3.2 HW2

3.2.1 Description of HW

1. Find solution to spring mass system with forcing function for different IC.
2. Compute IC for spring mass system to cause it to oscillate at some frequency.
3. Find EQM for spring-damper-mass system on incline surface.
4. Car and passengers and different speeds, deflection calculations.
5. The unbalanced mass rotating, the washing machine problem.

3.2.2 Problem 2.7

Solution sketch: Obtain the Lagrangian, find EQM, solve in terms of general initial conditions \(x\left ( 0\right ) =x_{0}\) and \(\dot {x}\left ( 0\right ) =v_{0}\), then solve parts (a) and (b) using this general solution.

This is one degree of freedom system. Using \(x\) as the generalized coordinates, we first obtain the Lagrangian \(L\)

\[ L=T-U \]

Where

\begin{align*} T & =\frac {1}{2}m\dot {x}^{2}\\ U & =\frac {1}{2}kx^{2}\\ L & =\frac {1}{2}m\dot {x}^{2}-\frac {1}{2}kx^{2}\\ \frac {\partial L}{\partial \dot {x}} & =m\dot {x}\\ \frac {d}{dt}\frac {\partial L}{\partial \dot {x}} & =m\ddot {x}\\ \frac {\partial L}{\partial x} & =-kx \end{align*}

Hence the EQM is (using the Lagrangian equation), and \(F=10\) and \(\omega =10\) rad/sec. (the forcing frequency)

\begin{align} \frac {d}{dt}\frac {\partial L}{\partial \dot {x}}-\frac {\partial L}{\partial x} & =F\sin \omega t\nonumber \\ m\ddot {x}+kx & =F\sin \omega t\nonumber \\ \ddot {x}+\frac {k}{m}x & =\frac {F}{m}\sin \omega t\nonumber \\ \ddot {x}+\omega _{n}^{2}x & =\frac {F}{m}\sin \omega t\tag {1}\end{align}

Where \(\omega _{n}^{2}=\frac {k}{m}\). The solution is

\begin{equation} x\left ( t\right ) =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \tag {2}\end{equation}

To obtain \(x_{h}t\left ( t\right ) \)

\[ \ddot {x}_{h}\left ( t\right ) +\omega _{n}^{2}x_{h}\left ( t\right ) =0 \]

Assume \(x_{h}\left ( t\right ) =e^{\lambda t}\) and substitute in the above ODE we obtain the characteristic equation

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

Hence

\begin{equation} x_h\left ( t\right ) =A\cos \omega _nt+B\sin \omega _nt \tag {3}\end{equation}

Guess

\begin{align*} x_{p}\left ( t\right ) & =c_{1}\cos \omega t+c_{2}\sin \omega t\\ \dot {x}_{p}\left ( t\right ) & =-\omega c_{1}\sin \omega t+\omega c_{2}\cos \omega t\\ \ddot {x}_{p}\left ( t\right ) & =-\omega ^{2}c_{1}\cos \omega t-\omega ^{2}c_{2}\sin \omega t \end{align*}

Notice, the above guess is valid only under the condition that \(\omega \neq \omega _{n}\) which is the case in this problem. Now, substitute the above 3 equations into (1) we obtain

\begin{align*} \left ( -\omega ^{2}c_{1}\cos \omega t-\omega ^{2}c_{2}\sin \omega t\right ) +\omega _{n}^{2}\left ( c_{1}\cos \omega t+c_{2}\sin \omega t\right ) & =\frac {F}{m}\sin \omega t\\ \sin \omega t\left ( -\omega ^{2}c_{2}+\omega _{n}^{2}c_{2}\right ) +\cos \omega t\left ( -\omega ^{2}c_{1}+\omega _{n}^{2}c_{1}\right ) & =\frac {F}{m}\sin \omega t \end{align*}

By comparing coefficients, we obtain

\begin{align*} c_{2}\left ( \omega _{n}^{2}-\omega ^{2}\right ) & =\frac {F}{m}\\ c_{2} & =\frac {F/m}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) }\end{align*}

and \(c_{1}=0\), hence

\[ \fbox {$x_p\left ( t\right ) =\frac {F/m}{\left ( \omega _n^2-\omega ^2\right ) }\sin \omega t$}\]

Then from (2) we obtain

\begin{align*} x\left ( t\right ) & =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \\ & =x_{h}\left ( t\right ) +\frac {F/m}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) }\sin \omega t \end{align*}

