The Differential Equation of Free Motion

In summary, a spring with a spring constant of 100 pounds per foot and a 1-pound weight is stretched an additional 1 inch and released. Using the equation mx"+kx=0, with m=1 lb and k=100 lb/ft, we get the characteristic equation r=+/- 10i. The general solution is y(t)=Acos(10t)+Bsin(10t), and using the given initial conditions of x(0)=1/12 and x'(0)=0, we can determine the constants A and B. After rethinking the mass and realizing that 1 pound equals 1/32 of a pound, the correct equation becomes 1/32x"+100x=
  • #1
gotpho
6
0

Homework Statement


A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then stretched an additional 1 inch and released. Find the equation of motion, the amplitude, and the period. Neglect friction.

Homework Equations


mx"+kx=0

The Attempt at a Solution



I understand that I need to set up a differential equation but I'm not quite sure if this is right.

K = 100 lb/ft m = 1 lb

thus plugging into the equation

x" + 100x = 0

use characteristic equation,

I get r= +/- 10i

y[tex]^{g}[/tex](t)= cos (10t) + sin (10t)So far is everything is alright?

When I tried to solve for period. I set the inside, 10t=2pi. At least that's how my professor does it but apparently it's wrong when I submit my answers online.

I'm also not understanding how the initial conditions work?
 
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  • #2
You need constants A and B in front of your trig terms to get the general solution. You are given the initial position is one inch (watch your units), plus or minus depending on your coordinates, and apparently the initial velocity is zero. Use these two facts to determine your A and B.
 
  • #3
Thanks anyway but I solved this problem.

I rethought about the mass and realized that 1 pound is equal to mg. Thus 1/32 is the actual mass.

So the equation should have been

1/32x" + 100x=0, x(0)=1/12, x'(0)=0

So yeah, now everything makes sense.
 

1. What is the Differential Equation of Free Motion?

The Differential Equation of Free Motion is a mathematical expression that describes the motion of a particle or object in space without any external forces acting on it. It is a second-order differential equation that relates the position, velocity, and acceleration of the particle.

2. How is the Differential Equation of Free Motion derived?

The Differential Equation of Free Motion is derived from Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. By setting the net force to zero, we can isolate the acceleration and create a differential equation that describes the object's motion.

3. What are the applications of the Differential Equation of Free Motion?

The Differential Equation of Free Motion has various applications in physics and engineering, such as predicting the motion of celestial bodies, analyzing the behavior of pendulums and springs, and designing control systems for robots and vehicles.

4. How is the Differential Equation of Free Motion solved?

The Differential Equation of Free Motion can be solved using various mathematical techniques, such as separation of variables, integrating factors, and Laplace transforms. The solution depends on the initial conditions, which are the object's position and velocity at a given time.

5. What are the limitations of the Differential Equation of Free Motion?

The Differential Equation of Free Motion assumes that the object is moving in a vacuum and that there are no external forces acting on it. In reality, most objects experience some form of resistance or friction, which can affect their motion. Additionally, the equation is only accurate for small displacements and short time intervals.

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