-1

Consider the mathematical pendulum

$$\dot{\theta}=\omega$$ $$\dot{\omega}=-\frac{g}{L}sin\left(\theta\right)$$

How can one prove that it is impossible that the time period $T$ depends only on the length $L$, and the mass $m$, i.e, that there is no such function as $f(T,L,m)=0$.

jarhead
  • 272
  • 2
  • 10
  • 1
    Obviously, there is no $m$ in the equation, and the dynamic or time scale depends only on the fraction $\frac{g}{L}$. One could then in the next stage find that the period also depends on the level of the energy function. – Lutz Lehmann Apr 07 '19 at 12:45
  • Intuitively, if you make the initial value of $\omega$ large enough, the pendulum reaches the angle $\theta = \pi$ with only a small percentage reduction to $\omega$ and it continues in the same direction indefinitely. Consider what happens if you increase the initial $\omega.$ On the other hand, there's some initial $\omega$ such that $\omega$ goes to zero as $\theta$ approaches $\pi.$ – David K Apr 07 '19 at 12:53
  • There is a nice 3Blue1Brown tutorial starting at 5:55. – g.kov Apr 08 '19 at 16:26

1 Answers1

2

You can directly compute the period given the largest angle $θ_\max$ as $$ T=4\int_0^{θ_\max}\frac{dθ}{\sqrt{2\frac{g}L(\cosθ-\cosθ_\max)}} =2\sqrt{\frac{L}g}\int_0^{θ_\max}\frac{dθ}{\sqrt{\sin^2(θ_\max/2)-\sin^2(θ/2)}} $$ So you get a dependence on $\frac{L}{g}$ and $θ_\max$, but not on $m$ and $L$ alone.

Lutz Lehmann
  • 126,666