How do you solve for A in a critically damped oscillator problem?

[derravaragh]

Homework Statement

(A) A damped oscillator is described by the equation
m x′′ = −b x′− kx .
What is the condition for critical damping? Assume this condition is satisfied.
(B) For t < 0 the mass is at rest at x = 0. The mass is set in motion by a sharp impulsive force at t = 0, so that the velocity is v0 at time t = 0. Determine the position x(t) for t > 0.
(C) Suppose k/m = (2π rad/s)2 and v0=10 m/s. Plot, by hand, an accurate graph of x(t). Use graph paper. Use an appropriate range of t.

Homework Equations

For critically damped, β² = w₀²
where β = b/(2m) and w₀ = √(k/m)

The Attempt at a Solution

Ok, for this problem, what I did initially was find the general form of position for a critically damped oscillator, which is:
x(t) = (A + B*t)*e^-β*t

and the velocity function is:
v(t) = -Aβe^-βt + (Be^-βt - Bβte^-βt)

Using the conditions given, I found:
x(0) = A (obviously) which we don't know x(0)
B = v₀ + Aβ
and x(t) can be rewritten as:
x(t) = A(e^-βt + βte^-βt) + v₀te^-βt

This is where I run into a wall. I can't seem to solve for A. I believe that x(0) should also be the max displacement since there is no driver for the impulse force, so A should be the max displacement, but this doesn't seem to get me anywhere. Any help on solving for A? I know how to do the rest other than that.

 

[TSny]

A "sharp impulsive force" is defined such that it instantaneously gives the mass an initial velocity without any displacement of the mass. So the mass is still at x = 0 immediately after the impulse.

 

[derravaragh]

Ok, that makes sense. Thank you for the help.