[tex]dane:\\A = 2 \ cm\\t = \frac{1}{6}T\\szukane:\\x = ?\\\\Rozwiazanie\\\\x = A\cdot sin(\omega\cdot t)\\\\\omega = \frac{2\pi}{T}\\\\x = A\cdot sin(\frac{2\pi}{T}\cdot\frac{T}{6})\\\\x = A\cdot sin\frac{\pi}{3}\\\\x = 2 \ cm\cdot sin60^{o}\\\\x = 2 \ cm\cdot\frac{\sqrt{3}}{2}\\\\x = \sqrt{3} \ cm[/tex]
Odp. Szukana odległość wynosi √3 cm.
