2.9 HW8

2.9.1 problem description

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2.9.2 problem 1

2.9.2.1 part(a)

Let initial length of the spring (un stretched length) be $L_0$ and when the mass $m$ has moved to the right by an amount $x$ then let the current length be $L_{cur}.$

Therefore the stretch in the spring is$$\mathrm{\Delta }=L_{cur}-L_0$$

Let the height of the bar by $H,$ where $\tan {\theta }_0=\frac{H}{L}$ or $H=L\tan {\theta }_0$

Hence from the above diagram we see that $L_0=\sqrt{H^2+L^2}$ and $L_{cur}=\sqrt{H^2+{(L+x)}^2}$, therefore$$\begin{array}{rl}\mathrm{\Delta }& =\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}\\ {\mathrm{\Delta }}^2& ={(\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2})}^2\end{array}$$

Now we can derive the equation of motion using energy methods.

Let $T$ be the current kinetic energy in the system, and let $V$ be the current potential energy.  This system is one degree of freedom, since we only need one generalized coordinate to determine the position of the mass $m$. This coordinate is $x.$$$\begin{array}{rl}T& =\frac{1}{2}m{\dot{x}}^2\\ V& =\frac{1}{2}k{\mathrm{\Delta }}^2\\ & =\frac{1}{2}k{(\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2})}^2\end{array}$$

Hence the Lagrangian $\mathrm{\Phi }$ is$$\begin{array}{rl}\mathrm{\Phi }& =T-V=T-V\\ & =\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{(\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2})}^2\end{array}$$

Now the equation of motion for coordinate $x$ is (using the standard Lagrangian form)$$\frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}\right)-\frac{\partial \mathrm{\Phi }}{\partial x}=Q_x$$

But $Q_x$, then generalized force, is zero since there is no external force and no damping. Now we just need to evaluate each part of the above expression to obtain the EOM.$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}& =m\dot{x}\\ \frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}\right)& =m\ddot{x}\end{array}$$

and$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial x}& =\frac{\partial }{\partial x}(\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{(\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2})}^2)\\ & =-k(\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2})\frac{1}{2}{(H^2+{(L+x)}^2)}^{\frac{-1}{2}}2(L+x)\\ & =-k\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)(L+x)\end{array}$$

Hence EOM becomes$$\begin{array}{rl}\frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}\right)-\frac{\partial \mathrm{\Phi }}{\partial x}& =0\\ m\ddot{x}+k\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)(L+x)& =0\end{array}$$

2.9.2.2 part(b)

For small $x$ we need to expand $f(x)=k\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)(L+x)$ around $x=0$ in Taylor series and let higher powers of $x$ go to zero.$$f(x)=f(0)+x{f}^{\prime }(0)+\frac{x^2{f}^{\prime \prime }(0)}{2!}+HOT.$$

$$\begin{array}{rl}f(0)& =k\left(\frac{\sqrt{H^2+{(L+0)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+0)}^2}}\right)(L+0)\\ & =k\left(\frac{\sqrt{H^2+L^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+L^2}}\right)L\\ & =0\end{array}$$

and now for $f^{\prime }(0)$$$\begin{array}{rl}{f}^{\prime }(0)& =\frac{d}{dx}f(x{)}_{x=0}\\ & =k\frac{d}{dx}{\left[\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)(L+x)\right]}_{x=0}\\ & =k{((L+x)\frac{d}{dx}\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)+\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)\frac{d}{dx}(L+x))}_{x=0}\\ & =k{((L+x)\left(\sqrt{H^2+L^2}\frac{L+x}{{(H^2+L^2+2Lx+x^2)}^{\frac{3}{2}}}\right)+\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right))}_{x=0}\end{array}$$

Now we evaluate it at $x=0$$$\begin{array}{rl}{f}^{\prime }(0)& =k((L+0)\left(\sqrt{H^2+L^2}\frac{L+0}{{(H^2+L^2+2L0+0)}^{\frac{3}{2}}}\right)+\left(\frac{\sqrt{H^2+{(L+0)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+0)}^2}}\right))\\ & =k(L\left(\sqrt{H^2+L^2}\frac{L}{{(H^2+L^2)}^{\frac{3}{2}}}\right)+\left(\frac{\sqrt{H^2+L^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+L^2}}\right))\\ & =k\left(L^2\left(\frac{{(H^2+L^2)}^{\frac{1}{2}}}{{(H^2+L^2)}^{\frac{3}{2}}}\right)\right)\\ & =k\left(\frac{L^2}{(H^2+L^2)}\right)\end{array}$$

Therefore, EOM of motion becomes (notice we ignored higher order terms, which contains $x^2$ in them)$$m\ddot{x}+(f(0)+x{f}^{\prime }(0))=0$$

