3.1 HW1

3.1.1 Description of HW

My grade is 66/70.

1. Equation of motion of spring mass on an incline surface.
2. Equation of motion of pendulum attached to spring.
3. Solving 2nd order ODE’s.
4. Equation of motion of mass attached to 2 pulley’s.

3.1.2 Problem 1.6

Figure 3.1: Problem

Taking displacement along the x-direction shown to be from the static equilibrium position, then applying \(\sum F_{x}=m\ddot {x}\) along the shown \(x\) direction, we obtain

\begin{align*} m\ddot {x} & =-kx\\ \ddot {x}+\frac {k}{m}x & =0 \end{align*}

which is the equation of motion.  To obtain the natural frequency, we consider free vibration \(\ddot {x}+\frac {k}{m}x=0\), which implies that \(\omega _{n}=\sqrt {\frac {k}{m}}\), hence we see that the natural frequency is independent of \(g\).

We see that gravity has no effect on the spring mass system, this is because we use \(x\) to be from the static equilibrium position of the spring.

3.1.3 Problem 1.16

Figure 3.2: Problem

First we need to derive the equation of motion. Considering the following diagram

Using as generalized coordinates \(\theta \), we obtain

\begin{align*} T & =\frac {1}{2}m\left ( L\dot {\theta }\right ) ^{2}\\ U & =\frac {1}{2}k\left ( L\sin \theta \right ) ^{2}+mg\left ( L-L\cos \theta \right ) \end{align*}

Notice that in the calculation of \(U\) above, we assumed that the spring stretches by \(L\sin \theta \) in the horizontal direction only, which we are allowed to do for small \(\theta \).

Now we can find Lagrangian

\begin{align*} L & =T-U\\ & =\frac {1}{2}m\left ( L\dot {\theta }\right ) ^{2}-\frac {1}{2}kL^{2}\sin ^{2}\theta -mgL\left ( 1-\cos \theta \right ) \end{align*}

Hence the equation of motion is

\begin{align*} \frac {d}{dt}\left ( \frac {\partial L}{\partial \dot {\theta }}\right ) -\frac {\partial L}{\partial \theta } & =0\\ \frac {d}{dt}\left ( mL^{2}\dot {\theta }\right ) -\left ( -kL^{2}\sin \theta \cos \theta -mgL\sin \theta \right ) & =0\\ mL^{2}\ddot {\theta }+kL^{2}\sin \theta \cos \theta +mgL\sin \theta & =0 \end{align*}

The above is nonlinear equation. Linearize around \(\theta =0\) (equilibrium point) using Taylor series, and for small \(\theta \) we obtain \(\sin \theta \approx \theta \) and \(\cos \theta \approx 1\), hence the above becomes

\begin{align*} mL\ddot {\theta }+kL\theta +mg\theta & =0\\ \ddot {\theta }+\left ( \frac {mg+kL}{mL}\right ) \theta & =0 \end{align*}

Hence effective \(\omega _{n}\)can be found from

\[ \omega _{n}^{2}=\frac {mg+kL}{mL}\]

Hence

\[ \omega _{n}=\sqrt {\frac {g}{L}+\frac {k}{m}} \]

Comparing the above to the natural frequency of pendulum with no spring attached which is \(\omega _{n}\)=\(\sqrt {\frac {g}{L}}\), we can see the effect of adding a spring on the natural frequency: The more stiff the spring is, in other words, the larger \(k\) is, the larger \(\omega _{n}\) will become, and the smaller the period of oscillation will be. We conclude that a pendulum with a spring attached to it will always oscillate with a period which is smaller than the same pendulum without the spring attached. This makes sense as a mass with spring alone has \(\omega _{n}=\sqrt {\frac {k}{m}}\)

3.1.4 Problem 1.32

We need to solve \(\ddot {x}+2\dot {x}+2x=0\) for \(x_{0}=0mm\) and \(v_{0}=1mm/s\)

