The equation of motion and initial conditions governing the vertical position of a vertical spring-mass system are given below. Transform this equation into a system of 1st order ODEs and solve the system using MATLAB's ode45 function. Provide a plot of y vs. t from 0 to 2 seconds. mj = −k(y – y) – mg y(0) = 1 m j(0) =4m/s k = 100 N/m m = 2 kg g = 9.81 m/s² Yo = 1m
The equation of motion and initial conditions governing the vertical position of a vertical spring-mass system are given below. Transform this equation into a system of 1st order ODEs and solve the system using MATLAB's ode45 function. Provide a plot of y vs. t from 0 to 2 seconds. mj = −k(y – y) – mg y(0) = 1 m j(0) =4m/s k = 100 N/m m = 2 kg g = 9.81 m/s² Yo = 1m
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![### Equation of Motion for a Vertical Spring-Mass System
The objective is to transform the given equation of motion into a system of first-order ordinary differential equations (ODEs) and solve it using MATLAB's `ode45` function. Then, provide a plot of displacement \( y \) versus time \( t \) from 0 to 2 seconds.
**Equation of Motion:**
\[ m\ddot{y} = -k(y - y_0) - mg \]
**Initial Conditions:**
- \( y(0) = 1 \, \text{m} \)
- \( \dot{y}(0) = 4 \, \text{m/s} \)
**Given Parameters:**
- Spring constant, \( k = 100 \, \text{N/m} \)
- Mass, \( m = 2 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
- Initial position, \( y_0 = 1 \, \text{m} \)
### Task:
- Transform the second-order ODE into a system of first-order ODEs.
- Utilize MATLAB's `ode45` for solving the system.
- Graph \( y \) against \( t \) for the duration of 0 to 2 seconds.
The solution involves converting the second-order ODE into a system by defining:
1. \( y_1 = y \)
2. \( y_2 = \dot{y} \)
This yields:
\[ \dot{y}_1 = y_2 \]
\[ \dot{y}_2 = \frac{-k(y_1 - y_0) - mg}{m} \]
With these equations, code up the system in MATLAB, use the `ode45` solver, and plot \( y_1 \) against \( t \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa487fd1f-659d-4c4f-ad85-f8cf24cc282d%2Fb127c548-bf0b-4269-ae7e-368fb2b36240%2Fhd5hmgr_processed.png&w=3840&q=75)
Transcribed Image Text:### Equation of Motion for a Vertical Spring-Mass System
The objective is to transform the given equation of motion into a system of first-order ordinary differential equations (ODEs) and solve it using MATLAB's `ode45` function. Then, provide a plot of displacement \( y \) versus time \( t \) from 0 to 2 seconds.
**Equation of Motion:**
\[ m\ddot{y} = -k(y - y_0) - mg \]
**Initial Conditions:**
- \( y(0) = 1 \, \text{m} \)
- \( \dot{y}(0) = 4 \, \text{m/s} \)
**Given Parameters:**
- Spring constant, \( k = 100 \, \text{N/m} \)
- Mass, \( m = 2 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
- Initial position, \( y_0 = 1 \, \text{m} \)
### Task:
- Transform the second-order ODE into a system of first-order ODEs.
- Utilize MATLAB's `ode45` for solving the system.
- Graph \( y \) against \( t \) for the duration of 0 to 2 seconds.
The solution involves converting the second-order ODE into a system by defining:
1. \( y_1 = y \)
2. \( y_2 = \dot{y} \)
This yields:
\[ \dot{y}_1 = y_2 \]
\[ \dot{y}_2 = \frac{-k(y_1 - y_0) - mg}{m} \]
With these equations, code up the system in MATLAB, use the `ode45` solver, and plot \( y_1 \) against \( t \).
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