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

In summary, the conversation is discussing a damped oscillator and how to find the position of the mass after an impulse force is applied. The condition for critical damping is when β^2 = w0^2, where β is equal to b/(2m) and w0 is equal to the square root of k/m. The position function for a critically damped oscillator is x(t) = (A + B*t)*e^-βt and the velocity function is v(t) = -Aβe^-βt + (Be^-βt - Bβte^-βt). To solve for A, the condition x(0) = A is used, where x(0) is the maximum displacement of the mass.
  • #1
derravaragh
24
0

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.
 
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  • #2
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.
 
  • #3
Ok, that makes sense. Thank you for the help.
 

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

What is a critically damped oscillator?

A critically damped oscillator is a type of system in which the damping force exactly balances out the restoring force, resulting in no oscillations or overshooting. It is the fastest type of damping, as it returns to equilibrium in the shortest amount of time.

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

In an overdamped oscillator, the damping force is greater than the restoring force, resulting in slow return to equilibrium without any oscillations. In an underdamped oscillator, the damping force is less than the restoring force, causing the system to oscillate and potentially overshoot before returning to equilibrium.

What are some real-life examples of critically damped oscillators?

Some examples include shock absorbers in cars, door closers, and certain types of electrical circuits.

How is the damping ratio related to a critically damped oscillator?

The damping ratio, denoted by the symbol ζ (zeta), is a dimensionless parameter that determines the type of damping in an oscillator. For a critically damped oscillator, the damping ratio is equal to 1.

How is a critically damped oscillator mathematically described?

The equation of motion for a critically damped oscillator is given by mx'' + bx' + kx = 0, where m is the mass, b is the damping coefficient, and k is the spring constant. This equation can be solved using techniques from differential equations to determine the position and velocity of the oscillator as a function of time.

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