Hence the linearized EOM is$$\boxed{m\ddot{x}+k\frac{L^2}{(H^2+L^2)}x=0}$$

Or in terms of ${\theta }_0$ the EOM can be written as$$\begin{array}{rl}m\ddot{x}+k\frac{L^2}{({(L\tan {\theta }_0)}^2+L^2)}x& =0\\ m\ddot{x}+k\frac{1}{1+{\tan }^2{\theta }_0}x& =0\end{array}$$

This is the linearized EOM around $x=0.$ Using numerical values given in the problem $L=1,m=1,k=1000N/m,{\theta }_0=\frac{\pi }{4}$, it becomes$$\begin{array}{rl}\ddot{x}+1000\frac{1}{1+{(\tan \frac{\pi }{4})}^2}x& =0\\ \ddot{x}+500x& =0\end{array}$$

Therefore the linearized stiffness is $500x$ while the nonlinearized stiffness is$$\begin{array}{rl}& {\left[k\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)(L+x)\right]}_{L=1,\theta ={45}^0}\\ & =1000\left(\frac{\sqrt{{(\tan \frac{\pi }{4})}^2+{(1+x)}^2}-\sqrt{{(\tan \frac{\pi }{4})}^2+1}}{\sqrt{{(\tan \frac{\pi }{4})}^2+{(1+x)}^2}}\right)(1+x)\\ & =1000\left(\frac{\sqrt{{(x+1.0)}^2+1.0}-1.4142}{\sqrt{{(x+1.0)}^2+1.0}}\right)(1+x)\end{array}$$

Here is a plot of linearized vs. non-linearized stiffness for $x=-1\cdots 1$

2.9.2.3 part(c)

The spring extension $\mathrm{\Delta }$ is first found by assuming there is a point $A$ at $x=0$ and point $B$ where the spring is attached to the ceiling. Hence

$$\begin{array}{rl}\dot{\mathrm{\Delta }}& =({\dot{u}}_A-{\dot{u}}_B)e_{A/B}\\ & =(\dot{x}\hat{ı}-0\hat{ȷ})\cdot (\cos {\theta }_0\hat{ı}-\sin {\theta }_0\hat{ȷ})\\ & =\dot{x}\cos {\theta }_0\end{array}$$

Therefore$$\mathrm{\Delta }=x\cos {\theta }_0$$

Now we repeat the same calculations but using $\mathrm{\Delta }=x\cos {\theta }_0$ for the spring extension.$$\begin{array}{rl}T& =\frac{1}{2}m{\dot{x}}^2\\ V& =\frac{1}{2}k{\mathrm{\Delta }}^2\\ & =\frac{1}{2}k{(x\cos {\theta }_0)}^2\end{array}$$

Hence the Lagrangian $\mathrm{\Phi }$ is$$\begin{array}{rl}\mathrm{\Phi }& =T-V\\ & =\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{(x\cos {\theta }_0)}^2\end{array}$$

Now the equation of motion for coordinate $x$ is (using the standard Lagrangian form)$$\frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}\right)-\frac{\partial \mathrm{\Phi }}{\partial x}=0$$

It is equal to zero above, since there is no generalized force associated with coordinate $x.$ Now we just need to evaluate each part of the above expression to obtain the EOM.$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}& =m\dot{x}\\ \frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}\right)& =m\ddot{x}\end{array}$$

and$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial x}& =\frac{\partial }{\partial x}(\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{(x\cos {\theta }_0)}^2)\\ & =-k(x\cos {\theta }_0)\cos {\theta }_0\\ & =-kx{\cos }^2{\theta }_0\end{array}$$

Hence EOM becomes$$\begin{array}{rl}\frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial \dot{x}}\right)-\frac{\partial \mathrm{\Phi }}{\partial x}& =0\\ m\ddot{x}+kx{\cos }^2{\theta }_0& =0\end{array}$$

But $\cos {\theta }_0=\frac{L}{\sqrt{H^2+L^2}}$ hence$$\boxed{m\ddot{x}+kx\frac{L^2}{H^2+L^2}=0}$$

This is the same as the EOM for the linearized case found in part(c)

2.9.2.4 part(d)

Now we need to solve numerically the nonlinear EOM found in part(a) which is$$m\ddot{x}+k\left(\frac{\sqrt{H^2+{(L+x)}^2}-\sqrt{H^2+L^2}}{\sqrt{H^2+{(L+x)}^2}}\right)(L+x)=0$$

using $m=1,k=1000,L=1,{\theta }_0={45}^0.$ For IC we use $x(0)=0.1,x^{\prime }(0)=0$ for first case, and for second case using $x(0)=0.5,x^{\prime }(0)=0.$ This is a plot showing both responses on same diagram

The period for the response for case of IC$$given by x$(0)=0.5$ is seen to be about $0.375$ seconds and for the case $x\left(0\right)=0.1$ it is $0.275$ sec.