The characteristic equation is \(\lambda ^{2}+2\lambda +2=0\) which has roots \(\lambda _{1,2}=\frac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\frac {-2\pm \sqrt {4-8}}{2}=-1\pm j\)

Hence the solution is

\[ x_{h}=e^{-t}\left ( A\cos t+B\sin t\right ) \]

is the general solution. Now we use I.C. to find \(A,B\). When \(t=0\)

\[ 0=A \]

Hence \(x_{h}=Be^{-t}\sin t\), and \(\dot {x}_{h}=Be^{-t}\cos t-Be^{-t}\sin t\) and at \(t=0\), we obtain \(0.01=B\)

Then

\[ x_{h}=0.01e^{-t}\sin t \]

This is a plot of the solution for \(t\) up to 50 seconds

3.1.5 Problem 1.43

We need to solve \(\ddot {x}-\dot {x}+x=0\) for \(x_{0}=1\) and \(v_{0}=0\)

The characteristic equation is \(\lambda ^{2}-\lambda +1=0\) which has roots \(\lambda _{1,2}=\frac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\frac {1\pm \sqrt {1-4}}{2}=\frac {1}{2}\pm j\frac {\sqrt {3}}{2}\)

Hence the solution is

\[ x_{h}=e^{\frac {1}{2}t}\left ( A\cos \frac {\sqrt {3}}{2}t+B\sin \frac {\sqrt {3}}{2}t\right ) \]

is the general solution. Now we use I.C. to find \(A,B\). When \(t=0\)

\[ 1=A \]

Hence \(x_{h}=e^{\frac {1}{2}t}\left ( \cos \frac {\sqrt {3}}{2}t+B\sin \frac {\sqrt {3}}{2}t\right ) \), and

\[ \dot {x}_{h}=\frac {1}{2}e^{\frac {1}{2}t}\left ( \cos \frac {\sqrt {3}}{2}t+B\sin \frac {\sqrt {3}}{2}t\right ) +e^{\frac {1}{2}t}\left ( -\frac {\sqrt {3}}{2}\sin \frac {\sqrt {3}}{2}t+B\frac {\sqrt {3}}{2}\cos \frac {\sqrt {3}}{2}t\right ) \]

and at \(t=0\), we obtain

\begin{align*} 0 & =\frac {1}{2}+B\frac {\sqrt {3}}{2}\\ B & =\frac {-1}{\sqrt {3}}\end{align*}

Hence

\[ x_{h}=e^{\frac {1}{2}t}\left ( \cos \frac {\sqrt {3}}{2}t-\frac {1}{\sqrt {3}}\sin \frac {\sqrt {3}}{2}t\right ) \]

This is a plot of the solution for \(t\) up to 12 seconds

3.1.6 Problem 1.62

This is a single degree of freedom linear system. Assume \(x\) from static equilibrium, then (using parallel springs) we obtain

\begin{align*} T & =\frac {1}{2}m\dot {x}^{2}\\ U & =\frac {1}{2}kx^{2}+\frac {1}{2}kx^{2}=kx^{2}\end{align*}

Hence \(L=T-U=\frac {1}{2}m\dot {x}^{2}-kx^{2}\) and the Lagrangian equation is

\begin{align*} \frac {d}{dt}\left ( \frac {\partial L}{\partial \dot {x}}\right ) -\frac {\partial L}{\partial x} & =0\\ \frac {d}{dt}\left ( m\dot {x}\right ) -\left ( -2kx\right ) & =0 \end{align*}

Hence equation of motion is

\[ m\ddot {x}+2kx=0 \]

And \(\omega _{n}=\sqrt {\frac {2k}{m}}\)

3.1.7 Problem 1.90

3.1.7.1 Part(a)

\begin{align*} T & =\frac {1}{2}m\left ( \frac {l}{2}\dot {\theta }\right ) ^{2}\\ & =\frac {1}{8}ml^{2}\dot {\theta }^{2}\end{align*}

\[ U_{springs}=\frac {1}{2}k\left ( l\sin \theta \right ) ^{2}+\frac {1}{2}k\left ( l\sin \theta \right ) ^{2}\]

