[tex]Dane:\\k = 20 \ \frac{N}{m}\\T = 1 \ s\\Szukane:\\m = ?[/tex]
Rozwiązanie
Korzystamy ze wzoru na okres wahadła sprężystego:
[tex]T = 2\pi\sqrt{\frac{m}{k}} \ \ |^{2}\\\\T^{2} = 4\pi^{2}\cdot\frac{m}{k} \ \ /\cdot k\\\\4\pi^{2}m = kT^{2} \ \ /:4\pi^{2}\\\\\underline{m = \frac{kT^{2}}{4\pi^{2}}}\\\\m = \frac{20\frac{N}{m}\cdot(1 \ s)^{2}}{4\cdot3,14^{2}} = \frac{20\frac{kg\cdot m}{s^{2}}\cdot\frac{1}{m}\cdot s^{2}}{39,4384}\\\\\boxed{m \approx0,5 \ kg}[/tex]