The linearized EOM is $\ddot{x}+500x=0$ and hence ${\omega }_n^2=500$ or ${\omega }_n=\sqrt{500}=22.361$ rad/sec, hence $T=\frac{2\pi }{{\omega }_n}=\frac{2\pi }{22.361}=$$0.281$ sec.

We notice this agrees well with the period of the response of the nonlinear equation for only the case $x=0.1$.This is because $x=0.1$ is very close to $x=0$ the point at which the linearization happened. Therefore, the linearized EOM gave an answer of $0.281$ sec that is very close the more exact value of $0.275$ seconds. But when the initial conditions changed to $x\left(0\right)=0.5$, then $T$ found from linearized EOM does not agree with the exact value of $0.375$ seconds.

This is because $x=0.5$ is far away from the point $x=0$ where the linearized was done. Hence the linearized EOM can be used for only initial conditions that are close to the point where the linearization was done.

Additionally, the nonlinearity manifests itself in the response of the system by noticing that the frequency of the free vibration response has actually changed depending on initial conditions. In a linear system, only the phase and amplitude of the free vibration response will change as initial conditions is changed, while the natural frequency of vibrations does not change.

2.9.3 problem 2

Use $y$ and $\theta$ as generalized coordinates as shown in this diagram in the positive direction

Using Lagrangian method, we start by finding the kinetic energy of the system, then the potential energy.$$T=\frac{1}{2}m{\dot{y}}^2+\frac{1}{2}\left(m{r}_G^2\right){\dot{\theta }}^2$$

For the potential energy, there will be potential energy due to $k_y$ spring extension and due to $k_T$ spring angle of rotation in system. From the diagram above, we see that, for small angle $\theta$$$V=\frac{1}{2}{k}_y{\mathrm{\Delta }}^2+\frac{1}{2}{k}_T{\theta }^2$$

To find $\mathrm{\Delta }$ we use the stiff spring approximation. Let the point the spring is attached at the top be $B$, then$$\begin{array}{rl}\dot{\mathrm{\Delta }}& =({\dot{u}}_E-{\dot{u}}_B)e_{E/B}\\ & =((l\dot{\theta }-\dot{y})\hat{ȷ}-0)\cdot (-\hat{ȷ})\\ & =(l\dot{\theta }-\dot{y})\hat{ȷ}\cdot (-\hat{ȷ})\\ & =(\dot{y}-l\dot{\theta })\end{array}$$

Hence$$\mathrm{\Delta }=y-l\theta$$

Therefore, the potential energy now can be found to be$$V=\frac{1}{2}{k}_y{(y-l\theta )}^2+\frac{1}{2}{k}_T{\theta }^2$$

Therefore, the Lagrangian $\mathrm{\Phi }$ is$$\begin{array}{rl}\mathrm{\Phi }& =T-V\\ & =\frac{1}{2}m{\dot{y}}^2+\frac{1}{2}\left(m{r}_G^2\right){\dot{\theta }}^2-\frac{1}{2}{k}_y{(y-l\theta )}^2-\frac{1}{2}{k}_T{\theta }^2\end{array}$$

We now find the equations for each coordinate. For $y$$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial \dot{y}}& =m\dot{y}\\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial \dot{y}}& =m\ddot{y}\\ \frac{\partial \mathrm{\Phi }}{\partial y}& =-k_y(y-l\theta )\end{array}$$

Hence EOM is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial \dot{y}}-\frac{\partial \mathrm{\Phi }}{\partial y}& =Q_y\\ m\ddot{y}+k_y(y-l\theta )& =Q_y\end{array}$$

We just need to find $Q_y$ the generalized force in the $y$ direction. Using virtual work, we make small virtual displacement $\delta y$ in positive $y$ direction while fixing all other generalized coordinates from moving (in this case $\theta$) and then find out the work done by external forces. In this case, there is only one external force which is $L$. Hence $$\delta W=L\delta y$$

Therefore $Q_y=L$ since that is the force that is multiplied by $\delta y.$ Hence EOM for $y$ is now found $$\boxed{m\ddot{y}+k_y(y-l\theta )=L}$$

verification: As $L$ increases, then we see that $y^{\prime \prime }$ gets larger. This makes sense since $y$ is upwards acceleration, so wing accelerates in the same direction.

