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I've been looking through my textbooks and I've found a number of different equations - so I wanted to confirm with you which it is.

What is the equation which determines the damping constant (gamma) in a spring executing shm?

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There are many different models for springs, and usually damping is added in by putting a term proportional to the velocity with a negative coefficient. Please elaborate. Also, – Sammy Black May 6 '11 at 7:28
Also, is it actually simple harmonic motion if the motion is damped? – Sammy Black May 6 '11 at 7:29
Besides, the damping constant is a parameter. It's something you put into your equations so that you can model your system correctly. Of course, if you assume the model to be a damped oscillator, there's a way to learn from the equation what measurement/experiment you should perform to determine the damping constant of the system. Maybe that is what you are after? – Raskolnikov May 6 '11 at 7:36
It is an underdamped spring. Basically I'm trying to find the equation which will calculate the damping constant given the mass, spring constant, amplitude or whichever other variables are relevant. There are a variety of formulas in the link Sammy provided, basically I'd just like some confirmation as to which will find the 'damping constant' (ie the constant in the exponent of the decaying exponential equation of the damped motion). – user10265 May 6 '11 at 8:03

In the equation for the motion of a damped spring

$$m\ddot{x} + c\dot{x} + kx = 0$$

the parameter $c$ is known as the viscous damping coefficient. From it you can determine the damping ratio

$$\zeta = \frac{c}{2\sqrt{mk}}$$

which determines if the spring is underdamped ($\zeta<1$), overdamped ($\zeta>1$) or critically damped. The parameter $c$ itself is a property of the spring, and is not determined by $m$ or $k$.

You solve this equation by postulating a solution of the form $x(t)=e^{\gamma t}$, giving a quadratic equation for $\gamma$:

$$m\gamma^2 + c\gamma + k = 0$$

which you solve with the usual quadratic formula:

$$\gamma_{\pm} = \frac{1}{2m} \left( -c \pm \sqrt{c^2 - 4mk} \right)$$

giving two possible values for the decay constant. If $c^2<4mk$ then your spring is underdamped, and the rate of decay is given by the real part of $\gamma$, ie $\mathrm{Re}(\gamma) = -c/(2m)$.

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Thanks Chris, that's exactly the confirmation I was looking for. Just one thing, how do I actually calculate the value of c? – user10265 May 6 '11 at 12:13
You can't calculate it given the information in the equation, ie. from the values of $m$ and $k$. It is an intrinsic property of the spring. To calculate it you'd either need to perform an experiment to measure the decay rate, and use $c=2m\mathrm{Re}(\gamma)$, or derive it from physical properties of the spring, the material that you're oscillating in (vacuum, air, water will all give different values of $c$) and possibly other factors. I'm not a physicist, though, so can't help much here! – Chris Taylor May 6 '11 at 12:36
Of course, that makes sense. Finally, can you explain the practical, physical reasons why and increase in mass/spring constant causes the damping constant to decrease? I mean is there additional friction between the spring coils at a higher spring constant? What is the reasoning? – user10265 May 7 '11 at 5:21
What are the units of the damping constant? (by dimensional analysis of the equation) – user10265 May 7 '11 at 10:57

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