Using (3) in the above

\begin{equation} x\left ( t\right ) =A\cos \omega _nt+B\sin \omega _nt+\frac {F/m}{\left ( \omega _n^2-\omega ^2\right ) }\sin \omega t \tag {4}\end{equation}

Now assume \(x\left ( 0\right ) =x_{0}\) and \(\dot {x}\left ( 0\right ) =v_{0}\) For the condition \(x\left ( 0\right ) =x_{0}\) we obtain

\[ \fbox {$x_0=A$}\]

For the condition \(\dot {x}\left ( 0\right ) =v_{0}\) we obtain

\begin{align*} \dot {x}\left ( t\right ) & =-A\omega _{n}\sin \omega _{n}t+B\omega _{n}\cos \omega _{n}t+\omega \frac {F/m}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) }\cos \omega t\\ \dot {x}\left ( 0\right ) & =v_{0}=B\omega _{n}+\omega \frac {F/m}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) }\end{align*}

Hence

\[ B=\frac {v_0}{\omega _n}-\frac {\omega }{\omega _n}\frac {F/m}{\left ( \omega _n^2-\omega ^2\right ) } \]

Hence (4) can be written as

\[ x\left ( t\right ) =x_{0}\cos \omega _{n}t+\left ( \frac {v_{0}}{\omega _{n}}-\frac {\omega }{\omega _{n}}\frac {F/m}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) }\right ) \sin \omega _{n}t+\frac {F/m}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) }\sin \omega t \]

Let \(\frac {\omega }{\omega _{n}}=r\), the above becomes

\[ x\left ( t\right ) =x_{0}\cos \omega _{n}t+\left ( \frac {v_{0}}{\omega _{n}}-\frac {\frac {F}{m}\ r}{\omega _{n}^{2}\left ( 1-r^{2}\right ) }\right ) \sin \omega _{n}t+\frac {F/m}{\omega _{n}^{2}\left ( 1-r^{2}\right ) }\sin \omega t \]

But \(\omega _{n}^{2}=\frac {k}{m}\) hence

\[ x\left ( t\right ) =x_{0}\cos \omega _{n}t+\left ( \frac {v_{0}}{\omega _{n}}-\frac {\frac {F}{m}\ r}{\frac {k}{m}\left ( 1-r^{2}\right ) }\right ) \sin \omega _{n}t+\frac {\frac {F}{m}}{\frac {k}{m}\left ( 1-r^{2}\right ) }\sin \omega t \]

Therefore, the general solution is

\begin{equation} x\left ( t\right ) =x_0\cos \omega _nt+\left ( \frac {v_0}{\omega _n}-\frac {F}{k}\frac {\ r}{\left ( 1-r^2\right ) }\right ) \sin \omega _nt+\frac {F}{k}\frac {1}{\left ( 1-r^2\right ) }\sin \omega t \tag {5}\end{equation}

3.2.2.1 Part(a)

When \(x_{0}=0,v_{0}=0\) we obtain from (5)

\begin{equation} x\left ( t\right ) =\left ( -\frac {F}{k}\frac {\ r}{\left ( 1-r^2\right ) }\right ) \sin \omega _nt+\frac {F}{k}\frac {1}{\left ( 1-r^2\right ) }\sin \omega t \tag {6}\end{equation}

Substitute numerical values, and plot the solution. \(F=10,\omega =10\) rad/sec.\(,k=2000,m=100,\omega _{n}=\sqrt {\frac {2000}{100}}=\allowbreak 4.472\,1,r=\frac {\omega }{\omega _{n}}=\frac {10}{\allowbreak 4.\,\allowbreak 472\,1}=\allowbreak 2.236\,1,\) then equation (6) becomes

\begin{align*} x\left ( t\right ) & =\left ( -\frac {10}{2000}\frac {\ 2.236\,1}{\left ( 1-2.236\,1^{2}\right ) }\right ) \sin 4.472\,1t+\frac {10}{2000}\frac {1}{\left ( 1-2.236\,1^{2}\right ) }\sin 10t\\ & =0.002795\sin 4.472\,1t-0.001250\sin 10t \end{align*}

In the following plot, we show the homogeneous solution and the particular solution separately, then show the general solution.

