The parameter of a single-degree-of-freedom system are given by m=1 kg, c=5 N-sec/m, and k=16 N/m. Find the response of the system for the following initial conditions:
The parameter of a single-degree-of-freedom system are given by m=1 kg, c=5 N-sec/m, and k=16 N/m. Find the response of the system for the following initial conditions:
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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ANSWER problem 2.149& 2.150.Use x(0)=.1m and x(dot)(0)=2 as a template for answering the problem stated problem. And note theta tells you which quadrant the tan inverse value will fall in so for example theta with two positive values so the angle inputted will be in quadrant 1 (in the unit circle).Negative value from tan inverse means subtract the inverse tan value from larger value and positive value means add to the smaller value. EX: quad 2 (in the unit circle) smaller value is 90 and larger value is 180
![### System Response Analysis
The parameters of a single-degree-of-freedom system are given by:
- Mass, \( m = 1 \, \text{kg} \),
- Damping coefficient, \( c = 5 \, \text{N-sec/m} \),
- Stiffness, \( k = 16 \, \text{N/m} \).
We need to find the response of the system for the following initial conditions:
#### Initial Conditions
**Condition (a):**
- \( x(0) = 0.1 \, \text{m} \)
- \( \dot{x}(0) = 2 \, \text{m/sec} \)
**Condition (b):**
- \( x(0) = -0.1 \, \text{m} \)
- \( \dot{x}(0) = -2 \, \text{m/sec} \)
---
### System Equations
The system equation is:
\[ m\ddot{x} + c\dot{x} + kx = 0 \]
Simplifying, we get:
\[ \ddot{x} + 5\dot{x} + 16x = 0 \]
The characteristic equation is:
\[ x = \frac{-5 \pm \sqrt{25 - 4(16)}}{2} = \frac{-5 \pm i\sqrt{39}}{2} \]
### General Solution
The response function is:
\[ x(t) = X_0 e^{-\zeta \omega_n t} \sin(\omega_d t + \phi_0) \]
Where:
- \( X_0 = \frac{\sqrt{x_0^2 \omega^2_n + \dot{x}_0^2 + 2\zeta \omega_n x_0 \dot{x}_0}}{\omega_d} \)
- \( \phi_0 = \tan^{-1}\left(\frac{x_0 \omega_d}{\dot{x}_0 + \zeta \omega_n x_0} \right) \)
### Natural Frequency and Damping
- Natural Frequency: \( \omega_n = \sqrt{\frac{k}{m}} = 4 \, \text{rad/sec} \)
- Critical Damping Coefficient: \( c_c = 2\sqrt{km} = 8 \, \text{N-sec/m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20000a63-5a8a-4b5e-adbd-85ecbf6c298b%2Fd4cfc9ff-530b-4553-97c3-29e0589c469a%2F01mn1pa_processed.png&w=3840&q=75)
Transcribed Image Text:### System Response Analysis
The parameters of a single-degree-of-freedom system are given by:
- Mass, \( m = 1 \, \text{kg} \),
- Damping coefficient, \( c = 5 \, \text{N-sec/m} \),
- Stiffness, \( k = 16 \, \text{N/m} \).
We need to find the response of the system for the following initial conditions:
#### Initial Conditions
**Condition (a):**
- \( x(0) = 0.1 \, \text{m} \)
- \( \dot{x}(0) = 2 \, \text{m/sec} \)
**Condition (b):**
- \( x(0) = -0.1 \, \text{m} \)
- \( \dot{x}(0) = -2 \, \text{m/sec} \)
---
### System Equations
The system equation is:
\[ m\ddot{x} + c\dot{x} + kx = 0 \]
Simplifying, we get:
\[ \ddot{x} + 5\dot{x} + 16x = 0 \]
The characteristic equation is:
\[ x = \frac{-5 \pm \sqrt{25 - 4(16)}}{2} = \frac{-5 \pm i\sqrt{39}}{2} \]
### General Solution
The response function is:
\[ x(t) = X_0 e^{-\zeta \omega_n t} \sin(\omega_d t + \phi_0) \]
Where:
- \( X_0 = \frac{\sqrt{x_0^2 \omega^2_n + \dot{x}_0^2 + 2\zeta \omega_n x_0 \dot{x}_0}}{\omega_d} \)
- \( \phi_0 = \tan^{-1}\left(\frac{x_0 \omega_d}{\dot{x}_0 + \zeta \omega_n x_0} \right) \)
### Natural Frequency and Damping
- Natural Frequency: \( \omega_n = \sqrt{\frac{k}{m}} = 4 \, \text{rad/sec} \)
- Critical Damping Coefficient: \( c_c = 2\sqrt{km} = 8 \, \text{N-sec/m
Transcribed Image Text:**Problem 2.149 & 2.150: Response of Damped System**
Determine the values of ζ (damping ratio), ωₙ (natural frequency), and the free-vibration response of the following viscously damped systems when:
\( x_0 = 0.1 \, \text{(m)}, \quad \dot{x}_0 = 10 \, \left(\frac{\text{m}}{\text{sec}}\right) \)
(a) \( m = 10 \, \text{kg}, \quad c = 150 \, \left(\frac{\text{N} \cdot \text{sec}}{\text{m}}\right), \quad k = 1{,}000 \, \left(\frac{\text{N}}{\text{m}}\right) \)
(b) \( m = 10 \, \text{kg}, \quad c = 200 \, \left(\frac{\text{N} \cdot \text{sec}}{\text{m}}\right), \quad k = 1{,}000 \, \left(\frac{\text{N}}{\text{m}}\right) \)
(c) \( m = 10 \, \text{kg}, \quad c = 250 \, \left(\frac{\text{N} \cdot \text{sec}}{\text{m}}\right), \quad k = 1{,}000 \, \left(\frac{\text{N}}{\text{m}}\right) \)
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