User:Egm4313.s12.team17.Li

=Problem R1.2 Equation of Motion Derivation=

Given: Two Spring-Mass-Damper Systems (Figure A and Figure B) Find: Derive the Equation of Motion for each of the systems.

A)

B)

=Solution=

Part A
For Figure A, we see that the system is fixed on both ends with an applied force $$ f(t) $$. The Free Body Diagram of the mass is as follows:

Note that for the forces, we have:
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\displaystyle f_c = cy'

$$ and
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\displaystyle f_k = ky $$
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Applying Newton's second law to the system:


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\sum F = f(t)-f_k-f_c=ma $$ By definition, the acceleration can be as follows:
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a=y'' $$ Substituting the known values, we can now obtain the equation of motion:
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\displaystyle my''+cy'+ky=f(t) $$
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Part 2
For Figure B, the system is only fixed on one end with an applied force, $$ f(t) $$. Therefore the free body diagram is as follows:

Since the spring and damper forces are along the same line and due to Newton's Third Law, we can make this definition:
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f_T=f_k=f_c $$ If Newtons Second Law is applied to the system the same way as in Figure A:
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\sum F = ma $$
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\sum F = f(t) - f_T = ma $$ Acceleration can be defined as:
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a=y'' $$ Where the equation of motion now becomes:
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f(t) - f_T = ma=my'' $$ Rearranging, the Equation of motion is:
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\displaystyle my'' + f_T = f(t) $$
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=Problem 5= Consider the spring-mass damper system on p.53-2, and the following cases for the roots of the characteristic equations.

1. $$ \lambda_1 = -0.5, \lambda_2 = -1.5 $$ 2. $$ \lambda_1 = \lambda_2 = -0.5 $$ 3. $$ \lambda_1 = -0.5 + i2, \lambda_2 = -0.5 - i2 $$

Initial Conditions: $$ y(0) = 1, y'(0) = 0 $$

Case 1: Overdamped

$$ \Delta = a^2 - 4b > 0 $$ $$ (\lambda + 0.5)(\lambda + 1.5) = 0 $$ $$ \lambda^2 + 4\lambda + 0.75 = 0 $$ Therefore L2-ODE-CC: $$ y_h'' + 4y_h' + 0.75y_h = 0 $$ Solution: $$ y = C_1e^{(\lambda_1x)} + C_2e^{(\lambda_2x)} $$ $$ y = C_1e^{(-0.5x)} + C_2e^{(-1.5x)} $$ ............... (1) $$ y' = -0.5C_1e^{(-0.5x)} + (-1.5)C_2e^{(-1.5x)} $$ .............(2) Using the two initial conditions and solving equation (1) and (2) the coefficients are as follows: $$ C_1 = 1.5, C_2 = -0.5 $$ Therefore, $$ y = 1.5e^{(-0.5x)} + -0.5e^{(-1.5x)} $$



Case 2: Critically damped

$$ \Delta = a^2 - 4b = 0 $$ $$ (\lambda + 0.5)(\lambda + 0.5) = 0 $$ $$ \lambda^2 + \lambda + 0.25 = 0 $$ Therefore L2-ODE-CC: $$ y_h'' + y_h' + 0.25y_h = 0 $$ Solution: $$ y = C_1e^{(\lambda_1x)} + C_2xe^{(\lambda_2x)} $$ $$ y = C_1e^{(-0.5x)} + C_2xe^{(-0.5x)} $$ ............... (3) $$ y' = -0.5C_1e^{(-0.5x)} + (-0.5)C_2xe^{(-0.5x)} + C_2e^{(-0.5x)} $$ .............(4) Using the two initial conditions and solving equation (3) and (4) the coefficients are as follows: $$ C_1 = 2/3, C_2 = 1/3 $$ Therefore, $$ y = (2/3)e^{(-0.5x)} + (1/3)xe^{(-0.5x)} $$



Case 3: Underdamped

$$ \Delta = a^2 - 4b < 0 $$ $$ \alpha = -0.5, \beta = 2 $$ $$ (\lambda - (-0.5 + 2i))(\lambda - (-0.5 - 2i)) = 0 $$ $$ \lambda^2 + 1\lambda + 2.25 = 0 $$ Therefore L2-ODE-CC: $$ y_h'' + y_h' + 2.25y_h = 0 $$ Solution: $$ y = C_1e^{(\alpha x)}cos(\beta x) + C_2e^{(\alpha x)}sin(\beta x) $$ $$ y = C_1e^{(-0.5x)}cos(2x) + C_2e^{(-0.5x)}sin(2x) $$ ......... (5) $$ y' = -0.5C_1e^{(-0.5x)}cos(2x) - 2C_1e^{(-0.5x)}sin(2x) - 0.5C_2e^{(-0.5x)}sin(2x) + 2C_2e^{(-0.5x)}cos(2x) $$ ......(6) Using the two initial conditions and solving equation (5) and (6) the coefficients are as follows: $$ C_1 = 1, C_2 = .25 $$ Therefore, $$ y =e^{(-0.5x)}cos(2x) + 0.25e^{(-0.5x)}sin(2x) $$