3.2.2.2 Part(b)

When \(x_{0}=0.05\) and \(v_{0}=0\) we obtain from (5)

\[ x\left ( t\right ) =0.05\cos \omega _{n}t+\left ( -\frac {F}{k}\frac {\ r}{\left ( 1-r^{2}\right ) }\right ) \sin \omega _{n}t+\frac {F}{k}\frac {1}{\left ( 1-r^{2}\right ) }\sin \omega t \]

Substitute numerical values found in part(a), then the solution becomes

\begin{align*} x\left ( t\right ) & =0.05\cos \allowbreak 4.472\,1t+\left ( -\frac {10}{2000}\frac {\ 2.236\,1}{\left ( 1-2.236\,1^{2}\right ) }\right ) \sin 4.472\,1t+\frac {10}{2000}\frac {1}{\left ( 1-2.236\,1^{2}\right ) }\sin 10t\\ & =0.05\cos \allowbreak 4.472\,1t+0.002795\sin 4.472\,1t-0.001250\sin 10t \end{align*}

In the following plot, we show the homogeneous solution and the particular solution separately, then show the general solution.

3.2.3 Problem 2.10

Following the approach taken in problem 2.7, the EQM is

\[ \ddot {x}+\omega _{n}^{2}x=\frac {F_{0}}{m}\cos \omega t \]

And \(x\left ( t\right ) =x_{h}\left ( t\right ) +x_{p}\left ( t\right ) \) where \(x_{h}\left ( t\right ) =A\cos \omega _{n}t+B\sin \omega _{n}t\). For \(x_{p}\left ( t\right ) ,\) guess \(x_{p}\left ( t\right ) =c_{1}\cos \omega t+c_{2}\sin \omega t\) and following the same steps in problem 2.7, we obtain

\[ \sin \omega t\left ( -\omega ^{2}c_{2}+\omega _{n}^{2}c_{2}\right ) +\cos \omega t\left ( -\omega ^{2}c_{1}+\omega _{n}^{2}c_{1}\right ) =\frac {F_{0}}{m}\cos \omega t \]

Notice that the above guess is valid only under the condition that \(\omega \neq \omega _{n}\) . Compare coefficients, we find \(c_{2}=0\) and

\[ c_{1}=\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\]

Hence

\[ x_{p}\left ( t\right ) =\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\cos \omega t \]

Then, the general solution is

\begin{equation} x\left ( t\right ) =A\cos \omega _{n}t+B\sin \omega _{n}t+\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\cos \omega t\tag {1}\end{equation}

Let, at \(t=0\), \(x\left ( 0\right ) =x_{0}\), and \(\dot {x}\left ( 0\right ) =v_{0}\),then from (1), we find

\begin{align*} x_{0} & =A+\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\\ A & =x_{0}-\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\end{align*}

And since

\[ \dot {x}\left ( t\right ) =-A\omega _{n}\sin \omega _{n}t+B\omega _{n}\cos \omega _{n}t-\omega \frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\sin \omega t \]

Then

\begin{align*} v_{0} & =B\omega _{n}\\ B & =\frac {v_{0}}{\omega _{n}}\end{align*}

Therefore, the general solution is (from (1))

\[ \fbox {$x\left ( t\right ) =\left ( x_0-\frac {F_0}{m}\frac {1}{\omega _n^2-\omega ^2}\right ) \cos \omega _nt+\frac {v_0}{\omega _n}\sin \omega _nt+\frac {F_0}{m}\frac {1}{\omega _n^2-\omega ^2}\cos \omega t$}\]

To make the response oscillate at frequency \(\omega \) only, we can set \(v_{0}=0\) to eliminate the \(\sin \omega _{n}t\), and set \(x_{0}=\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\) to eliminate the \(\cos \omega _{n}t\) term. Hence, the initial conditions are

\begin{align*} v_{0} & =0\\ x_{0} & =\frac {F_{0}}{m}\frac {1}{\omega _{n}^{2}-\omega ^{2}}\end{align*}

3.2.4 Problem 2.29

This is one degree of freedom system. Using \(x\) along the inclined surface as the generalized coordinates, we first obtain the Lagrangian \(L\)

We first note that \(k\Delta =mg\cos \theta \) and the mass will lose potential as it slides down the surface. We measure everything from the relaxed position (not the static equilibrium.) This is done to show more clearly that the angle do not affect the solution.

\[ L=T-U \]

Where

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

Hence

\begin{align*} L & =\frac {1}{2}m\dot {x}^{2}-\left ( \frac {1}{2}k\left ( x+\Delta \right ) ^{2}-mg\left ( x+\Delta \right ) \cos \theta \right ) \\ & =\frac {1}{2}m\dot {x}^{2}-\frac {1}{2}k\left ( x+\Delta \right ) ^{2}+mgx\cos \theta +mg\Delta \cos \theta \\ \frac {\partial L}{\partial \dot {x}} & =m\dot {x}\\ \frac {d}{dt}\frac {\partial L}{\partial \dot {x}} & =m\ddot {x}\\ \frac {\partial L}{\partial x} & =-k\left ( x+\Delta \right ) +mg\cos \theta \\ & =-kx-k\Delta +mg\cos \theta \end{align*}

