# Damped drive oscillating systems

I am currently looking at the theory of find the viscosity of and object through damped harmonic motion, and tho it can be done there is obviously a limitation with regrades to the medium. If the medium is too viscous then critical dampening will occur and we lose the ability to find the viscosity.

I was looking at the equation for motion of a driven system and was wondering if you could use a driven oscillator to work out the viscosity of a more viscous medium.

• No I do mean critically damped. So with a under damped system you get an exponential decay, and I was just wondering if you could take a critically damped system and turn into a under damped system – james2018 Jan 23 at 23:11
• Yes sorry, I have now rephrased the question more fully. – james2018 Jan 23 at 23:25

## 2 Answers

The solution to $$m \ddot{x} + k x + c \dot{x} = 0$$ can be characterized with the following parameter substitutions

• $$k = m \omega_n^2$$
• $$c =2 \zeta m \omega_n$$

Where $$\omega_n$$ is the natural frequency of the system, and $$\zeta$$ the damping ratio. When $$\zeta=1$$ the system is critically damped when over 1 it is overdamped and under 1 it is underdamped.

The equation of motion is now

$$\ddot{x} + \omega_n^2 x + 2 \zeta \omega_n \dot{x} = 0$$

The solution is

$$x(t) = \begin{cases} \exp(-\omega_n \zeta t) \left( C_1 \sin (\omega t) + C_2 \cos (\omega t) \right) & \zeta \leq 1, \;\omega = \omega_n \sqrt{1+\zeta^2} \\ \exp(-\omega_n \zeta t) \left( C_1 \sinh (\omega t) + C_2 \cosh (\omega t) \right) & \zeta > 1, \;\omega = \omega_n \sqrt{\zeta^2-1} \end{cases}$$

So if you find the exponential decay rate $$\beta = \omega_n \zeta$$ experimentally, you can then solve for $$\zeta$$ and then for $$c$$ if you know the mass $$m$$, and the stiffness $$k$$.

I assume it is a mechanical system that you are using and possibly a spring-mass system?

Having made the assumption that the resistive force is proportional to the speed of the mass, the equation of motion for a system undergoing free oscillations is $$m\ddot x + b\dot x + kx =0$$.

If $$b^2 < 4km$$ the system is under damped and so will perform oscillatory motion.
If at present your system is over damped you might consider increasing the mass $$m$$ and $$k$$ which often is the spring constant to move the system into the under damped regime.

The exponential envelope will be of the form $$\exp\left( -\frac {b}{2m}t\right)$$

If the system is to be driven then the equation of motion changes to something like $$m\ddot x + c \dot x + kx =F_0\cos (\omega t)$$.
Having a lot of damping is good news in that the driven system will reach a steady state (constant amplitude) in a relatively short period of time.

The equation for the amplitude of the driven system $$A$$ as a function of driver frequency $$\omega$$ is quite complicated $$A(\omega) = \dfrac{F_0}{\sqrt{m^2(\omega^2-\omega^2_0)+b^2\omega^2_0}}$$ where $$\omega_0$$ is the natural frequency of the system when it is undamped.

At resonance (maximum amplitude of the driven system) $$A = \dfrac {F_0}{b\omega_0}$$ however it does need you to know the amplitude of the force $$F_0$$ produced by the driver if you want to find $$b$$.

Now you can get away with not knowing what $$F_0$$ is by taking a number of readings of amplitude of the driven system as a function of driver frequency but to eliminate $$F_0$$ you will have to make sure that $$F_0$$ does not depend on the frequency of the driver.