The good news for modeling this response is that
- Damping is minimal, eliminating the need for an additional term in the equation of motion that opposes movement, and
- The displacements are small; since $\cos(15^\circ)$ is nearly 1, the weights continue to act nearly perpendicular to the beam (i.e., with a nearly constant downward force).
In other words, you have some downward forces from gravity and some inertias—that's all.
The bad news is that—judging from the no-added-mass and added-mass frequencies—the beam can't be idealized as either a tuning-fork geometry (i.e., a clamped–free cantilever beam with distributed mass but no end mass) or a massless cantilever with an end load. That's a shame, because both of those idealizations are quite accessible and well studied; Niels has linked the tuning-fork natural-frequency equation above:
$$f = \frac{1.875^2}{2\pi l^2} \sqrt\frac{EI}{\rho A}.$$
In addition, the natural frequency of a weightless spring (where the spring is the bending cantilever beam) is just $\sqrt{k/m}$, where $k$ is the spring constant (obtainable from the load–displacement relationship for a weightless cantilever beam) and $m$ is the mass of the hanging weight:
$$f = \sqrt\frac{3EI}{ml^3}.$$
In your experiment, however, you've added enough weight at the end of the beam to change the system frequency but not so much that the beam's self-weight is negligible in comparison. This takes the solution out of undergraduate textbooks (and probably even many graduate textbooks).
That's not to say that the problem hasn't been considered in the literature; one treatment appears in Laura et al.'s "A note on the vibrations of a clamped-free beam with a mass at the free end" Journal of Sound and Vibration (1974), for instance. (See also its references and citing articles.)
Laura et al. provide a table of frequencies for various ratios of end mass to beam mass (for example, if the weight at the end is 20% as heavy as the beam itself, the 1.875 factor above changes to 1.616, so I'm going to predict from your frequency measurements that your ruler has a mass of ~20 g) and you could certainly plot your results against these tabulated predictions, even if the mathematical machinery is more complex than you wish to explore right now. Of course, you're also free to solve the (nonlinear differential) equation of motion numerically yourself if that sounds interesting and feasible.
Specifically, Laura et al. start with the beam vibration equation
$$EI\frac{\partial^4 w}{\partial x^4}+\rho A\frac{\partial^2 w}{\partial t^2}=0,$$
with flexural stiffness $EI$, deformation $w$, axial distance $x$, density $\rho$, cross-sectional area $A$, and time $t$. They then apply, in addition to the free-end and clamped-end boundary conditions, the boundary condition corresponding to a lumped mass $M$:
$$-\left[-EI\frac{\partial^3 w}{\partial x^3}(L,t)\right]=M\frac{\partial^2 w}{\partial t^2}(L,t).$$
They ultimately end up with the transcendental equation
$$\frac{1}{y}\left(\frac{1+\cos y\cosh y}{\sin y\cosh y-\cos y\sinh y}\right)=\frac{M}{M_v},$$
where $M_v$ is the beam mass. The first-mode resonant frequency is then
$$f_1=\frac{1}{2\pi}\left(\frac{y_1}{L}\right)^2\sqrt{\frac{EI}{\rho A}}.$$
For $M=0$, $y_1\approx 1.875$, as given above.
These days, one can solve the transcendental equation with ease by visiting Wolfram Alpha, say, and entering Solve[(1/y)(1+Cos[y]Cosh[y])/(Sin[y]Cosh[y]-Cos[y]Sinh[y])==(0.2),y]
for the case of $M/M_v=0.2$, which yields a root of about 1.616. One can confirm Laura et al.'s table this way and—much more importantly—easily fill in intermediate values.
Good luck, and please let me know if anything's unclear.