2nd order ODE: modeling a spring

In summary, the spring with a constant 6N/m has a mass attached at one end and the other end is attached to a wall. The friction/damping constant is 1 N s/m. If there is no spring, the equation of motion reads ##F = ma = m\ddot x##. However, if there is a spring, the equation of motion becomes 3x'' + 1x' + 6x =0.
  • #1
ReidMerrill
66
2
Thread moved from the technical forums, so no Homework Template is shown
Suppose a spring with spring constant 6N/m is horizontal and has one end attached to the wall and the other end attached to a 3 kg mass. Suppose the friction/damping constant is 1 N s/m

Set up a differential equation that describes this system with x denoting displacement of the mass from equilibrium position and give your answer in terms of x,x', and x''. Assume that positive displacement means the mass is farther from the wall than equalibrium

I'm not sure what to do here. I'm guessing I'll need the equation

mx'' + bx' + kx = F Where mx'' = Fdamping + F spring + F inertia b is damping coefficient: 1 N m/s k is spring coefficient: 6 N/m

Where do I get m? Fdamping is 1 and Fspring is 6 right? Would I get Finertia from F=ma? Because a is not given.
 
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  • #2
Hi,

You already have the acceleration a when you write ##\ddot x##.

If there is no spring, your equation of motion reads ##F = ma = m\ddot x##. In that case, what is F ?

Are there other (external) forces at work, except for the spring and the damping force ?
 
  • #3
Why do you think you need a "Finertia" term? Newton's Second Law only involves the sum of real forces; no fictional forces are to be included. Draw the FBD and include in your force sum on the actual forces present.
 
  • #4
BvU said:
Hi,

You already have the acceleration a when you write ##\ddot x##.

If there is no spring, your equation of motion reads ##F = ma = m\ddot x##. In that case, what is F ?

Are there other (external) forces at work, except for the spring and the damping force ?

I don't think so.
 
  • #5
You asked "where do I get m?" but the problem statement said that one end of the spring was attached to a 3 kg mass. That is where you get m.
 
  • #6
You missed the 'in that case, what is F?' :smile:. I was fishing for the answer: 'the externally applied force'
ReidMerrill said:
I don't think so.
So that means in your differential equation the right-hand side can be set to 0 and ##m\ddot x = ## the two forces in the story (##\beta\dot x## and ##kx##) , each with the proper sign
 
  • #7
BvU said:
You missed the 'in that case, what is F?' :smile:. I was fishing for the answer: 'the externally applied force'
So that means in your differential equation the right-hand side can be set to 0 and ##m\ddot x = ## the two forces in the story (##\beta\dot x## and ##kx##) , each with the proper sign
So the equation would be 3x'' + 1x' + 6x =0 ?
 
  • #8
Only if you willingly omit the dimensions -- generally not a wise thing to do :smile:. Any doubt remaining ?
 

Related to 2nd order ODE: modeling a spring

1. What is a 2nd order ODE?

A 2nd order ODE, or second-order ordinary differential equation, is a mathematical equation that relates the dependent variable, usually denoted as y, to its independent variable, usually denoted as x, and its first and second derivatives. It is typically written in the form y'' = f(x,y,y'). This type of equation is commonly used in physics and engineering to model and analyze systems that involve acceleration, such as springs.

2. How is a spring modeled using a 2nd order ODE?

A spring can be modeled using a 2nd order ODE by considering the forces acting on the spring. The equation of motion for a spring is given by the second law of motion, F = ma, where F is the force, m is the mass, and a is the acceleration. For a spring, the force is equal to the spring constant, k, multiplied by the displacement, x. This results in the equation m(d^2x/dt^2) = -kx, which is a 2nd order ODE that can be solved to determine the motion of the spring.

3. What are the initial conditions for a 2nd order ODE modeling a spring?

The initial conditions for a 2nd order ODE modeling a spring are typically the initial displacement, x(0), and the initial velocity, dx/dt(0). These values represent the starting position and velocity of the spring, and are necessary to solve the equation and determine the behavior of the spring over time.

4. How is the solution to a 2nd order ODE for a spring interpreted?

The solution to a 2nd order ODE for a spring is typically a function of time, x(t), that represents the position of the spring at any given time. This solution can be interpreted as the displacement of the spring from its equilibrium position at any point in time. The velocity of the spring can also be determined from the solution by taking the first derivative, dx/dt.

5. What are some real-world applications of modeling a spring using a 2nd order ODE?

There are many real-world applications of modeling a spring using a 2nd order ODE. One common example is in car suspension systems, where springs are used to absorb and dampen the vibrations caused by bumps on the road. Another application is in mechanical watches, where springs are used to power the movement of the watch hands. Additionally, springs are used in many other engineering and physics systems, such as shock absorbers, trampolines, and door hinges.

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