- #1

mishima

- 570

- 36

- Homework Statement
- Boas Chapter 11, Section 12, Problem 22. Show that the exact solution when ##\alpha## is not small is

$$sin \frac \theta 2=sin\frac \alpha 2 sn \sqrt \frac g l t$$

where k = ##sin\frac \alpha 2## is the modulus of the elliptic function.

- Relevant Equations
- $$sn(u) = \int_0^\phi \frac{d\theta}{\sqrt{1-k^{2}sin^{2}\theta}}=sin\phi$$

I understand how to reach

$$\int_0^\phi \frac{d\theta}{\sqrt{1-k^{2}sin^{2}\theta}}=\sqrt \frac g l t$$

from physics but from there I don't get how to turn that into this new (for me) sn(u) form.

$$\int_0^\phi \frac{d\theta}{\sqrt{1-k^{2}sin^{2}\theta}}=\sqrt \frac g l t$$

from physics but from there I don't get how to turn that into this new (for me) sn(u) form.