Now we find EOM for $\theta$$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial \dot{\theta }}& =m{r}_G^2\dot{\theta }\\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial \dot{\theta }}& =m{r}_G^2\ddot{\theta }\\ \frac{\partial \mathrm{\Phi }}{\partial \theta }& =k_{y}l(y-l\theta )-k_T\theta \end{array}$$

Therefore the EOM is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial \dot{\theta }}-\frac{\partial \mathrm{\Phi }}{\partial \theta }& =Q_{\theta }\\ m{r}_G^2\ddot{\theta }-k_{y}l(y-l\theta )+k_T\theta & =Q_{\theta }\\ m{r}_G^2\ddot{\theta }-k_{y}ly+k_y{l}^2\theta +k_T\theta & =Q_{\theta }\\ m{r}_G^2\ddot{\theta }-k_{y}ly+(k_T+k_y{l}^2)\theta & =Q_{\theta }\end{array}$$

We just need to find $Q_{\theta }$ the generalized force in the $\theta$ direction. Using virtual work, we make small virtual displacement $\delta \theta$ in positive $\theta$ direction (i.e. anticlock wise) while fixing all other generalized coordinates from moving (in this case $y$) and then find out the work done by external forces. In this case, there is only one external force which is $L$. When we make $\delta \theta$ rotation in the positive $\theta$ direction, the displacement where the force $L$ acts is $(l+s)\delta \theta$ for small angle. But this displacement is in the downward direction, hence it is negative, since we are using $y$ as positive upwards. Hence $$\delta W=-L(l+s)\delta \theta$$

Therefore $Q_{\theta }=-L(l+s)$ since that is the force that is multiplied by $\delta \theta .$ Hence EOM for $\theta$ is now found $$m{r}_G^2\ddot{\theta }-k_{y}ly+(k_T+k_y{l}^2)\theta =-L(l+s)$$

Verification: As $L$ gets larger, then $\ddot{\theta }$ gets negative (since $L$ has negative sign). This makes sense, since as $L$ gets larger, the rotation as shown in the positive direction will change sign and the wing will now swing the opposite direction (i.e. anticlockwise).

Now we can make the matrix of EOM$$\begin{array}{rl}M{X}^{\prime \prime }+kX& =Q\\ \left(\begin{array}{cc}m& 0\\ 0& m{r}_G^2\end{array}\right)\left(\begin{array}{c}\ddot{y}\\ \ddot{\theta }\end{array}\right)+\left(\begin{array}{cc}{k}_y& -l{k}_y\\ -l{k}_y& k_T+k_y{l}^2\end{array}\right)\left(\begin{array}{c}y\\ \theta \end{array}\right)& =\left(\begin{array}{c}L\\ -L(l+s)\end{array}\right)\end{array}$$

Notice that for $\left[k\right]$ the matrix is symmetric as expected, and also positive on the diagonal as expected. The mass matrix $\left[m\right]$ is symmetric and positive definite as well.

2.9.4 problem 3

Let $\theta$ be the small angle of rotation that the rod rotates by in the anti clockwise direction. Let the point the spring is fixed be $B$ and the moving point where the spring is attached to the rod be $A.$To find spring extension $\mathrm{\Delta }$ we use the stiff spring approximation. Let the angle $\alpha ={53.13}^0$, hence$$\begin{array}{rl}\dot{\mathrm{\Delta }}& =({\dot{u}}_A-{\dot{u}}_B)\cdot {\mathbf{e}}_{A/B}\\ & =(\left(\frac{L}{3}\dot{\theta }\right)\hat{ȷ}-0)\cdot (\cos \alpha \hat{ı}+\sin \alpha \hat{ȷ})\\ & =\frac{L}{3}\dot{\theta }\sin \alpha \end{array}$$

Hence$$\mathrm{\Delta }=\frac{L}{3}\theta \sin \alpha$$

Using Lagrangian method, we start by finding the kinetic energy of the system, then the potential energy. $\theta$ is the only generalized coordinate. Assume bar has mass $m$ and hence $I=$ $\frac{m{L}^2}{3}$$$T=\frac{1}{2}I{\dot{\theta }}^2$$

For the potential energy, there will be potential energy due to $k$ spring extension. From the diagram above, we see that $$V=\frac{1}{2}k{(\frac{L}{3}\theta \sin \alpha )}^2$$

Therefore, the Lagrangian $\mathrm{\Phi }$ is$$\begin{array}{rl}\mathrm{\Phi }& =T-V\\ & =\frac{1}{2}I{\dot{\theta }}^2-\frac{1}{2}k\frac{L^2}{9}{\theta }^2{\sin }^2\alpha \end{array}$$

Now we find EOM for $\theta$$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial \dot{\theta }}& =I\dot{\theta }\\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial \dot{\theta }}& =I\ddot{\theta }\\ \frac{\partial \mathrm{\Phi }}{\partial \theta }& =-\frac{k{L}^2{\sin }^2\alpha }{9}\theta \end{array}$$

Therefore the EOM is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial \dot{\theta }}-\frac{\partial \mathrm{\Phi }}{\partial \theta }& =Q_{\theta }\\ I\ddot{\theta }+\frac{k{L}^2{\sin }^2\alpha }{9}\theta & =Q_{\theta }\end{array}$$

We now need to find the generalized force due to virtual $\delta \theta$ rotation using the virtual work method. There are 2 external forces, the damping force which will have negative sign since it takes energy away from the system, and the external force $F$ which will add energy hence will have positive sign.

