What is the solution for a critical damped SHM equation?

In summary: Omega*a*e-\Omega*t.In summary, if the particle is initially released from rest at x=a, its subsequent motion is given by x=a*(e^-(\Omega*t))*(1+\Omega*t).
  • #1
Benzoate
422
0

Homework Statement


Find the general solution of the damped SHM equation(5.9) for the special case of critical damping , that is , when K = [tex]\Omega[/tex]. Show that , if the particle is initially relaeased from rest at x= a , then the subsequent motion is given by

x=a*(e^-([tex]\Omega[/tex]*t))*(1+[tex]\Omega[/tex]*t)


Homework Equations



x''+2Kx'+([tex]\Omega[/tex])2*x=0
x=A*cos([tex]\Omega[/tex]*t) + B*sin([tex]\Omega[/tex]*t)


The Attempt at a Solution



x=e[tex]\lambda[/tex]*t

x'= [tex]\lambda[/tex]*e[tex]\lambda[/tex]*t

x''=[tex]\lambda[/tex]2*e[tex]\lambda[/tex]*t

x'= -A*[tex]\Omega[/tex]*sin([tex]\Omega[/tex]*t) + B*[tex]\Omega[/tex]*cos([tex]\Omega[/tex]*t)

x''= -A*[tex]\Omega[/tex]2*cos([tex]\Omega[/tex]*t) - B*[tex]\Omega[/tex]2*sin([tex]\Omega[/tex]*t)

Not sure how to proceed...
 
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  • #2
Take a look at the characteristic polynomial of the DE:

[tex]x''(t)+2 \Omega x'(t) + \Omega ^2 x(t)=0 \Rightarrow \lambda ^2 +2 \Omega \lambda + \Omega ^2 =0[/tex]

where [tex] \lambda ^n x(t) \equiv \frac{d^n x(t)}{dt^n}[/tex].

What are the solutions for [itex]\lambda[/itex]? Are they linearly independent? What does that tell you about the general solution for [itex]x(t)[/itex]?
 
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  • #3
gabbagabbahey said:
Take a look at the characteristic polynomial of the DE:

[tex]x''(t)+2 \Omega x'(t) + \Omega ^2 x(t)=0 \Rightarrow \lambda ^2 +2 \Omega \lambda + \Omega ^2 \lambda ^2=0[/tex]

where [tex] \lambda ^n \equiv \frac{d^n x(t)}{dt^n}[/tex].

What are the solutions for [itex]\lambda[/itex]? Are they linearly independent? What does that tell you about the general solution for [itex]x(t)[/itex]?

That [tex]\lambda[/tex]=-[tex]\Omega[/tex]; its a double root

I remember from my ODE class that if a polynomial had more than two root, then x(t)=c1*e-r1*t+c2-r2*t; Otherwise , if the polynomial has only one solution, then x(t)=c1*e-r1*t+c2*t*e-r1*t

How would I go about showing that x(t)=c1*e-r1*t+c2*t*e-r1*t
rather than loosely remembering what I learned about the conditions for ODE equations certain solutions
 
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  • #4
I assume you mean [itex]x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}[/itex]?

If so, you can show this by substituting this general solution into the ODE and showing that it satisfies it.

Then, just use the fact that x(0)=a to show that the specific solution is the one given in the problem.
 
  • #5
gabbagabbahey said:
I assume you mean [itex]x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}[/itex]?

If so, you can show this by substituting this general solution into the ODE and showing that it satisfies it.

Then, just use the fact that x(0)=a to show that the specific solution is the one given in the problem.

Should I start off by assuming that the original ODE for a double root is [itex]x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}[/itex]? or do I need to go about proving that I arrived at this solution ?
 
  • #6
I don't think you'll need to prove that it's the general solution, the solutions to constant coefficient ODE's are well known and can be found in almost any text on DE's; but you may want to check with your prof to make sure.
 
  • #7
gabbagabbahey said:
I don't think you'll need to prove that it's the general solution, the solutions to constant coefficient ODE's are well known and can be found in almost any text on DE's; but you may want to check with your prof to make sure.

x(0)=a

I think the initial conditions for v is zero as well , since the damped motion is undriven.

x' = -[tex]\Omega[/tex]*a*e-[tex]\Omega[/tex]*t + c2*(e-[tex]\Omega[/tex]*t-x(0)=a

I think the initial conditions for v is zero as well , since the damped motion is undriven.

x' = -[tex]\Omega[/tex]*a*e-[tex]\Omega[/tex]*t + c2*(e-[tex]\Omega[/tex]*t-[tex]\Omega[/tex]*t*ex(0)=a

I think the initial conditions for v is zero as well , since the damped motion is undriven.

x' = -[tex]\Omega[/tex]*a*e-[tex]\Omega[/tex]*t + c2*(e-[tex]\Omega[/tex]*t-[tex]\Omega[/tex]*t*e*t*[tex]\Omega[/tex]

How would I go about finding c2?
 
  • #8
Your last post didn't make much sense.

You can determine [itex]c_1[/itex] by using x(0)=a, and [itex]c_2[/itex] by using x'(0)=0 (since the particle is released from rest at t=0)
 
  • #9
gabbagabbahey said:
Your last post didn't make much sense.

You can determine [itex]c_1[/itex] by using x(0)=a, and [itex]c_2[/itex] by using x'(0)=0 (since the particle is released from rest at t=0)

sorry for the confusion , but I was essentially saying what you said in your last post: Should I assume for an undriven damped oscillator , that the initial conditions for x' and x are zero at t=0.
 
  • #10
Benzoate said:
sorry for the confusion , but I was essentially saying what you said in your last post: Should I assume for an undriven damped oscillator , that the initial conditions for x' and x are zero at t=0.

There's no need to assume anything here; you are given the initial conditions:

Show that , if the particle is initially released from rest at x= a , then the subsequent motion is given by ...

that means x(t=0)=a and x'(t=0)=0.
 

What is critical damped motion?

Critical damped motion is a type of motion in which a system returns to equilibrium as quickly as possible without overshooting or oscillating around the equilibrium point.

How is critical damped motion different from overdamped and underdamped motion?

Critical damped motion falls in between overdamped and underdamped motion. In overdamped motion, the system takes longer to return to equilibrium, while in underdamped motion, the system overshoots or oscillates before returning to equilibrium.

What factors determine the critical damping coefficient?

The critical damping coefficient is determined by the mass, stiffness, and damping properties of the system. It is the minimum amount of damping required for critical damped motion to occur.

What are some real-life examples of critical damped motion?

Examples of critical damped motion can be seen in shock absorbers, car suspensions, and door closers. In these systems, critical damped motion allows for a smooth and efficient return to equilibrium without any bouncing or oscillations.

How is critical damped motion useful in engineering and design?

Critical damped motion is useful in engineering and design because it allows for precise control and stability in systems. It ensures a quick and smooth return to equilibrium, which can be important in applications such as aircraft landing gear, elevator systems, and earthquake-resistant structures.

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