Critically Damped oscillator

In summary, a critically damped oscillator with a damping constant b=2 m ω0 has a motion given by x=(A+Bt)e^(−βt), where A and B are constants. When the oscillator is at rest at equilibrium, it can be given a sharp impulse I at t=0. This impulse only affects the initial conditions of the motion, and does not significantly impact the oscillation itself. The maximum displacement of the oscillator can be calculated using the given data for I, m, and k.
  • #1
kraigandrews
108
0

Homework Statement


If the damping constant of a free oscillator is given by b=2 m ω0, the oscillator is said to be critically damped. Show by direct substitution that in this case the motion is given by
x=(A+Bt)e^(−βt)
where A and B are constants.

A critically damped oscillator is at rest at equilibrium. At t = 0 the mass is given a sharp impulse I. Sketch the motion. Calculate the maximum displacement.
Data: I = 11.1 Ns; m = 1.1 kg; k = 18.2 N/m.


Homework Equations



[itex]\beta[/itex]=b/(2m)


The Attempt at a Solution


Two things I find wrong here:
1: since x(0)=0 and v(0)=0 it implies that A and B= 0 which is wrong so that must mean I plays a role, obviously.
2. I do not know how to incorportate the impulse into x(t)
 
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  • #2
1: since x(0)=0 and v(0)=0 it implies that A and B= 0 which is wrong so that must mean I plays a role, obviously.

You can't set both v=0 and x=0 as initial conditions. If I put a spring at its equilibrium position and don't make it move, it's not surprising that the spring's going to stay there. Any non-zero value of either v or x will give you the same period, just different amplitudes and phases.

2. I do not know how to incorportate the impulse into x(t)

The impulse makes the block move at some initial speed before the spring has time to react, so this initial speed is essentially v(0).
 
  • #3
ok so the impulse function just has an impact in the initial condition? For some reason I waas thinking of dirac's delta function
 
  • #4
Yes, it only impacts the initial condition because its duration is too short to affect the oscillation. You can think of it as a Dirac's delta function, with the integral being I instead of 1.
 
  • #5
or the general solution.

I understand your concerns and will provide a response to the content and address the issues you have raised.

Firstly, a critically damped oscillator is a type of damped oscillator where the damping force is equal to the critical damping force. This means that the system will return to equilibrium without any oscillations, but in the shortest possible time. This type of damping is often desired in engineering applications where a quick return to equilibrium is important.

To show that in the case of a critically damped oscillator, the motion is given by x=(A+Bt)e^(−βt), we can start by considering the equation of motion for a damped oscillator:

m\ddot{x}+b\dot{x}+kx=0

where m is the mass, b is the damping constant, k is the spring constant, and x is the displacement from equilibrium.

Substituting in the given value for b=2m ω0, we get:

m\ddot{x}+2m ω0\dot{x}+kx=0

Dividing by m and rearranging, we get:

\ddot{x}+2 ω0\dot{x}+\frac{k}{m}x=0

Comparing this to the general solution for a damped oscillator:

x(t)=Ae^(−αt)cos(ωt+ϕ)

where α is the damping constant and ϕ is the phase angle, we can see that in the case of a critically damped oscillator, α=2 ω0. Substituting this into the general solution, we get:

x(t)=Ae^(−2 ω0t)cos(ωt+ϕ)

Using the identity e^(−αt)=cos(ωt)+sin(ωt), we can rewrite this as:

x(t)=Ae^(−2 ω0t)(cos(ωt)+sin(ωt))

Expanding this, we get:

x(t)=Ae^(−2 ω0t)cos(ωt)+Ae^(−2 ω0t)sin(ωt)

Since we are only interested in the real part of this solution, we can ignore the term with the sine function. Thus, we are left with:

x(t)=Ae^(−2 ω0t)cos(ωt)

Using the identity cos(
 

1. What is a critically damped oscillator?

A critically damped oscillator is a type of damped harmonic oscillator in which the damping force is equal to the critical damping coefficient. This means that the system returns to equilibrium in the shortest possible time without any oscillations or overshooting.

2. How does a critically damped oscillator differ from an overdamped or underdamped oscillator?

In an overdamped oscillator, the damping force is greater than the critical damping coefficient, resulting in slower return to equilibrium and no oscillations. In an underdamped oscillator, the damping force is less than the critical damping coefficient, leading to oscillatory motion that decreases over time.

3. What are the applications of critically damped oscillators?

Critically damped oscillators have many practical applications, such as shock absorbers in vehicles, vibration control in buildings and bridges, and in electronic circuits for filtering or signal processing.

4. How is the behavior of a critically damped oscillator described mathematically?

The motion of a critically damped oscillator can be described by a second-order differential equation of the form mx'' + bx' + kx = 0, where m is the mass, b is the damping coefficient, and k is the spring constant. The solution to this equation results in a decaying exponential function.

5. Can a critically damped oscillator be unstable?

No, a critically damped oscillator is always stable and will always return to equilibrium without oscillations. This is because the damping force is equal to the critical damping coefficient, providing enough energy dissipation to prevent any oscillatory behavior.

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