Question

un corp de masă m=0,1 Kg fixat de un resort, executa o miscare oscilatorie armonica cu frecventa v=20Hz si amplitudinea A=0,15m. Scrieti legea elongatiei, vitezei. Calculati energia totala a oscilatorului. Considerati faza initiala egala cu zero.

Answer 1

Elongația:  $y=A\sin(\omega t)=A\sin(2\pi\nu t)=0.15\sin(40\pi t)$

Viteza:  $v=\omega A\cos(\omega t)=2\pi\nu A\cos(2\pi\nu t)=40\pi\cdot 0.15\cos(40\pi t)$

Energia:  $E=\frac{mv^2}{2}+\frac{ky^2}{2}$

Dar  $k=m\omega ^2$

Deci $E=\frac{1}{2}mv^2+\frac{1}{2}m\omega ^2 y^2=\frac{1}{2}m\omega ^2 A^2\sin^2(\omega t)+\frac{1}{2}m\omega^2 A^2 \cos ^2 (\omega t)= \\ \\ =\frac{1}{2}m\omega^2A^2\left[\sin^2(\omega t)+\cos^2(\omega t)\right]= \\ \\ =\frac{1}{2}m\omega^2A^2$