We start by making $\delta \theta$ and then find the work done by these 2 forces.

Work done by $F$ is $FL\delta \theta$ since the displacement is $L\delta \theta$ for small angle.  Now the work done by damping is$\left(c\frac{L}{3}\dot{\theta }\right)\frac{L}{3}\delta \theta$ hence total work is$$\begin{array}{rl}\delta W& =FL\delta \theta -\left(c\frac{L}{3}\dot{\theta }\right)\frac{L}{3}\delta \theta \\ & =(FL-c\frac{L^2}{9}\dot{\theta })\delta \theta \end{array}$$

Notice that work due to damping was added with negative sign since damping removes energy from the system.

Hence $Q_{\theta }=(FL-c\frac{L^2}{9}\dot{\theta })$ therefore the EOM is$$\begin{array}{rl}I\ddot{\theta }+\frac{k{L}^2{\sin }^2\alpha }{9}\theta & =FL-c\frac{L^2}{9}\dot{\theta }\\ I\ddot{\theta }+c\frac{L^2}{9}\dot{\theta }+\frac{k{L}^2{\sin }^2\alpha }{9}\theta & =FL\end{array}$$

Hence the damping coefficient is $c\frac{L^2}{9}.$

2.9.5 problem 4

Let $x_1$ and $x_2$ be the generalized coordinates as shown in this diagram

Let mass of cart be $m_1$ and mass of small sliding block be $m_2$ (at the end, they will be replaced by values given). Let $k$ for spring attached to wall be $k_1$ and $k$ for spring for small block be $k_2.$We start by finding the kinetic energy of the system$$T=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{v}^2$$

where $v$ is the velocity of the block. To find this $v$ it is easier to resolve components on the $x$ and $y$ direction. Therefore we find that $$\overrightarrow{v}={\dot{x}}_2\sin \theta j+({\dot{x}}_2\cos \theta +{\dot{x}}_1)i$$

Hence $$\begin{array}{rl}{\left|\overrightarrow{v}\right|}^2& ={({\dot{x}}_2\sin \theta )}^2+{({\dot{x}}_2\cos \theta +{\dot{x}}_1)}^2\\ & =({\dot{x}}_2^2{\sin }^2\theta )+({\dot{x}}_2^2{\cos }^2\theta +{\dot{x}}_1^2+2{\dot{x}}_2{\dot{x}}_1\cos \theta )\\ & ={\dot{x}}_2^2({\sin }^2\theta +{\cos }^2\theta )+{\dot{x}}_1^2+2{\dot{x}}_2{\dot{x}}_1\cos \theta \\ & ={\dot{x}}_2^2+{\dot{x}}_1^2+2{\dot{x}}_2{\dot{x}}_1\cos \theta \end{array}$$

Therefore $$T=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2({\dot{x}}_2^2+{\dot{x}}_1^2+2{\dot{x}}_2{\dot{x}}_1\cos \theta )$$

Now we find the potential energy.$$V=\frac{1}{2}{k}_1{x}_1^2+\frac{1}{2}{k}_2{x}_2^2-m_{2}g(x_2\sin \theta )$$

There are no external forces, hence generalized forces $Q_{x_1},Q_{x_2}$ are zero.  The Lagrangian $\mathrm{\Phi }$ is$$\begin{array}{rl}\mathrm{\Phi }& =T-V\\ & =\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2({\dot{x}}_2^2+{\dot{x}}_1^2+2{\dot{x}}_2{\dot{x}}_1\cos \theta )-\frac{1}{2}{k}_1{x}_1^2-\frac{1}{2}{k}_2{x}_2^2+m_{2}g(x_2\sin \theta )\end{array}$$

Now we find EOM for $x_1$ is$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial {\dot{x}}_1}& =m_1{\dot{x}}_1+m_2{\dot{x}}_1+m_2{\dot{x}}_2\cos \theta \\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{x}}_1}& =m_1{\ddot{x}}_1+m_2{\ddot{x}}_1+m_2{\ddot{x}}_2\cos \theta \\ & =(m_1+m_2){\ddot{x}}_1+m_2{\ddot{x}}_2\cos \theta \\ \frac{\partial \mathrm{\Phi }}{\partial x_1}& =-k_1{x}_1\end{array}$$

Therefore the EOM for $x_1$ is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{x}}_1}-\frac{\partial \mathrm{\Phi }}{\partial x_1}& =0\\ (m_1+m_2){\ddot{x}}_1+m_2{\ddot{x}}_2\cos \theta +k_1{x}_1& =0\end{array}$$

