m4tx45w

2022-01-27

Why the periodicity of solution for $\theta +\gamma \theta =0$ implies $\sqrt{\gamma }=n\in \mathbb{N}$
One solution for $\theta +\gamma \theta =0$
is, for $\gamma >0$
$\mathrm{\Theta }\left(\theta \right)=A\mathrm{cos}\sqrt{\gamma }\theta +B\mathrm{sin}\sqrt{\gamma }\theta$

Tyrn7i

Expert

The minimal period of the solution is $\frac{2\pi }{\gamma }$. You want that $2\pi$ is also a period of the solution. Thus you need it to be an integer multiple of the minimal period,
$2\pi =\frac{2\pi }{\sqrt{\gamma }}\cdot n$
This is directly equivalent to $\sqrt{\gamma }=n$

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