How does one find out the viscosity of a fluid, using viscous damping of an oscillating spring? Any help would be greatly appreciated.

  • $\begingroup$ Welcome to Physics Stack Exchange. This is a question and answer site, but it's not a problem solving help site. With questions like this it is required that you show your own effort and ask a specific conceptual question about whatever has you stuck. $\endgroup$
    – DanielSank
    Oct 1, 2015 at 23:19
  • 1
    $\begingroup$ Possible duplicate of How to calculate viscous damping coefficient? $\endgroup$
    – nluigi
    Oct 2, 2015 at 0:15

1 Answer 1


You have to make an additional assumption about how the friction force is affected by the viscosity. For small speeds you can assume Stokes flow around the oscillating mass. If the mass is a ball with radius $a$, then the friction force $F$ is given by velocity $v$ and dynamic viscosity $\mu$ by this relation: $$ F = -6 \pi \mu a v $$

(see https://en.wikipedia.org/wiki/Stokes_flow)

From the solution to the harmonic oscillator equation we know, that for the object on a spring damped by the force given by $F = -bv$ (so in our stokes flow, $b = 6 \pi \mu a$), the resulting motion $x(t)$ will obey this equation:

$$ x(t) = A e^{-\frac{2b}{m}t} \sin(\omega t + \phi_0) $$

(this is valid for under-damped case, when the damping is small enough that the object can oscillate, generally see https://en.wikipedia.org/wiki/Harmonic_oscillator). In this equation $m$ is the mass of the object, $\omega$ is the natural frequency of the oscillator, determined by the $b$ and spring stiffness, and $A$ and $\phi_0$ are constants determined by initial conditions.

For us it means, that the amplitudes of the oscillation decrease with time exponentially as $e^{-\frac{2b}{m}t}$. Therefore, if we measure the times and amplitudes of several oscillations, we can determine the $\frac{2b}{m}$ constant an in turn, compute the viscosity $\mu$ using the equation $b = 6 \pi \mu a$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.