Math Problem Statement

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). mequals1​, cequals10​, kequals25​, x 0equals4​, v 0equalsnegative 5 Question content area bottom Part 1 ​x(t)equals    4 e Superscript negative 5 t Baseline plus 15 t e Superscript negative 5 t​, which means the system is 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​.) Part 2 The undamped position is ​u(t)equals    4.1231 cosine left parenthesis 5 t minus 6.0382 right parenthesis. ​(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​.)

Solution

To solve this problem, we start with the given parameters:

  • Mass, m=1m = 1
  • Damping constant, c=10c = 10
  • Spring constant, k=25k = 25
  • Initial position, x0=4x_0 = 4
  • Initial velocity, v0=5v_0 = -5

Step 1: Determine the Type of Damping

First, we find the damping ratio ζ\zeta using the formula:

ζ=c2mk\zeta = \frac{c}{2\sqrt{mk}}

Substituting the values:

ζ=102125=1010=1\zeta = \frac{10}{2\sqrt{1 \cdot 25}} = \frac{10}{10} = 1

Since ζ=1\zeta = 1, the system is critically damped.

Step 2: Position Function x(t)x(t)

For a critically damped system, the position function can be expressed as:

x(t)=(C1+C2t)eζω0tx(t) = (C_1 + C_2 t)e^{-\zeta \omega_0 t}

where ω0=km\omega_0 = \sqrt{\frac{k}{m}}.

Calculating ω0\omega_0:

ω0=251=5\omega_0 = \sqrt{\frac{25}{1}} = 5

Thus, x(t)x(t) becomes:

x(t)=(C1+C2t)e5tx(t) = (C_1 + C_2 t)e^{-5t}

Step 3: Finding Constants C1C_1 and C2C_2

Using the initial conditions to find C1C_1 and C2C_2:

  1. At t=0t = 0:

x(0)=C1=4x(0) = C_1 = 4

  1. Velocity is the derivative of position:

v(t)=dxdt=ddt[(C1+C2t)e5t]v(t) = \frac{dx}{dt} = \frac{d}{dt}[(C_1 + C_2 t)e^{-5t}]

Using the product rule:

v(t)=(C2e5t+(C1+C2t)(5)e5t)=e5t(C25C15C2t)v(t) = (C_2 e^{-5t} + (C_1 + C_2 t)(-5)e^{-5t}) = e^{-5t}(C_2 - 5C_1 - 5C_2 t)

At t=0t = 0:

v(0)=C25C1=5    C220=5    C2=15v(0) = C_2 - 5C_1 = -5 \implies C_2 - 20 = -5 \implies C_2 = 15

Thus, the position function is:

x(t)=(4+15t)e5tx(t) = (4 + 15t)e^{-5t}

Step 4: Undamped Position Function u(t)u(t)

For the undamped system (where c=0c = 0), the position function is:

u(t)=C0cos(ω0tα0)u(t) = C_0 \cos(\omega_0 t - \alpha_0)

Using x(0)=4x(0) = 4 and v(0)=5v(0) = -5:

  1. At t=0t = 0:

u(0)=C0cos(α0)=C0    C0=4u(0) = C_0 \cos(-\alpha_0) = C_0 \implies C_0 = 4

  1. Velocity is:

v(t)=C0ω0sin(ω0tα0)v(t) = -C_0 \omega_0 \sin(\omega_0 t - \alpha_0)

At t=0t = 0:

v(0)=C0ω0sin(α0)=45sin(α0)=20sin(α0)v(0) = -C_0 \omega_0 \sin(-\alpha_0) = -4 \cdot 5 \sin(-\alpha_0) = -20 \sin(-\alpha_0)

Setting v(0)=5v(0) = -5:

20sin(α0)=5    sin(α0)=14    α0=arcsin(14)0.2527-20 \sin(-\alpha_0) = -5 \implies \sin(-\alpha_0) = \frac{1}{4} \implies \alpha_0 = -\arcsin\left(\frac{1}{4}\right) \approx -0.2527

Thus, the undamped position function is:

u(t)=4cos(5t+0.2527)u(t) = 4 \cos(5t + 0.2527)

Summary of Results

  1. Position function x(t)=(4+15t)e5tx(t) = (4 + 15t)e^{-5t} (critically damped).
  2. Undamped position u(t)=4cos(5t+0.2527)u(t) = 4 \cos(5t + 0.2527).

Step 5: Graph Comparison

To illustrate the effect of damping, we would plot both x(t)x(t) and u(t)u(t) on the same graph.

If you need the graph created, I can assist with that as well. Do you have any questions or need further details on any step? Here are a few follow-up questions:

  1. Would you like to see the graphical representation of the functions?
  2. Do you want a deeper explanation of the damping types?
  3. Would you like to learn how to derive these formulas step-by-step?
  4. Are you interested in examples of real-world applications of these concepts?
  5. Do you need assistance with related differential equations?

Tip: When analyzing damping in physical systems, remember that the damping ratio ζ\zeta plays a crucial role in determining the behavior of the system, guiding the design of systems like car suspensions or mechanical watches.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Dynamics
Damping Systems

Formulas

Damping ratio: ζ = c / (2√(mk))
Position function for critically damped: x(t) = (C1 + C2 t)e^(-ζω0t)
Undamped position function: u(t) = C0 cos(ω0t - α0)

Theorems

Critically Damped Motion Theorem

Suitable Grade Level

College Level