I'm a high school student doing a research paper on the vibration of cantilever beams. Specifically, I am investigating the relationship between the mass on a cantilever beam and the time period of its oscillation. I attached several masses to the fixed end of a cantilever beam (a 100 cm ruler) and gave it an initial displacement, causing an oscillation.

My experimental data shows a strong correlation between the mass on the fixed end of the cantilever and the period. However, I am not sure how to verify this without a mathematical model. I am trying to find an equation or at least derive one that relates these two variables.

I have been researching Euler-Bernoulli beam theory and have come across the Euler-Bernoulli equation and the dynamic beam equation. I am unfamiliar with the advanced calculus for the latter so kindly bear with me for any theoretical mistakes in my question.

I did find the following thread: https://physics.stackexchange.com/a/171260/340452 but I am not sure if the result stands for oscillation with a load still on the beam. Can an equation be derived for this situation? And further, what difference would it make if the mass was instead attached at the center of the beam?

Thank you.

  • $\begingroup$ Look up "tuning fork" on wikipedia for the equation you seek. -NN $\endgroup$ Commented Jul 12, 2022 at 6:13
  • $\begingroup$ @nielsnielsen I see, thank you. I'm not sure how this would apply to the situation of a point load on a beam while it vibrates, though. Is it a variable that is relevant? $\endgroup$
    – Emp1
    Commented Jul 12, 2022 at 8:24
  • $\begingroup$ The resonant frequency of a cantilever beam is derived from the equations of motion and the deflection equations for that cantilever beam. I thought the tuning fork case might furnish some useful clues on how to model the system. -NN $\endgroup$ Commented Jul 12, 2022 at 16:00
  • $\begingroup$ Welcome to Physics SE! (1) Do the oscillations continue for at least several cycles without much attenuation, or do they damp out quickly after a few cycles or less? (2) Is the frequency with a mass attached very different from the frequency with no mass attached? (3) What is the maximum angle (from the horizontal) of the tip during oscillation? That is, are the tip displacements small or large? The answers will determine the appropriate model to explore. $\endgroup$ Commented Jul 12, 2022 at 17:21
  • $\begingroup$ Thank you @Chemomechanics here are the answers: (1) There is little damping, the beam keeps oscillating at frequencies ranging from 2-4 Hz for several seconds (~10 seconds). (2) The beam oscillates at a frequency of 2.2 Hz with no mass attached, 2.14 Hz with 20 g attached (incrementing by 20 g for each run) until the frequency is 1.74 Hz with 140 g attached. (3) I would say about 10-15 degrees. $\endgroup$
    – Emp1
    Commented Jul 13, 2022 at 1:03

1 Answer 1


The good news for modeling this response is that

  1. Damping is minimal, eliminating the need for an additional term in the equation of motion that opposes movement, and
  2. 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.

  • $\begingroup$ Thank you so much! Laura et al. is very informative. $\endgroup$
    – Emp1
    Commented Jul 13, 2022 at 17:23
  • $\begingroup$ I'm very glad this was helpful! Since that paper is behind a paywall, I might as well reproduce the key equations, as the result is easily numerically obtained with today's computers, even online portals. One example is given. $\endgroup$ Commented Jul 13, 2022 at 18:07

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