# Verifying that a critical damped oscillator approaches zero the fastest.

1. Jan 7, 2009

### Quantumpencil

1. The problem statement, all variables and given/known data

Not actually a homework problem, just something from my book I'm trying to verify.

2. Relevant equations

The general form of the equation for damped oscillations...

$$\ x(t) = e^{-\gamma t}(Ae^{\sqrt{\gamma^{2}-\omega^{2}}t}+Be^{-\sqrt{\gamma^{2}-\omega^{2}}t})$$

Here gamma = b/m, where b is the constant associated with the strength of a velocity dependent damping force. (A and B are constants determined by the initial conditions)

and omega is the angular frequency, root (k/m).

And, in the critically damped case, the solution is

$$\ x(t) = e^{-\gamma t}(A+Bt)$$

3. The attempt at a solution

It's obvious that it b<k, that is gamma<omega, then when the oscillation is critically damped (gamma = omega), it will reach zero more quickly. When the oscillation is underdamped, then the exponents in the larger expression in parenthesis are imaginary, so we have simple harmonic motion falling off as e^-gamma(t), and since gamma is less than omega, this expression reaches zero faster than in the critically damped case, which falls off as e^-omega(t).

However when the oscillation is overdamped, and we have b>k, gamma>omega, how can we show that the exponents (which are then positive) are still less than omega? That is, how can we show

$$\gamma-\sqrt{\gamma^{2}-\omega^{2}}\leq\omega$$

2. Jan 7, 2009

### Thaakisfox

Rearranging the inequality etc.:

$$\gamma -\omega \leq \sqrt{\gamma^2-\omega^2}$$

But we know that: $$\gamma-\omega = \sqrt{\gamma-\omega}\sqrt{\gamma-\omega}$$.

So we have:

$$\sqrt{\gamma-\omega}\sqrt{\gamma-\omega} \leq \sqrt{\gamma-\omega}\sqrt{\gamma+\omega}$$

Wich is obviously true, since $$\gamma > \omega$$