User:Egm6321.f11.team4.shin.js/HW3

= Problem 3.10 - Free vibration of coupled pendulums = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]

Given
The equations of motion of coupled pendulums: Parameters: $$\displaystyle \begin{align} \mbox{Pendulums: } &a=0.3, \mbox{ } l=1, \mbox{ } k = 0.2 \\ &m_1 g = 3, \mbox{ } m_2 g = 6 \end{align} $$ $$\displaystyle \mbox{No applied forces: } u_1=u_2=0 $$ $$\displaystyle \begin{align} \mbox{Ini}&\mbox{tial conditions: } \\ &\theta_1 (0)=0, \mbox{ } \dot{\theta_1}(0)=-2 \\ &\theta_2 (0)=0, \mbox{ } \dot{\theta_2}(0)=+1 \\ \end{align} $$

Find
1. Integrate the system in matrix form for $$\displaystyle t \in [0,7] $$ 2. Use (2) p.15-2 to find the solution 3. Plot the results

Integrate the system in matrix form
In order to represent the Eq. (10.1) in matrix form (1) From [[media:Pea1.f11.mtg14.djvu|p.14-4]], the following state variables are defined. Then, the Eq. (10.1) can be rewritten as following. Rearrange the terms in the Eq. (10.3), Finally, the follow matrix equation is obtained.

The following is a matlab code to integrate the system.
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The following figure is the plot of the integration results.
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Use (2) p.15-2 to find the solution
Since no input is applied to the system, the second term in the Eq. (10.6) is zero. The matlab code shows the problem-solving using the Eq. (10.7).
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The following figure is the plot of the results solved using the Eq. (10.7).
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Plot the results
Refer to Figure 10.1 and Figure 10.2 $$\displaystyle x_1 $$ in Figure 10.1 and Figure 10.2 are $$\displaystyle \theta_1(t) $$ $$\displaystyle x_3 $$ in Figure 10.1 and Figure 10.2 are $$\displaystyle \theta_2(t) $$

Author
Contributed by Shin

= Problem 3.13 - The 2nd exactness condition = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]

Given
2nd exactness condition for N2-ODEs in the class note [[media:Pea1.f11.mtg16.djvu|16-5]] And, the N2-ODE equation

Find
1. Derive the Eq. (13.2), i.e., the 2nd relation in the 2nd exactness condition. 2. Derive the Eq. (13.1), i.e., the 1st relation in the 2nd exactness condition. 3. Verify that the Eq. (13.3) satisfies the 2nd exactness condition.

Derive the 2nd relation in the 2nd exactness condition.
From the equation (3), (4) in the class note [[media:Pea1.f11.mtg16.djvu|16-4]], g(x,y,p) and f(x,y,p) are as below. Differentiating the Eq. (13.4) with respect to p, Differentiating the Eq. (13.6) with respect to p, Rewriting the Eq. (13.7) using the symmetric property of mixed 2nd partial derivatives of the function $$\displaystyle \phi(x,y,p) $$ as below, Substituting the Eq. (13.5) into the Eq. (13.8),

- The Eq. (13.9) shows the derivation of the 2nd relation in the 2nd exactness condition.

Derive the 1st relation in the 2nd exactness condition.
To solve this problem, 3 relations (Eq. 13.10) for the symmetry of mixed 2nd partial derivatives are exploited. From the first relation in the Eq. (13.10), Express $$\displaystyle \phi_y $$ the Eq. (13.12) in terms of $$\displaystyle f, g, p $$, From the second relation in the Eq. (13.10), Rewrite the Eq. (13.14) using the Eq. (13.4), (13.5), Express $$\displaystyle \phi_x $$ the Eq. (13.15) in terms of $$\displaystyle f, g, p $$, Take the partial derivative of the Eq. (13.13) with respect to x, Take the partial derivative of the Eq. (13.16) with respect to y, Use the third relation in the Eq. (13.10), Move the terms with f to the LHS and move the terms with g to the RHS, - The Eq. (13.20) shows the derivation of the 1st relation in the 2nd exactness condition.

Verify that the Eq. (13.3) satisfies the 2nd exactness condition.
From the 1st exactness condition for N2-ODEs in the class note [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]], The 1st relation in the 2nd exactness condition is satisfied as below. Also, the 2nd relation in the 2nd exactness condition is satisfied as below.

Author
Contributed by Shin

= Problem 3.14 - Find h(x,y) = From [[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]

Given
The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|17-1]]. where $$\displaystyle p(x) := y^{\prime}(x) $$. The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|18-1]]

Find
Find $$\displaystyle h(x,y) $$ without assuming a priority that h=constant.

Solution
The Eq. (14.2) can be rewritten in the following form. Since y is the function of x, the Eq (14.3) is the partial derivative of the function h(x,y) with respect to x. So, the LHS term in the Eq. (14.3) is rewritten as follows.

Then, the following is true from the Eq. (14.3), (14.4).

The Eq. (14.5) implies that h(x,y) is the function of only y.

Substituting the Eq. (14.5) into the Eq. (14.2) would result the following. It is concluded that $$\displaystyle h_y = 0 $$ because if $$\displaystyle y^{\prime} = 0 $$ the solution would be trivial.

Author
Contributed by Shin

= Problem 3.15 - Verification of the 2nd exactness condition = From [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]

Given
The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|18-3]].

2nd exactness condition for N2-ODEs

Find
Verify whether the Eq. (15.1) satisfies the 2nd exactness condition (Eq. 15.2).

Solution
As proven in the class note [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]], the Eq. (15.1) satisfies the 1st exactness condition for N2-ODEs.

The first condition (Eq. 15.2)
Computing each term in the Eq. (15.2) $$\displaystyle f_{xx} :$$ $$\displaystyle 2pf_{xy} :$$ $$\displaystyle f_{xy} = 0$$, since $$\displaystyle f_x$$ is a function of $$\displaystyle x$$ and $$\displaystyle y^{\prime}$$. $$\displaystyle p^2 f_{yy} :$$ $$\displaystyle f_{y} = 0, f_{yy} = 0 $$, since $$\displaystyle f$$ is a function of $$\displaystyle x$$ and $$\displaystyle y^{\prime}$$. $$\displaystyle g_{xp} :$$ $$\displaystyle g_{yp} :$$ $$\displaystyle g_{y} :$$ Substituting the Eq. (15.5)-(15.8) into the LHS of the Eq. (15.2), Substituting the Eq. (15.9)-(15.12) into the RHS of the Eq. (15.2), - Since the Eq. (15.13) and the Eq. (15.14) are the same equations, the first condition is satisfied.

The second condition (Eq. 15.3)
Computing each term in the Eq. (15.3) $$\displaystyle f_{xp} :$$ $$\displaystyle f_{yp} :$$ $$\displaystyle f_{y} :$$ $$\displaystyle g_{pp} :$$ Substituting the Eq. (15.13)-(15.15) into the LHS of the Eq. (15.3), Substituting the Eq. (15.16) into the RHS of the Eq. (15.3), - Since the Eq. (15.17) and the Eq. (15.18) are the same equations, the second condition is satisfied.

Author
Contributed by Shin