Question

6. Un oscilator liniar armonic trece prin punctele de coordonate x1 = 3 cm şi x2 = 4 cm cu vitezele v1 = 4 m/s, respectiv v2 = 3 m/s. Dacă masa oscilatorului este m = 20 g, calculaţi constanta elastică a resortului.

Answer 1

$x_1=Asin(\omega t_1+\varphi)=>sin(\omega t_1+\varphi)=\frac{x_1}{A} \\v_1=\omega Acos(\omega t_1+\varphi)=>cos(\omega t_1+\varphi)=\frac{v_1}{\omega A} \\sin^2(\omega t_1+\varphi)+cos^2(\omega t_1+\varphi)=1<=>(\frac{x_1}{A})^2+(\frac{v_1}{\omega A})^2=1<=>\frac{x_1^2}{A^2}+\frac{v_1^2}{\omega^2 A^2}=1 (1)$

$x_2=Asin(\omega t_2+\varphi)=>sin(\omega t_2+\varphi)=\frac{x_2}{A} \\v_2=\omega Acos(\omega t_2+\varphi)=>cos(\omega t_2+\varphi)=\frac{v_2}{\omega A} \\sin^2(\omega t_2+\varphi)+cos^2(\omega t_2+\varphi)=1<=>(\frac{x_2}{A})^2+(\frac{v_2}{\omega A})^2=1<=>\frac{x_2^2}{A^2}+\frac{v_2^2}{\omega^2 A^2}=1 (2)$

$(1)-(2)=>\frac{x_1^2}{A^2}+ \frac{v_1^2}{\omega^2A^2}- \frac{x_2^2}{A^2}+ \frac{v_2^2}{\omega^2A^2}=0<=>\frac{\omega^2x_1^2+v_1^2}{\omega^2A^2}-\frac{\omega^2x_2^2+v_2^2}{\omega^2A^2}=0|*\omega^2A^2=>\omega^2(x_1^2-x_2^2)=v_2^2-v_1^2=>\omega=\sqrt{\frac{v_2^2-v_1^2}{x_1^2-x_2^2} }=\sqrt{\frac{3^2-4^2}{3^2-4^2} }=\sqrt{1}=1 rad/s$

$(1)=>\frac{x_1^2}{A^2}+ \frac{v_1^2}{\omega^2A^2}}=1<=>\frac{3^2}{A^2}+\frac{4^2}{A^2}=1<=>25=A^2=>A=\sqrt{25}=>A=5 cm=0,05 m\\ E=\frac{m\omega^2A^2}{2}=\frac{0,02*1*0,0025}{2}= 0,000025 J\\ E=\frac{kA^2}{2}=>k=\frac{2E}{A^2}=\frac{2*0,000025}{0,0025}= 0,02 N/m =>k=0,02 N/m$