How do we determine the damping coefficient given the acceleration vs time graph of a 1D mechanical system? Consider a mechanical system about which the only data we have is a graph that shows acceleration vs time. I would like to figure out what the damping coefficient $c$ is. 
Instead of displacement as shown in the attached image, the $Y$ axis value would be acceleration. The mass being damped is moving horizontally and does not move at all vertically.
How would I go about this?
 A: How to Fit

We have data points the oscillate curve $~x(t)~$.
Step 1: obtain the blue points
The period  of this curve is $~T=2~$, hence the points are
$$X=T_0+n\,T~,Y=~x(T_0+n\,T)~,n=0,(1),N_p,$$
where $~T_0=\frac T4.$
Step II
Fit the points $[~X~,Y]~$ to the curve $~a\,e^{-\beta\,t}~$ where $~a=x(T_0)$ and you obtain $~\beta~[1/s]$ from the  fitting process.

The blue curve is your result.
A: The question in the OP is solved by tools referred to as second order system identification (ID) techniques in the control theory.

 Consider a general second order system with the constant parameters of the constant force, mass, spring constant and damping coefficient denoted by $\kappa \in \mathbf{R}, 0<m, 0\leq k, c$ respectively determining the second order one dimensional linear differential (ODE) equation $m\ddot{x}=\kappa-kx-c\dot{x}+u$ which is equivalent to the system $m\ddot{x}+c\dot{x}+kx=\tilde{u}$ with the time-varying proportional-integral-derivative (PID) and additive affine control function $\tilde{u}=u-\text{gain}_1x-\text{gain}_2\dot{x}-\text{gain}_3\ddot{x}$, wherein $0\leq \text{gain}_1,\text{gain}_2,\text{gain}_3$. Let $y:=x-\frac{\kappa}{k}$ so that the equivalent ODE is $m\ddot{y}=-ky-c\dot{y}+u$ which is equivalent to the equation $\ddot{y}+\frac{c}{m}\dot{y}+\frac{k}{m}y=\ddot{y}+2\xi\omega_n\dot{y}+\omega_n^2y=\frac{u}{m}$ where $0<\omega_n:=\sqrt{\frac{k}{m}}$ and $0<\xi:=\frac{c}{2m\omega_n}=\frac{c}{2\sqrt{km}}$. The Laplace transform of the ODE with vanishing initial condition is $(s^2+2\xi\omega_n s+\omega_n^2)Y(s)=\frac{U(s)}{m}$ where $s^2+2\xi\omega_n s+\omega_n^2=0$ is the characteristic equation of the transfer function $T(s)=\frac{Y(s)}{U(s)}=\frac{1}{k}\frac{\omega_n^2}{s^2+2\xi\omega_n s+\omega_n^2}$ of the system represented by the ODE with no zeros and poles $s_{1,2}:=-\xi\omega_n\pm i\omega_d=-\omega_n(\xi\mp i\sqrt{1-\xi^2})$ where $\omega_d=\omega_n\sqrt{1-\xi^2}$, so that $T(s)=\frac{1}{k}\frac{\omega^2}{(s-s_1)(s-s_2)}=\frac{\text{const}_1}{s-s_1}+\frac{\text{const}_2}{s-s_2}$, wherein $\text{const}_1,\text{const}_2$ are constants, the inverse transform of which readily yields the general solution to the ODE. Note that the Laplace transform of the ODE without the vanishing initial condition assumption $(s^2+2\xi\omega_n s+\omega_n^2)Y(s)=\frac{U(s)}{m}+y(0)s+(y(0)+\dot{y}(0))$ or $\frac{Y(s)}{U(s)+my(0)s+m(y(0)+\dot{y}(0)}=\frac{1}{k}\frac{\omega_n^2}{s^2+2\xi\omega_n s+\omega_n^2}$, in the unforced case $u=0$, yields $Y(s)=\frac{1}{k}\frac{(m y(0) s+m (y(0)+\dot{y}(0)))\omega_n^2}{(s^2+2\xi\omega_n s+\omega_n^2)}=\frac{y(0) s+ (y(0)+\dot{y}(0))}{(s^2+2\xi\omega_n s+\omega_n^2)}$, wherein it is usually possible to identify the time instants to initialize the data such that $y(0)=0$ or $\dot{y}(0)=0$. If we choose the former assumption, the response in this case is $Y(s)=\frac{\dot{y}(0)}{s^2+2\xi\omega_n s+\omega_n^2}$. Here, we note the correspondence to the fact that the transfer function of an linear time invariant (LTI) single-input-single-output (SISO) system is the impulse response of the output, that is, in other words the velocity acts as the impulse magnitude in this case.



 A relevant aside is that the general solution of the $n^{th}$ order linear ODE $\sum_k \alpha_k y^{(k)}=0$ where $y^{k}:=\frac{d^k y}{dt}$, $0\leq k$, with real valued coefficients $\alpha_k$ is given by $y=\sum_k \beta_k e^{-s_k t}$ in the case that ${s_k}_{0\leq k\leq n}$ are distinct roots of the characteristic equation of the ODE. We now present the explicit analytical solution to the unforced dynamics wherein the control vanishes $u=0$ which have been shown to have the response Laplace function $Y(s)=\frac{\gamma}{s^2+2\xi\omega_n s+\omega_n^2}$, $\gamma\in\mathbf{R}$. Note that this is not a restrictive assumption and is in fact representative of data available for most systems which require system ID. This is because it is convenient to obtain the time domain data of the unforced system $u=0$ by exciting it initially with an impulse or short duration step input. Indeed, the step response of the system enables further analysis and related system ID relationships associated with the rise time, settling time and maximum overshoot. However, the analysis presented here enables us to obtain the parameters $\omega_n, \xi$ which determine the physical characteristics of the system using time domain data of the unforced system without knowledge of the type of input excitation by using response data after the excitation has ceased.


