Question

Question #927f2

Answer 1

It would be inversely proportional. That's easy to imagine, a system that is damped more will have its fluctuations damped more strongly.

Explanation:

Lets get to the basics, lets start from the basic differential equation describing a damped simple harmonic oscillator.
`\frac{d^2x}{dt^2}+b\frac{dx}{dt}+\omega^2x=0`
where `b` is the damping constant, `\omega` is its natural frequency, and `x` dynamical parameter being described by this equation.

Let us use a trial solution of the form `A\exp(\lambda t)`, substituting we get,
`\lambda^2+b\lambda+\omega^2=0`
solving this we get the solution
`\lambda=\frac{-b+\sqrt{b^2-4\omega^2}}{2}`

Let us for a moment assume a oscillatory solution, ie `\lambda` should be imaginary, remember, oscillations imply solutions of the form `\exp (i\alpha t)`, so lets just rearrange the terms inside the square root.
`\lamda=\frac{-b+i\sqrt{4\omega^2-b^2}}{2}` all I did was change the the sign inside the square root and pull a `i` out, this is what is called a under damped solution. This implies the solution is
`x(t)=Aexp(-\frac{b}{2}t)\exp(\frac{i\sqrt{4\omega^2-b^2}}{2})`
`A` is determined by an initial condition.

The complex exponential represents oscillations but the first part represents an decaying exponential. A decaying process is often written in the form
`x(t)=A\exp(-\frac{t}{\tau})`, where `\tau` is referred to as the relaxation time. Comparing with the solution we got, we get
`\tau=\frac{2}{b}`.

Indeed, as per our intuition the relaxation time is indeed inversely proportional to damping constant. This particular form is particular for an underdamped oscillator, but you can be pretty sure that larger the damping constant, smaller the relaxation time.

Hope this helps!