Math Problem Statement

The motion of a​ mass-spring system with damping is governed by y double prime left parenthesis t right parenthesis plus 6 y prime left parenthesis t right parenthesis plus ky left parenthesis t right parenthesis equals 0​; y left parenthesis 0 right parenthesisequals1​, and y prime left parenthesis 0 right parenthesisequals0. Find the equation of motion and sketch its graph for kequals7​, 9​, and 11. Question content area bottom Part 1 What is the equation of motion for kequals7​? ​y(t)equals    e Superscript negative 3 t Baseline left parenthesis cosine left parenthesis StartRoot 2 EndRoot t right parenthesis plus StartFraction 3 StartRoot 2 EndRoot Over 2 EndFraction sine left parenthesis StartRoot 2 EndRoot t right parenthesis right parenthesis

Solution

To solve for the equation of motion for the given second-order linear differential equation with damping:

y(t)+6y(t)+ky(t)=0y''(t) + 6y'(t) + ky(t) = 0 with initial conditions: y(0)=1,y(0)=0y(0) = 1, \quad y'(0) = 0

Step 1: Find the characteristic equation The characteristic equation for this differential equation is obtained by assuming solutions of the form y(t)=erty(t) = e^{rt}. Substituting this into the differential equation yields:

r2+6r+k=0r^2 + 6r + k = 0

For k=7k = 7, the characteristic equation becomes:

r2+6r+7=0r^2 + 6r + 7 = 0

Step 2: Solve the characteristic equation We can solve the characteristic equation using the quadratic formula:

r=b±b24ac2ar = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For our equation r2+6r+7=0r^2 + 6r + 7 = 0, a=1a = 1, b=6b = 6, and c=7c = 7. Substituting these values into the quadratic formula:

r=6±624(1)(7)2(1)=6±36282=6±82=6±222r = \frac{-6 \pm \sqrt{6^2 - 4(1)(7)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 28}}{2} = \frac{-6 \pm \sqrt{8}}{2} = \frac{-6 \pm 2\sqrt{2}}{2}

This simplifies to:

r=3±2r = -3 \pm \sqrt{2}

Thus, the roots are complex and given by:

r1=3+2,r2=32r_1 = -3 + \sqrt{2}, \quad r_2 = -3 - \sqrt{2}

Step 3: Write the general solution Since the roots are complex, the general solution to the differential equation is of the form:

y(t)=e3t(C1cos(2t)+C2sin(2t))y(t) = e^{-3t} \left( C_1 \cos(\sqrt{2}t) + C_2 \sin(\sqrt{2}t) \right)

Step 4: Apply initial conditions

We are given y(0)=1y(0) = 1 and y(0)=0y'(0) = 0.

At t=0t = 0:

y(0)=e3(0)(C1cos(0)+C2sin(0))=1C1=1y(0) = e^{-3(0)} \left( C_1 \cos(0) + C_2 \sin(0) \right) = 1 \Rightarrow C_1 = 1

Now, differentiate the solution to find y(t)y'(t):

y(t)=e3t(3C1cos(2t)2C1sin(2t)3C2sin(2t)+2C2cos(2t))y'(t) = e^{-3t} \left( -3C_1 \cos(\sqrt{2}t) - \sqrt{2}C_1 \sin(\sqrt{2}t) - 3C_2 \sin(\sqrt{2}t) + \sqrt{2}C_2 \cos(\sqrt{2}t) \right)

At t=0t = 0, applying the initial condition y(0)=0y'(0) = 0:

y(0)=e3(0)(3C1cos(0)+2C2cos(0))=03C1+2C2=0y'(0) = e^{-3(0)} \left( -3C_1 \cos(0) + \sqrt{2}C_2 \cos(0) \right) = 0 \Rightarrow -3C_1 + \sqrt{2}C_2 = 0

Substitute C1=1C_1 = 1:

3+2C2=0C2=32-3 + \sqrt{2}C_2 = 0 \Rightarrow C_2 = \frac{3}{\sqrt{2}}

Step 5: Write the final solution

Thus, the equation of motion for k=7k = 7 is:

y(t)=e3t(cos(2t)+322sin(2t))y(t) = e^{-3t} \left( \cos(\sqrt{2}t) + \frac{3\sqrt{2}}{2} \sin(\sqrt{2}t) \right)

This is the required equation of motion for k=7k = 7.

Would you like help with the equations for k=9k = 9 and k=11k = 11, or with any other part of the problem?

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Second-Order Linear Equations
Damped Harmonic Motion
Quadratic Equations

Formulas

Characteristic equation: r^2 + 6r + k = 0
General solution: y(t) = e^(rt) (C1 * cos(ωt) + C2 * sin(ωt))
Quadratic formula: r = (-b ± √(b² - 4ac)) / 2a

Theorems

Solution to second-order linear differential equations with constant coefficients

Suitable Grade Level

Grades 11-12