Now we replace the actual values for $m_1=2m,m_2=m,k_1=3k$ hence$$\boxed{3m{\ddot{x}}_1+m{\ddot{x}}_2\cos \theta +3k{x}_1=0}$$

Now we find EOM for $x_2$ is$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial {\dot{x}}_2}& =m_2({\dot{x}}_2+{\dot{x}}_1\cos \theta )\\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{x}}_1}& =m_2({\ddot{x}}_2+{\ddot{x}}_1\cos \theta )\\ & =m_2\cos \theta {\ddot{x}}_1+m_2{\ddot{x}}_2\\ \frac{\partial \mathrm{\Phi }}{\partial x_2}& =-k_2{x}_2+m_{2}g\sin \theta \end{array}$$

Therefore the EOM for $x_2$ is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{x}}_2}-\frac{\partial \mathrm{\Phi }}{\partial x_2}& =0\\ m_2\cos \theta {\ddot{x}}_1+m_2{\ddot{x}}_2+k_2{x}_2-m_{2}g\sin \theta & =0\end{array}$$

Now we replace the actual values for $m_1=2m,m_2=m,k_2=k$ hence$$m\cos \theta {\ddot{x}}_1+m{\ddot{x}}_2+k{x}_2=m_{2}g\sin \theta$$

Now we can make the matrix of EOM$$\begin{array}{rl}M{X}^{\prime \prime }+kX& =Q\\ \left(\begin{array}{cc}3m& m\cos \theta \\ m\cos \theta & m\end{array}\right)\left(\begin{array}{c}{\ddot{x}}_1\\ {\ddot{x}}_2\end{array}\right)+\left(\begin{array}{cc}3k& 0\\ 0& k\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)& =\left(\begin{array}{c}0\\ m_{2}g\sin \theta \end{array}\right)\end{array}$$

Hence$$\boxed{m\left(\begin{array}{cc}3& \cos \theta \\ \cos \theta & 1\end{array}\right)\left(\begin{array}{c}{\ddot{x}}_1\\ {\ddot{x}}_2\end{array}\right)+k\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ m_{2}g\sin \theta \end{array}\right)}$$

Notice that there zeros now off diagonal in the $\left[K\right]$ matrix, which means the springs are not coupled. (which is expected, as motion of one is not affected by the other). But mass matrix $\left[m\right]$ has non-zeros off the diagonal. So the masses are coupled. i.e. EOM is coupled. This means we can’t solve on EOM on its own and both have to be solved simultaneously.

2.9.6 problem 5

There are 2 degrees of freedom, ${\theta }_1$ and ${\theta }_2$ as shown in this diagram, using anticlock wise rotation as positive

The Lagrangian $\mathrm{\Phi }=T-V$ where$$T=\frac{1}{2}{I}_1{\dot{\theta }}_1^2+\frac{1}{2}{I}_2{\dot{\theta }}_2^2$$

Where $I_1=\frac{m_1{L}^2}{3}$ and $I_2=\frac{m_2{L}^2}{12}+m_2{\left(\frac{L}{4}\right)}^2$ (using parallel axis theorem). Hence $I_2=\frac{m_2{L}^2}{12}+m_2\frac{L^2}{16}=\frac{7}{48}{L}^2{m}_2$

Now we find the potential energy, assuming springs remain straight (stiff spring assumption) and assuming small angles$$V=\frac{1}{2}{k}_1{\left(\stackrel{{\mathrm{\Delta }}_1}{\overbrace{\frac{3L}{4}{\theta }_1+\frac{3L}{4}{\theta }_2}}\right)}^2+\frac{1}{2}{k}_2{\left(\stackrel{{\mathrm{\Delta }}_2}{\overbrace{L{\theta }_1+\frac{L}{2}{\theta }_2}}\right)}^2+m_{1}g\frac{L}{2}{\theta }_1-m_{2}g\frac{L}{4}{\theta }_2$$

Hence $$\begin{array}{rl}\mathrm{\Phi }& =T-V\\ & =(\frac{1}{2}{I}_1{\dot{\theta }}_1^2+\frac{1}{2}{I}_2{\dot{\theta }}_2^2)-(\frac{1}{2}{k}_1{(\frac{3L}{4}{\theta }_1+\frac{3L}{4}{\theta }_2)}^2+\frac{1}{2}{k}_2{(L{\theta }_1+\frac{L}{2}{\theta }_2)}^2+m_{1}g\frac{L}{2}{\theta }_1-m_{2}g\frac{L}{4}{\theta }_2)\end{array}$$

