$ \frac{14}{3} = \frac{28}{6} $, $ \frac{7}{2} = \frac{21}{6} $, LCM of periods $ \frac{14}{3} $ and $ \frac{7}{2} $: write as $ \frac{28}{6}, \frac{21}{6} $, GCD of 28 and 21 is 7, so period is $ \frac{14}{\gcd(14,21)} \cdot \frac{1}{\text{some factor}} $. Better: find least $ T $ such that $ \frac{3\pi}{7}T = 2\pi m $, $ \frac{4\pi}{7}T = 2\pi n $, so $ T = \frac{14m}{3} = \frac{7n}{2} \Rightarrow \frac{14m}{3} = \frac{7n}{2} \Rightarrow 28m = 21n \Rightarrow 4m = 3n $. Small

Sat, 02 May 2026 20:15:32 +0000

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