How does a metronome allow such a wide range of tempos in such a short distance? I've been working on a question for metronomes asking the following:

Suppose a metronome can be treated as a double-weighted pendulum. Let $m_1$ be the mass of the metronome's movable weight, $l_1$ the distance of $m_1$'s center of mass from the rotation point, $m_2$ the mass of the fixed counterweight, and $l_2$ the distance of $m_2$'s center of mass from the rotation point.  Derive an equation for the natural frequency of such a system in beats per minute.  Assume the mass of the rod is negligible and oscillations are small, and treat the two masses as point masses.     


I first set up the angular equation of motion as:
$$m_2g\sin\theta l_2-m_1g\sin\theta l_1=(m_1l_1^2+m_2l_2^2)\frac{d^2\theta}{dt^2}$$
Using that for small angles, $\sin \theta \approx \theta$ and rearranging, I get
$$\frac{d^2\theta}{dt^2}+\frac{m_1l_1-m_2l_2}{m_1l_1^2+m_2l_2^2}g\theta=0$$
from which I get the natural angular frequency
$$\omega_0=\sqrt{\frac{m_1l_1-m_2l_2}{m_1l_1^2+m_2l_2^2}g}$$
which I can convert to beats per minute by noting that a metronome beats twice per cycle, resulting in:
$$bpm=\frac{60}{\pi}\sqrt{\frac{m_1l_1-m_2l_2}{m_1l_1^2+m_2l_2^2}g}$$  
The range of tempos for a metronome is typically 40-208 bpm, with 208 bpm being the closest marked tempo to the rotation point, and 40 bpm the farthest.  Since the counterweight is fixed, the bpm equation is essentially a function of $l_1$, and so if we fix particular values of $m_1$, $m_2$, and $l_2$, then we can find how long of a rod we need (i.e. the range of values for $l_1$ for tempos between 40 and 208 bpm).  If I set bpm=208 (and $m_1$=20 grams, $m_2$=22 grams, and $l_2$ around 20 mm, although these values are not particularly special) and solve for $l_1$, I get reasonable answers on the order of a few centimeters, but for bpm=40, I'm getting values on the order of meters, which is clearly much longer than an actual metronome rod.  So overall my questions are:  


*

*Is my analysis correct based on the assumptions I've made?

*If so, what is causing such a dramatic error in the length of the rod?


My guesses on potentially spurious assumptions I've made are:  


*

*assuming the rod has negligible mass, although this seems reasonable looking at a typical metronome.

*approximating the two masses as point masses, since the moment of inertia of the system is the denominator of the $\omega_0$ term.

*Neglecting frictional losses/the type of frictional loss present.  Including a loss term proportional to the speed doesn't seem to help much, but I've seen that metronomes are typically modeled with a Van der Pol loss term, which could make a difference, but I don't know enough about the topic to say.  

*Neglecting the effect of the escapement, the gear-spring mechanism that regulates the amplitude of the metronome.  


Any advice on this would be greatly appreciated, as well as any criticisms on formatting, as I am still learning.
 A: I suspect that your "order of meters" results are stemming from forgetting the sign in the  l.h.s. of your very first equation and subsequently jumping into the realm of purely imaginary frequencies.
The correct first equation is
$$
-m_2g\, l_2 \sin\theta+m_1g\,l_1\sin\theta =(m_1l_1^2+m_2l_2^2)\frac{d^2\theta}{dt^2}.
$$
To check it, we can set lighter mass to zero and assume $\theta \ll 1$. We get $-  g \theta =   l_2 \ddot\theta$, that is a normal equation for a pendulum.
One suggestion: your frequency $\omega$ remains the same if you multiply both masses by  the same number. So in order to reduce the number of parameters you have to fit, eliminate one mass (say, $m_2$) from the equation by introducing ratio $x=\frac{m_1 }{m_2 }$:
$$
\omega_0=\sqrt{\frac{l_2 -x\,l_1}{l_2^2+x\,l_1^2}\,g}.
$$
Signs were corrected w.r.t. your expression. 
Some analysis:


*

*Since $m_1$ is the lighter mass we have $x<1$.

*For the frequency to be real we have the maximum allowed value of $l_1$: $l_1 < x^{-1}\,l_2$. The frequency would tend to zero in that case.

*Maximum frequency achievable for a given $l_2$ is obtained by setting $l_1$ to zero: $\omega_\text{max}=\sqrt{\frac g {l_2}}$

