harmonic-oscillatorhomework-and-exercisesmassspring
I'm trying to determine the viscous damping coefficient of a spring $c$. Read about it on Wikipedia here.
The two equations which I have are:
$f=-cv$ and $ma+cv = -kx$
I know the spring constant $k=5$, the mass is $50\text{ }\mathrm{g}$ and the initial amplitude of the spring is $10\text{ }\mathrm{cm}$.
OK, I will assume you have the under-damped case.
If you continue reading the wikipedia article in question you'll find the solution for a underdamped oscillator writen as $$ x(t) = e^{- \zeta \omega_0 t} (A \cos(\omega_\mathrm{d}\,t) + B \sin(\omega_\mathrm{d}\,t )) $$ with $A$ and $B$ constant.
So, take you data, and plot all the maxima (or minima) as a function of time, fit an exponential {*} to that and $\zeta \omega_0$ pops right out.
If you also need to get $\omega_o$ from the data use $$ \omega_\mathrm{d} = \omega_0 \sqrt{1 - \zeta^2 } $$ where you get $\omega_\mathrm{d}$ by extracting the average period (i.e. time from peak to peak) in the data and noting that the period is $T_d = \frac{2 \pi}{\omega_\mathrm{d}}$.
Now you have two equations for two unknowns, so all you have left is a bit of algebra.
{*} Or plot amplitude versus time on semi-log paper if you are doing this the old-school way. Or plot log(amplitude) versus time on linear--linear graph paper. Then extract the slope.
Yes, one can calculate an equivalent mass. Take the following coordinates, that describe the position of the particles' centers with respect to the center of mass of the set, $G$. In the sketch that follows I have arbitrarily chosen $m_1 < m_2$, and so $G$ falls closer to $m_2$:
Since $G$ will remain fixed in the absence of external forces, it makes sense to use the coordinates above. The equations of motion of the two particles in these coordinates read $$m_1\frac{\mathrm{d}^2 x_1}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0 \\ m_2\frac{\mathrm{d}^2 x_2}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$ where $l = x_1 + x_2$ (the distance between centers). Adding these two equations and deviding through by $2$ one obtains $$\frac{m_1 + m_2}{2}\frac{\mathrm{d}^2 l}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$ which is of the form of the original equation (the one from Wikipedia) if one takes $$m = \frac{m_1 + m_2}{2}$$
Thus, the equivalent mass is just the arithmetic average of the two masses.
Best Answer
with the given $m$ and $k$ you indeed cannot calculate the damping coefficient $c$.
Remember that you just have a model where you put some constants in and you can only derive other constants which somehow depend on them.
The question concerning an actual measurement was answered in Investigating the dampening of a spring.
Greets