Now we find EOM for ${\theta }_1$$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial {\dot{\theta }}_1}& =I{\dot{\theta }}_1\\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{\theta }}_1}& =I{\ddot{\theta }}_1\\ \frac{\partial \mathrm{\Phi }}{\partial {\theta }_1}& =-k_1(\frac{3L}{4}{\theta }_2+\frac{3L}{4}{\theta }_1)\left(\frac{3L}{4}\right)-k_2(\frac{L}{2}{\theta }_2+L{\theta }_1)(L)-m_{1}g\frac{L}{2}\\ & =-\frac{3L}{4}{k}_1(\frac{3L}{4}{\theta }_2+\frac{3L}{4}{\theta }_1)-k_{2}L(\frac{L}{2}{\theta }_2+L{\theta }_1)-m_{1}g\frac{L}{2}\end{array}$$

Therefore the EOM for ${\theta }_1$ is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{\theta }}_1}-\frac{\partial \mathrm{\Phi }}{\partial {\theta }_1}& =Q_{{\theta }_1}\\ I_1{\ddot{\theta }}_1+\frac{3L}{4}{k}_1(\frac{3L}{4}{\theta }_2+\frac{3L}{4}{\theta }_1)+k_{2}L(\frac{L}{2}{\theta }_2+L{\theta }_1)+m_{1}g\frac{L}{2}& =0\end{array}$$

The generalized force is zero, since there is no direct external force acting on top rod.

Hence EOM for ${\theta }_1$ is from above$$\frac{m_1{L}^2}{3}{\ddot{\theta }}_1+{\theta }_1(k_1{\left(\frac{3L}{4}\right)}^2+k_2{L}^2)+{\theta }_2(k_1{\left(\frac{3L}{4}\right)}^2+k_2\frac{L^2}{2})=-m_{1}g\frac{L}{2}$$

Now we find EOM for ${\theta }_2$$$\begin{array}{rl}\frac{\partial \mathrm{\Phi }}{\partial {\dot{\theta }}_2}& =I{\dot{\theta }}_2\\ \frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{\theta }}_2}& =I{\ddot{\theta }}_2\\ \frac{\partial \mathrm{\Phi }}{\partial {\theta }_2}& =-k_1(\frac{3L}{4}{\theta }_2+\frac{3L}{4}{\theta }_1)\left(\frac{3L}{4}\right)-k_2(\frac{L}{2}{\theta }_2+L{\theta }_1)\left(\frac{L}{2}\right)+m_{2}g\frac{L}{4}\end{array}$$

Therefore the EOM for ${\theta }_2$ is$$\begin{array}{rl}\frac{d}{dt}\frac{\partial \mathrm{\Phi }}{\partial {\dot{\theta }}_2}-\frac{\partial \mathrm{\Phi }}{\partial {\theta }_2}& =Q_{{\theta }_2}\\ I_2{\ddot{\theta }}_2+k_1(\frac{3L}{4}{\theta }_2+\frac{3L}{4}{\theta }_1)\left(\frac{3L}{4}\right)+k_2(\frac{L}{2}{\theta }_2+L{\theta }_1)\left(\frac{L}{2}\right)-m_{2}g\frac{L}{4}& =Q_{{\theta }_1}\end{array}$$

Now $Q_{{\theta }_2}$is found by virtual work. Making a virtual displacement $\delta {\theta }_2$ while fixing ${\theta }_1$ and finding the work done by all external forces.$$\delta W=F\frac{L}{2}\delta {\theta }_2$$

Hence $Q_{{\theta }_2}=F\frac{L}{2}$ with positive sign since it add energy to the system. Hence EOM for ${\theta }_2$ is$$\frac{7}{48}{L}^2{m}_2{\ddot{\theta }}_2+{\theta }_1(k_1{\left(\frac{3L}{4}\right)}^2+k_2\frac{L^2}{2})+{\theta }_2(k_1{\left(\frac{3L}{4}\right)}^2+k_2{\left(\frac{L}{2}\right)}^2)=m_{2}g\frac{L}{4}+F\frac{L}{2}$$

Now we can make the matrix of EOM$$\begin{array}{rl}M{X}^{\prime \prime }+kX& =Q\\ \left(\begin{array}{cc}\frac{m_1{L}^2}{3}& 0\\ 0& \frac{7}{48}{L}^2{m}_2\end{array}\right)\left(\begin{array}{c}{\ddot{\theta }}_1\\ {\ddot{\theta }}_2\end{array}\right)+\left(\begin{array}{cc}{k}_1\frac{9L^2}{16}+k_2{L}^2& k_1{\left(\frac{3L}{4}\right)}^2+k_2\frac{L^2}{2}\\ k_1{\left(\frac{3L}{4}\right)}^2+k_2\frac{L^2}{2}& k_1{\left(\frac{3L}{4}\right)}^2+k_2{\left(\frac{L}{2}\right)}^2\end{array}\right)\left(\begin{array}{c}{\theta }_1\\ {\theta }_2\end{array}\right)& =\left(\begin{array}{c}-m_{1}g\frac{L}{2}\\ m_{2}g\frac{L}{4}+F\frac{L}{2}\end{array}\right)\end{array}$$

The matrix $\left[k\right]$ is coupled but the mass matrix $\left[m\right]$ is not.