But \(k\Delta =mg\cos \theta \), hence the above reduces to

\[ \frac {\partial L}{\partial x}=-kx \]

Hence the EQM is (using the Lagrangian equation)

\begin{align} \frac {d}{dt}\frac {\partial L}{\partial \dot {x}}-\frac {\partial L}{\partial x} & =F\cos \omega t\nonumber \\ m\ddot {x}+kx & =F\cos \omega t\nonumber \\ \ddot {x}+\omega _{n}^{2}x & =\frac {F}{m}\cos \omega t\tag {1}\end{align}

Where \(\omega _{n}^{2}=\frac {k}{m}\). We see that the angle \(\theta \) is not in the EQM. Hence the solution does not involve \(\theta \) and the oscillation magnitude is not affected by the angle. Intuitively, the reason for this is because the angle effect is already counted for to reach the static equilibrium. Once the system is in static equilibrium, the angle no longer matters as far as the solution is concerned.

3.2.5 Problem 2.46

From example 2.4.1, we note the following table

Also, from example 2.4.1, the mass of car 1 is \(1007kg\) and the mass of car 2 is \(1585kg\). Hence we write

\begin{align*} m_{1} & =1007kg\\ m_{2} & =1585kg \end{align*}

To find the deflection of the car, we use equation 2.70 in the book, which is

\[ X=Y\sqrt {\frac {1+\left ( 2\xi r\right ) ^{2}}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\]

Where \(X\) is the magnitude of the steady state deflection and \(Y\) is the magnitude of the base deflection, which is given as \(0.01\) meters in the example.

Hence, for each speed, we calculate \(\omega _{p}\) and then we find \(\omega _{n}=\sqrt {\frac {k}{m_{1}+m_{p}}}\) and then find \(r=\frac {\omega _{p}}{\omega _{n}}\) and then find \(\xi =\frac {c}{2\sqrt {k\left ( m_{1}+m_{p}\right ) }}\) and then using equation(1), we calculate \(X\). This is done for each different speed (all for car \(m_{1}).\) Next, we do the same for car \(m_{2}\). These calculation are shown in the following table. Note also that \(c=2000\) N s/m as given in the example and \(k=4\times 10^{4}\) N/m

car 1

\(v\ (km/h)\)	\(\omega _{p}=0.2909\ v\)	\(\omega _{n}=\sqrt {\frac {k}{m_{1}+m_{p}}}\)	\(r=\frac {\omega _{p}}{\omega _{n}}\)	\(\xi =\frac {c}{2\sqrt {k\left ( m_{1}+m_{p}\right ) }}\)	\(X=Y\sqrt {\frac {1+\left ( 2\xi r\right ) ^{2}}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\ (cm)\)
\(20\)	\(\allowbreak 5.818\)	\(5.756\,7\)	\(1.010\,6\)	\(0.143\,92\)	\(\allowbreak 3.571\)
\(80\)	\(23.\,\allowbreak 272\)	\(5.756\,7\)	\(4.042\,6\)	\(0.143\,92\)	\(0.0997\)
\(100\)	\(29.09\)	\(5.756\,7\)	\(\allowbreak 5.053\,2\)	\(0.143\,92\)	\(0.0718 \)
\(150\)	\(43.635\)	\(5.756\,7\)	\(\allowbreak 7.579\,9\)	\(0.143\,92\)	\(0.0425\,\)

car 2

\(v\ (km/h)\)	\(\omega _{p}=0.2909\ v\)	\(\omega _{n}=\sqrt {\frac {k}{m_{2}+m_{p}}}\)	\(r=\frac {\omega _{p}}{\omega _{n}}\)	\(\xi =\frac {c}{2\sqrt {k\left (m_{2}+m_{p}\right ) }}\)	\(X=Y\sqrt {\frac {1+\left ( 2\xi r\right )^{2}}{\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\ (cm)\)
\(20\)	\(5.818\)	\(4.7338\)	\(1.229\)	\(0.11835\)	\(1.7726\)
\(80\)	\(23.272\)	\(4.7338\)	\(4.9161\)	\(0.11835\)	\(0.066141\)
\(100\)	\(29.09\)	\(4.7338\)	\(6.1452\)	\(0.11835\)	\(0.04797\)
\(150\)	\(43.635\)	\(4.7338\)	\(9.2178\)	\(0.11835\)	\(0.02857\)

