User:Egm4313.s12.team18.gibson.mg

=Michelle Gibson=

Useful Links
Team 18 Course Wiki Course Website Report Table

Problem
Derive the equation of motion of the spring-mass-dashpot in fig. 5.3 in K 2011 p 85 with an applied force r(t) on the ball.



Solution
Find: The equation of motion with the applied force, r(t).




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$$  \displaystyle {F_c = cy'} $$     (1)
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$$  \displaystyle {F_k = ky} $$     (2)
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Using Newton's Second Law:
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$$  \displaystyle {\Sigma\vec{F}=m\vec{a}} $$
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$$  \displaystyle {\vec{a}=y''} $$
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From the Free Body Diagram above:


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$$  \displaystyle -ma=F_c+F_k-r(t) $$
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Or


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$$  \displaystyle r(t)=F_c+F_k+ma $$     (3)
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Plug Equation 1 and Equation 2 into Equation 3.


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$$  \displaystyle {r(t)=cy'+ky+my} $$
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Or


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$$  \displaystyle {r(t)=my''+cy'+ky} $$
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Problem
K 2011 p. 60 problems 8 and 15. These problems give you an equation and ask for the general solution. Once the general solution has been found, use substitution to check the answer.

Problem 8) $$\displaystyle y^{''}+y^{'}+3.25y=0 $$

Problem 15) $$\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi)y=0 $$

Solution
Problem 8)
 * $$ \displaystyle y^{''}+y^{'}+3.25y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+\lambda+3.25 =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:


 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12 $$

Or
 * $$\displaystyle a^{2}-4b=-12 $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:


 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{3.25-\frac{1}{4}(1)^{2}}) \rightarrow \omega=\sqrt{3} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{3}$$
 * $$ \displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

Checking the solution by using substitution:


 * $$\displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$


 * $$\displaystyle y^{'}=\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{-2}+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{'}=-\frac{1}{2}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$


 * $$\displaystyle y^{''}=-\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{4}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{''}=-\frac{1}{4}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$


 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$


 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-x/2}Bsin(\sqrt{3}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

Problem 15)
 * $$ \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+0.54\lambda+(0.0729+\pi ) =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:


 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi)=-4\pi $$

Or
 * $$\displaystyle a^{2}-4b=-4\pi $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:


 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{(0.0729-\pi)-\frac{1}{4}(0.54)^{2}}) =\sqrt{\pi} \rightarrow \omega=\sqrt{\pi} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{\pi}$$ $$ \displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

Checking the solution by using substitution:


 * $$\displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$


 * $$\displaystyle y^{'}=-0.27{e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))}+e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)) $$


 * $$\displaystyle y^{''}=0.0729e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$


 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$


 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$


 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-0.27}Bsin(\sqrt{\pi}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.