How Does a Critically Damped Oscillator Behave After a Sharp Impulse?

[kraigandrews]

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

$\beta$=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)

 

[ideasrule]

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).

 

[kraigandrews]

ok so the impulse function just has an impact in the initial condition? For some reason I waas thinking of dirac's delta function

 

[ideasrule]

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.