Assuming small angle oscillation, \(\sin \theta \simeq \theta \), hence

\[ U_{springs}=kl^{2}\theta ^{2}\]

and for the mass, since it losses potential, we have

\[ U_{mass}=-mg\left ( \frac {l}{2}-\frac {l}{2}\cos \theta \right ) \]

Hence Lagrangian \(L\) is

\begin{align*} L & =T-\left ( U_{springs}+U_{mass}\right ) \\ & =\frac {1}{8}ml^{2}\dot {\theta }^{2}-\left ( kl^{2}\theta ^{2}-mg\frac {l}{2}\left ( 1-\cos \theta \right ) \right ) \\ & =\frac {1}{8}ml^{2}\dot {\theta }^{2}-kl^{2}\theta ^{2}+mg\frac {l}{2}-mg\frac {l}{2}\cos \theta \end{align*}

Now find the Lagrangian equation

\begin{align*} \frac {\partial L}{\partial \dot {\theta }} & =\frac {1}{4}ml^{2}\dot {\theta }\\ \frac {d}{dt}\left ( \frac {\partial L}{\partial \dot {\theta }}\right ) & =\frac {1}{4}ml^{2}\ddot {\theta }\\ \frac {\partial L}{\partial \theta } & =-2kl^{2}\theta +mg\frac {l}{2}\sin \theta \end{align*}

Hence

\begin{align*} \frac {d}{dt}\left ( \frac {\partial L}{\partial \dot {\theta }}\right ) -\frac {\partial L}{\partial \theta } & =\frac {1}{4}ml^{2}\ddot {\theta }-\left ( -2kl^{2}\theta +mg\frac {l}{2}\sin \theta \right ) \\ & =\frac {1}{4}ml^{2}\ddot {\theta }+2kl^{2}\theta -mg\frac {l}{2}\sin \theta \end{align*}

And the equation of motion is

\begin{align*} \frac {1}{4}ml^{2}\ddot {\theta }+2kl^{2}\theta -mg\frac {l}{2}\sin \theta & =0\\ \ddot {\theta }+\frac {8k}{m}\theta -2\frac {g}{l}\sin \theta & =0 \end{align*}

Linearize by setting \(\sin \theta \simeq \theta \) we obtain equation of motion

\begin{equation} \ddot {\theta }+\theta \left ( \frac {8k}{m}-2\frac {g}{l}\right ) =0 \tag {1}\end{equation}

Hence

\[ \omega _{n}=\sqrt {2\left ( 4\frac {k}{m}-\frac {g}{l}\right ) } \]

3.1.7.2 Part (b)

To discuss stability, we need to determine the location of the roots of the characteristic equation of the homogeneous EQM, hence from equation (1), we see that

\[ \ddot {\theta }+\omega _{n}^{2}\theta =0 \]

And assuming solution \(\theta \left ( t\right ) =e^{\lambda t}\) leads to the characteristic equation

\begin{align*} \lambda ^{2}+\omega _{n}^{2} & =0\\ \lambda ^{2} & =-\omega _{n}^{2}\\ \lambda & =\pm \sqrt {-\omega _{n}^{2}}\\ & =\pm j\sqrt {\omega _{n}^{2}}\end{align*}

Since \(\omega _{n}^{2}>0\), then

\[ \lambda =\pm j\omega _{n} \]

Since roots of the characteristic equation on the imaginary axis, this is a marginally stable system regardless of the values of \(m,l,k\).

Since we are looking at the linearized system, there is only one equilibrium point, and the system is either stable or not. Here we found it is marginally stable. The effect of changing \(k,l,m\,\ \)is to change the period of oscillation around the equilibrium point.

3.1.8 Key for HW1

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