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?