Verifying that a critical damped oscillator approaches zero the fastest.

[Quantumpencil]

Homework Statement

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

Homework 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

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

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

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

 

[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 $$\gamma > \omega$$

 

[thomate1]

One way to show that the exponents are still less than omega in the overdamped case is to use the quadratic formula to solve for the roots of the expression under the square root. This will give you two possible values for the exponents, one positive and one negative. Since gamma is greater than omega, the positive exponent will be smaller in magnitude than omega, and thus the overdamped case will still approach zero faster than the critically damped case.