 Note that the general solution to the second order linear ODE with the Laplace transform $Y(s)=\frac{\gamma}{s^2+2\xi\omega_n s+\omega_n^2}$ is $y=c_1 e^{s_1 t}+c_2 e^{s_2 t}$ where $c_1,c_2$ are constants which can be obtained from the initial conditions (two of three $y(0),\dot{y}(0),\ddot{y}(0)$) if the roots of the characteristic equation are distinct.


- In the case that $0<\xi<1$ which represents the case of the under-damped system, $s_1$ and $s_2$ given by $s_{1,2}=-\xi\omega_n \pm i\omega_n \sqrt{1-\xi^2}=-\xi\omega_n \pm i \omega_d\in\mathbf{C}$ are complex conjugates in the left half of the $\mathbf{C}$ plane and the solution can be written as $y=e^{-\xi\omega_nt}(c_1\cos(\omega_d t)+c_2\sin(\omega_d t))=Ce^{-\xi\omega_nt}\sin(\omega_d t+\tilde{\phi})$ where $0<\omega_d:=\omega_n\sqrt{1-\xi^2}$ denotes the damped angular frequency and $\tilde{\phi}:=\arctan2(c_1,c_2)$ denotes the phase angle, wherein $c_1,c_2,C,\tilde{\phi}$ are constants which can be obtained from the initial conditions (two of three $y(0),\dot{y}(0),\ddot{y}(0)$). Note that the condition $\xi=0$ implies that $\omega_d=\omega_n$ and $y=C\sin(\omega_n t+\tilde{\phi})$, which represents the case of the un-damped system.
- In the case that $\xi=1$ which represents the case of the critically-damped system, $s_{1,2}=-\xi\omega_n$ and the solution can be written as $y=Cte^{-\xi\omega_n t}$, wherein $C$ is a constant which can be obtained from the initial conditions (one of three $y(0),\dot{y}(0),\ddot{y}(0)$). In this case the behavior of the system is identical to that of a system represented by a first order linear ODE.
- In the case that $1<\xi$ which represents the case of the over-damped system, $s_{1,2}=-\omega_n(\xi\pm\sqrt{\xi^2-1})<0$ and the solution can be written as $y=c_1 e^{s_1 t}+c_2 e^{s_2 t}$,  wherein $c_1,c_2$ are constants which can be obtained from the initial conditions (two of three $y(0),\dot{y}(0),\ddot{y}(0)$).-->

The following details the system ID technique of a second order system with the vanishing initial condition from the time domain data of it's unforced response $u=0$, rather than from the step or impulse response data. Even in the case of the step response, the unforced response data is available after the initial time interval (incremental in the case of the impulse input) in which the step or impulse input is applied.

*

*The data response of the system provided as accelerations as a function of time for a system can be time integrated to obtain data of the velocity and position as functions of time. The damped angular frequency of the system with the position response akin to that shown in the plot in the OP (which is readily determined to be an under-damped second order system based on the analysis presented), can be obtained by $$\omega_d=\frac{2\pi}{T_d}\text{ where } T_d:=\frac{t_k-t_1}{k-1},$$ wherein $t_k$ is the time instant at which the $k^{th}$ maximum or peak or crest (or minimum or negative peak or trough) of the $k^{th}$ oscillation (in other words $t_k-t_1$ is the time duration which is the sum of time periods of $k-1$ oscillations). Next, the difference between natural logarithm of two (say) peaks (or negative peaks) or maxima (or minima) $k$ peaks (or negative peaks) apart, given by $\ln(y^p(t_k))=-\xi\omega_n t_{k} + \ln(C\sin(\omega_d t_k+\tilde{\phi}))$ and $\ln(y^p(t_{k-1}))=-\xi\omega_n t_{k-1} + \ln(C\sin(\omega_d t_{k-1}+\tilde{\phi}))$ is $\ln(y^p(t_k))-\ln(y^p(t_{k-1}))=-\xi\omega_n (t_k-t_{k-1})$ since $\sin(\omega_d t_{k-1})+\tilde{\phi})=\sin(\omega_d t_{k}+\tilde{\phi}))$, so that $$\xi\omega_n=\frac{\ln(y^p(t_{k-1}))-\ln(y^p(t_k))}{t_k-t_{k-1}}=\frac{\ln(y^p(t_{k-1}))-\ln(y^p(t_k))}{\frac{2\pi}{\omega_d}},$$ where the time interval $[t_{k-1},t_k], \;0<t_{k-1}<t_k$ indicate the time period $T_d$ of the damped oscillatory response. Note that if we instead used the time instants $t_1$ and $t_k$ for this calculation, the relationship between the values of the logarithm of the two peak (or negative peak) values is given as $\xi\omega_n=\dfrac{\ln(y^p(t_{1}))-\ln(y^p(t_k))}{\frac{2\pi(k-1)}{\omega_d}}$. The natural angular frequency and subsequently, the damping coefficient, are then determined using the values of the expressions obtained above and the definition which states their relationship as $$\omega_d=\omega_n\sqrt{1-\xi^2}.$$

*An alternate solution is to use the method of least squares to estimate the parameters of the dynamical model by directly using the time domain data.

