A mass m is attached to both a spring​ (with given spring constant​ k) and a dashpot​ (with given damping constant​ c). The mass is set in motion with initial position x 0 and initial velocity v 0. Find the position function​ x(t) and determine whether the motion is​ overdamped, critically​ damped, or underdamped. If it is​ underdamped, write the position function in the form ​x(t)equalsUpper C 1 e Superscript negative pt Baseline cosine left parenthesis omega 1 t minus alpha 1 right parenthesis. ​Also, find the undamped position function ​u(t)equalsUpper C 0 cosine left parenthesis omega 0 t minus alpha 0 right parenthesis that would result if the mass on the spring were set in motion with the same initial position and​ velocity, but with the dashpot disconnected​ (so cequals​0). ​Finally, construct a figure that illustrates the effect of damping by comparing the graphs of​ x(t) and​ u(t).
mequals4​, cequals20​, kequals169​, x 0equals5​, v 0equals14
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Part 1
​x(t)equals
  
enter your response here​, which means the system is 
▼ 
underdamped.
overdamped.
critically damped.
​(Use integers or decimals for any numbers in the expression. Round to four decimal places as needed. Type any angle measures in radians. Use angle measures greater than or equal to 0 and less than 2pi​.)


To analyze the motion of the mass attached to a spring and dashpot, we start with the parameters provided:

- Mass \( m = 4 \)
- Damping constant \( c = 20 \)
- Spring constant \( k = 169 \)
- Initial position \( x_0 = 5 \)
- Initial velocity \( v_0 = 14 \)

### Part 1: Determine the Type of Damping

The characteristic equation of the system is given by:

\[
m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0
\]

This translates to:

\[
4 \frac{d^2x}{dt^2} + 20 \frac{dx}{dt} + 169x = 0
\]

Dividing through by \( m = 4 \):

\[
\frac{d^2x}{dt^2} + 5 \frac{dx}{dt} + \frac{169}{4} x = 0
\]

The damping ratio \( \zeta \) is defined as:

\[
\zeta = \frac{c}{2\sqrt{mk}}
\]

Calculating \( \zeta \):

\[
\sqrt{mk} = \sqrt{4 \cdot 169} = \sqrt{676} = 26
\]
\[
\zeta = \frac{20}{2 \cdot 26} = \frac{20}{52} = \frac{5}{13} \approx 0.3846
\]

Since \( \zeta < 1 \), the system is **underdamped**.

### Position Function \( x(t) \)

For an underdamped system, the position function has the form:

\[
x(t) = C_1 e^{-pt} \cos(\omega_1 t - \alpha_1)
\]

where:
- \( p = \frac{c}{2m} = \frac{20}{8} = 2.5 \)
- \( \omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5 \)
- \( \omega_1 = \sqrt{\omega_0^2 - p^2} = \sqrt{(6.5)^2 - (2.5)^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6 \)

To find \( C_1 \) and \( \alpha_1 \), we use the initial conditions:

1. \( x(0) = 5 \):
   \[
   x(0) = C_1 \cos(-\alpha_1) = C_1 \cos(\alpha_1) = 5
   \]

2. \( v(0) = 14 \):
   \[
   v(t) = \frac{dx}{dt} = C_1 e^{-pt} \left(-p \cos(\omega_1 t - \alpha_1) - \omega_1 \sin(\omega_1 t - \alpha_1)\right)
   \]
   Evaluating at \( t = 0 \):
   \[
   v(0) = C_1 \left(-p \cos(\alpha_1) - \omega_1 \sin(\alpha_1)\right) = 14
   \]

Setting up the equations:

1. \( C_1 \cos(\alpha_1) = 5 \)
2. \( -2.5 C_1 \cos(\alpha_1) - 6 C_1 \sin(\alpha_1) = 14 \)

Substituting \( \cos(\alpha_1) = \frac{5}{C_1} \) into the second equation yields:

\[
-2.5(5) - 6 C_1 \sin(\alpha_1) = 14
\]
\[
-12.5 - 6 C_1 \sin(\alpha_1) = 14
\]
\[
-6 C_1 \sin(\alpha_1) = 26.5 \implies C_1 \sin(\alpha_1) = -\frac{26.5}{6} \approx -4.4167
\]

To solve for \( C_1 \) and \( \alpha_1 \):

Using Pythagorean identity \( \cos^2(\alpha_1) + \sin^2(\alpha_1) = 1 \):

Let \( C_1 = \sqrt{(5)^2 + (-4.4167)^2} \):

Calculating:

\[
C_1 \approx \sqrt{25 + 19.5} \approx \sqrt{44.5} \approx 6.6733
\]

Using this, find \( \alpha_1 \):

\[
\cos(\alpha_1) = \frac{5}{C_1}, \quad \sin(\alpha_1) = -\frac{4.4167}{C_1}
\]

Now we can write:

\[
x(t) = 6.6733 e^{-2.5t} \cos(6t - \alpha_1)
\]

### Part 2: Undamped Position Function \( u(t) \)

The undamped position function without the dashpot is:

\[
u(t) = C_0 \cos(\omega_0 t - \alpha_0)
\]

With \( C_0 = 5 \) and \( \omega_0 = 6.5 \).

### Summary

1. **Position Function:**
   \[
   x(t) = 6.6733 e^{-2.5t} \cos(6t - \alpha_1)
   \]
   The system is **underdamped**.

2. **Undamped Position Function:**
   \[
   u(t) = 5 \cos(6.5t - \alpha_0)
   \]

### Graphical Comparison

To compare the graphs of \( x(t) \) and \( u(t) \), a plot can be created showing both functions over time, illustrating the effect of damping.

Would you like assistance with the plot or any other details? Here are some follow-up questions you might consider:

1. What are the specific values of \( \alpha_1 \) and \( \alpha_0 \)?
2. Can you explain how to derive the parameters for the undamped system?
3. How would changing the damping constant affect the motion?
4. What are real-world applications of such a damped harmonic oscillator?
5. How does the motion change if the mass is increased?

A mass m is attached to both a spring with given spring constant k and a dashpot with given damping constant c. The mass is set in motion with initial position x0 and initial velocity v0. Find the position function x(t) and determine whether the motion is overdamped, critically damped, or underdamped. If it is underdamped, write the position function in the form x(t)=C1e^(-pt)cos(ω1t−α1). Also, find the undamped position function u(t)=C0cos(ω0t−α0) that would result if the mass on the spring were set in motion with the same initial position and velocity, but with the dashpot disconnected. Finally, construct a figure that illustrates the effect of damping by comparing the graphs of x(t) and u(t). m=4, c=20, k=169, x0=5, v0=14

Explore the dynamics of a mass-spring-dashpot system. This problem covers the determination of the position function x(t), the type of damping (underdamped), and the undamped position function u(t), using given parameters m=4, c=20, k=169, x0=5, v0=14.

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