+\phi _{1}\right ) +r_{2}A_{1}^{\left ( 2\right ) }\cos \left ( \omega _{2}t+\phi _{2}\right ) \end {pmatrix} +\begin {pmatrix} \frac {1}{\left ( m_{1}\omega ^{2}+k_{1}\right ) }\sin \omega t\\ 0 \end {pmatrix} \tag {8} \end {equation}

In simulink, we will directly solve the system from the original formulation\begin {align*} m_{1}\ddot {x}_{1}-k_{1}x_{2}+k_{1}x_{1} & =F\left ( t\right ) \\ m_{2}\ddot {x}_{2}+\left ( k_{1}+k_{2}\right ) x_{2}-k_{1}x_{1} & =0 \end {align*}

or\begin {align*} \ddot {x}_{1}-\frac {k_{1}}{m_{1}}x_{2}+\frac {k_{1}}{m_{1}}x_{1} & =\frac {F\left ( t\right ) }{m_{1}}\\ \ddot {x}_{2}+\frac {\left ( k_{1}+k_{2}\right ) }{m_{2}}x_{2}-\frac {k_{1}}{m_{2}}x_{1} & =0 \end {align*}

Hence\begin {align*} \ddot {x}_{1} & =\frac {F\left ( t\right ) }{m_{1}}+\frac {k_{1}}{m_{1}}x_{2}-\frac {k_{1}}{m_{1}}x_{1}\\ \ddot {x}_{2} & =-\frac {\left ( k_{1}+k_{2}\right ) }{m_{2}}x_{2}+\frac {k_{1}}{m_{2}}x_{1} \end {align*}

The simulink block diagram will be as follows for the unit step input

For an initial run with parameters $$m_{1}=m_{2}=k_{1}=k_{2}=1$$ I get this warning below

EDU>> simulink
Warning: Using a default value of 0.2 for maximum step size.  The simulation
step size will be equal to or less than this value.  You can disable this
diagnostic by setting 'Automatic solver parameter selection' diagnostic to
'none' in the Diagnostics page of the configuration parameters dialog.



And this is the output for $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ for the unit step response

To verify the above output from Simulink, I solved the same coupled diﬀerential equations for zero initial conditions numerically (using a numerical diﬀerential equation solver) and plotted the solution for $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ and the result matches that shown above by simulink. Here is the code the plot as a result of this veriﬁcation

And the output for $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ is as follows

To verify the above output from Simulink, The same coupled diﬀerential equations were solved numerically for zero initial conditions numerically and the solution plotted for $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ and the result was found to match that shown above by simulink. Here is the code used to do the veriﬁcation.

The simulink block diagram will be as follows for the $$\sin \omega t$$ input

For an initial run with parameters $$m_{1}=m_{2}=k_{1}=k_{2}=1$$ this is the output for $$x_{1}\left ( t\right )$$ and $$x_{2}\left ( t\right )$$ and showing the input signal at the same time

### 5 Discussion

A coupled system of two masses and springs was analyzed using Simulink. The simulation was done for one set of parameters (masses and stiﬀness). Simulink made the simulation of this system under diﬀerent loading conditions easy to do. The 2 masses response were recorded using simulink scope and the signals captured on the same plot to make it easy to compare the response of the ﬁrst mass to the second mass.

The analytical analysis was more time consuming than actually making the simulation in simulink. The ability to easily change diﬀerent sources to the system was useful as well as the ability to change the frequency of the input and immediately see the eﬀect on the response.

This was my ﬁrst project using Simulink, and I can see that this tool will be useful to learn more as it allows one to easily analyze engineering problems.