Observations: The heavier car (car 2) has smaller defection (\(X\) values) for all speeds.  Adding passengers, causes \(\omega _{n}\) to change. This results in making the deflection smaller when passengers are in the car as compared without them. Heaver cars and heavier passenger results in smaller deflection values. For the lighter car however, adding the passenger did not result in smaller deflection for all speeds. For speed \(v=20\), adding the passenger caused a larger defection (\(3.19\) vs. \(3.571\)). As car 1 speed became larger, the deflection became smaller for both cars.

So, in conclusion: lighter cars have larger deflections at bumps, and the faster the car, the smaller the deflection.

3.2.6 Problem 2.57

Given \(m=120\ kg,k=800\times 10^{3}\)N/M, \(c=500\ \)kg/s, and mass \(m_{0}\) has angular speed of \(\omega _{r}=\frac {3000\times 2\pi }{60}\) \(=100\pi \) radians per seconds

3.2.6.1 Part(a)

The rotating mass will cause a downward force as the result of the centripetal force \(m_{0}e\omega _{r}^{2}\sin \left ( \omega _{r}t\right ) \). Hence the reaction to this force on the machine will be in the upward direction. Hence

\[ F_{r}=m_{0}e\omega _{r}^{2}\sin \left ( \omega _{r}t\right ) \]

Hence the machine equation of motion is

\begin{align} m\ddot {x}+c\dot {x}+kx & =m_{0}e\omega _{r}^{2}\sin \left ( \omega _{r}t\right ) \nonumber \\ \ddot {x}+2\xi \omega _{n}\dot {x}+\omega _{n}^{2}x & =\frac {m_{0}}{m}e\omega _{r}^{2}\sin \left ( \omega _{r}t\right ) \tag {1}\end{align}

By guessing \(x_{p}=X\sin (\omega _{r}t-\theta )\) then we find that (The method of undetermined coefficients is used, derivation is show in text book at page 115)

\begin{equation} X=\frac {m_{o}e}{m}\frac {r^{2}}{\sqrt {\left ( 1-r^{2}\right ) ^{2}+\left ( 2\xi r\right ) ^{2}}}\tag {2}\end{equation}

This is the maximum magnitude of motion in steady state. In the above, \(r=\frac {\omega _{r}}{\omega _{n}}\). Hence to find \(X\) we substitute the given values in the above expression. We first note that we are told that \(m_{0}e\omega _{r}^{2}=374N\), hence \(m_{o}e=\frac {374}{\omega _{r}^{2}}\) but we found that \(\omega _{r}=100\pi \) rad/sec, hence

\[ m_{o}e=\frac {374}{\left ( 100\ \pi \right ) ^{2}}=0.00\allowbreak 3789\,4 \]

And

\[ r=\frac {\omega _{r}}{\omega _{n}}=\frac {100\pi }{\sqrt {\frac {k}{m}}}=\frac {100\ \pi }{\sqrt {\frac {800\times 10^{3}}{120}}}=\allowbreak 3.\,\allowbreak 847\,6 \]

And

\[ \xi =\frac {c}{2\sqrt {km}}=\frac {500}{2\sqrt {800\times 10^{3}\times 120}}=\allowbreak 0.02551\,6 \]

Substituting into (2) gives

\begin{align*} X & =\left ( \frac {0.00\allowbreak 3789\,4}{120}\right ) \frac {3.\,\allowbreak 847\,6^{2}}{\sqrt {\left ( 1-3.\,\allowbreak 847\,6^{2}\right ) ^{2}+\left ( 2\times 0.02551\,6\times 3.\,\allowbreak 847\,6\right ) ^{2}}}\\ & =3.\,\allowbreak 386\,3\times 10^{-5}\text { }meter \end{align*}

3.2.6.2 Part(b)

We are told that \(m_{o}=0.01\times m\), hence \(m_{o}=0.01\times 120=1\). And since we are told that \(m_{0}e\omega _{r}^{2}=374N\) then

\[ e=\frac {374}{m_{0}\omega _{r}^{2}}=\frac {374}{1. \times \left ( 100\pi \right ) ^{2}}=3.8\times 10^{-3}\text { meter}\]

3.2.7 Key for HW2

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