# How to work out the damping or deflection curve from deflection angles?

I have a crane boom that is rotating and when it is stopped suddenly, the boom flexes back and fourth several times. I have measured the angles between each deflection and have $3.16 ^\circ$, $2.42 ^\circ$, $1.77 ^\circ$ and $1.31 ^\circ$, which if plotted in Excel forms a logarithmic curve equal to $$y = -1.33ln(x) + 3.2221$$ The goal now is to find how to plot this as a sin curve with the period of 3.5 seconds so that it it looks like a deflection vs time chart (or damping curve), similar to the one in red below. Can someone please explain how I can do this?

# The model

The equation for a damped periodic oscillation is $$y(t) = A e^{-\lambda t} \cos(\omega t + \phi)$$ where $A$ is the amplitude, $\lambda$ is the decay constant, $\omega = 2\pi/\tau$ where $\tau$ is the period, and $\phi$ is the phase angle at $t=0$. In your figure, $\phi = -\pi/2$, and therefore your damping curve has the form $$y(t) = A e^{-\lambda t} \sin(\omega t)$$

Instead of using $$y(t) = A e^{-\lambda t} \quad \quad \text{(Correct)}$$ you have fit the portion of the curve where $\sin(\omega t) = 1$ using the relation $$y(t) = B \ln(t) + C \quad \quad \text{(Wrong)}\,.$$ This model predicts an infinite displacement $y$ at time $t=0$.
If we fit the correct model to data extracted from your log model, we get the following curves (note that we can calculate the full deflection-time curve after fitting $A$ and $\lambda$ because $\omega$ can be calculated from the period):