2.9.7 problem 6

The inertia and stiffness matrices for a system are $\left[M\right]=\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]$ kg, $\left[K\right]=\left[\begin{array}{cc}200& 200\\ 200& 800\end{array}\right]$N/m. determine the corresponding natural frequencies and modes of free vibration.$$[\left[k\right]-{\omega }^2\left[M\right]]\left\{\mathrm{\Phi }\right\}=\left\{0\right\}$$

Solving for eigenvalues$$\begin{array}{rl}det(\left[\begin{array}{cc}200& 200\\ 200& 800\end{array}\right]-{\omega }^2\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right])& =0\\ det\left[\begin{array}{cc}200-4{\omega }^2& 200\\ 200& 800-2{\omega }^2\end{array}\right]& =0\\ (200-4{\omega }^2)(800-2{\omega }^2)-{200}^2& =0\\ 8{\omega }^4-3600{\omega }^2+120000& =0\end{array}$$

Hence, taking the positive square root only we find$$\begin{array}{rl}{\omega }_1& =20.341\text{ rad/sec}\\ {\omega }_2& =6.0211\text{ rad/sec}\end{array}$$

When $\omega ={\omega }_1$$$\left[\begin{array}{cc}200-4{\omega }_1^2& 200\\ 200& 800-2{\omega }_1^2\end{array}\right]\left\{\begin{array}{c}{\mathrm{\Phi }}_{11}\\ {\mathrm{\Phi }}_{21}\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$$

Let ${\mathrm{\Phi }}_{11}$ be the arbitrary value $1$ hence$$\begin{array}{rl}\left[\begin{array}{cc}200-4{\omega }_1^2& 200\\ \times & \times \end{array}\right]\left\{\begin{array}{c}1\\ {\mathrm{\Phi }}_{21}\end{array}\right\}& =\left\{\begin{array}{c}0\\ \times \end{array}\right\}\\ 200-4{\omega }_1^2+200{\mathrm{\Phi }}_{21}& =0\\ {\mathrm{\Phi }}_{21}& =\frac{-200+4{\omega }_1^2}{200}=\frac{-200+4{\left(20.341\right)}^2}{200}=7.2751\end{array}$$

Hence the first mode associated with $\omega =20.341$ rad/sec is $$\left\{\begin{array}{c}1\\ 7.2751\end{array}\right\}$$

When $\omega ={\omega }_2$$$\left[\begin{array}{cc}200-4{\omega }_2^2& 200\\ 200& 800-2{\omega }_2^2\end{array}\right]\left\{\begin{array}{c}{\mathrm{\Phi }}_{12}\\ {\mathrm{\Phi }}_{22}\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$$

Let ${\mathrm{\Phi }}_{12}$ be the arbitrary value $1$ hence$$\begin{array}{rl}\left[\begin{array}{cc}200-4{\omega }_2^2& 200\\ \times & \times \end{array}\right]\left\{\begin{array}{c}1\\ {\mathrm{\Phi }}_{22}\end{array}\right\}& =\left\{\begin{array}{c}0\\ \times \end{array}\right\}\\ 200-4{\omega }_2^2+200{\mathrm{\Phi }}_{22}& =0\\ {\mathrm{\Phi }}_{22}& =\frac{-200+4{\omega }_2^2}{200}=\frac{-200+4{\left(6.0211\right)}^2}{200}=-0.27493\end{array}$$

Hence the first mode associated with $\omega =6.0211$ rad/sec is $$\left\{\begin{array}{c}1\\ -0.27493\end{array}\right\}$$

Summary$$\text{begin {tabular} [c]{|l|l|}hline \$omega \_n\$ (rad/sec) \& mode shapehline \$6.0211\$ \& \$begin {Bmatrix} 1 -0.27493 end {Bmatrix} \$hline \$20.341\$ \& \$begin {Bmatrix} 1 7.2751 end {Bmatrix} \$hline end {tabular} }$$

Verification using Matlab:

EDU>> M=[4 0;0 2]; K=[200 200;200 800];
EDU>> [phi,omega]=eig(K,M);
EDU>> sqrt(omega)

    6.0211         0
         0   20.3407
EDU>> phi(:,1)/abs(phi(1,1))

   -1.0000
    0.2749

EDU>> phi(:,2)/abs(phi(1,2))

    1.0000
    7.2749

Which matches the result derived. One mode shape has both displacement in phase, and the other mode shape shows the displacements to be out of phase.

2.9.8 Key solution for HW 8

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