# Why do lighter pendulums come to rest faster than heavier ones?

I've been reading this website, which stated:

All pendulums eventually come to rest with the lighter ones coming to rest faster.

I have been taught the mass of a pendulum does not affect its frequency/period, but why does mass affect the time taken for a pendulum to come to rest?

Since we're talking about simple pendulums, we can assume the damping force is somewhat small and say $$x(t)=Ae^{-(b/2m)t}\cos(\omega't+\phi)$$ Where $$x$$ is the position as a function of time, $$A$$ is the initial amplitude of the pendulum, $$b$$ is a constant such that the frictional force is $$F=-bv$$, and $$\omega'=\sqrt{\omega^2_0-\frac{b^2}{4m^2}}$$ (on a side point, note that this indicates that for the same pendulum length and mass, the damped one will have a lower frequency than an undamped one, and this difference in frequencies between damped and undamped oscillators is dependent upon the mass). Clearly, decreasing the mass will decrease the exponent of $$e$$, so you'll see a 'faster' decay.

This makes sense intuitively: as mentioned, the damping friction is $$F=-bv$$, which is indpendet of the mass, and since $$F=ma$$, we visualize that the same force will cause a lower deceleration/acceleration for a smaller mass.

Because I like diagrams, here's a chart for you:

The continuous lines plot the exponential decay; the dotted lines are the actual displacements for a simple pendulum with $$A=0.15\ \rm m$$, $$l=1\ \rm m$$ (small angle approximations are valid), and $$b=5$$. The black line corresponds to a mass of $$15\ \rm kg$$, and the cyan line is for $$5\ \rm kg$$.

• Thank you. Could I ask of the source of the graph? – George Tian Oct 11 '18 at 12:19
• @GeorgeTian I made it with mathplotlib/python. But I threw it together while doing my homework, so the code isn't very clean, I'm afraid. If you'd like, I could send it to you over a chat, but I don't want to add that to the post itself. – Chair Oct 11 '18 at 12:31
• Well I have a variable graph on desmos: desmos.com/calculator/zstwnu0a7z – harshit54 Oct 11 '18 at 15:46
• @HarshitJoshi Hmm looks like you've clubbed together the $b/2m$ term, which makes it difficult to use while answering the question about the effect of the mass. – Chair Oct 11 '18 at 15:58
• I see your point. The graph might be useful to someone else though. – harshit54 Oct 11 '18 at 16:02

Suppose that you have two pendulums including the bobs of the same dimensions but made of materials with different densities.
If the bobs are displaced by the same amount then the bob made of the material with the larger density will have more gravitational potential energy and therefore will need to lose more energy before coming to rest.
Given that the major loss of energy will be due to air resistance and the aerodynamic shape of the two pendulums is the same the pendulum with starts with the greater amount of energy will take a longer time to come to rest.

The relaxation to equilibrium is predominantly a frictional effect, and the most obvious source is air resistance. The frictional force is written $$-\gamma v$$ where $$v$$ is the velocity and $$\gamma$$, the friction coefficient, depends on the dimensions of the pendulum bob and the viscosity of air (not on the particle mass). In the equations of motion (the Langevin equation) $$m\dot{v}=-\gamma v + \text{conservative forces} + \text{random force terms}$$ this naturally leads to a relaxation timescale $$\tau=m/\gamma$$. For simple damped harmonic motion, it is usually possible to neglect the random force terms, and the conservative forces are the usual harmonic ones.