Przy jakim wychyleniu względem amplitudy w ruchu harmonicznym: a) Ek=Eps b)Ek=2Eps c) 2Ek=Eps

Przy jakim wychyleniu względem amplitudy w ruchu harmonicznym: a) Ek=Eps b)Ek=2Eps c) 2Ek=Eps
Odpowiedź

[latex]a) E_{k}=E_{s} \ x(t)=x_{max} sin (omega t) \ V(t)=frac{dx}{dt}=omega x_{max}cos( omega t ) \ E_{k}=E_{s} \ frac{1}{2} mV(t)^{2}=frac{1}{2}x(t)^{2} \ [/latex]   [latex]momega^{2} x_{max}^{2}cos^{2}( omega t )=m omega^{2} x_{max}^{2}sin^{2}( omega t) \ cos^{2}( omega t)=sin^{2}( omega t) \ omega t= alpha \ [/latex]   [latex]left { {{sin^{2} alpha -cos^{2} alpha =0 } atop {sin^{2} alpha +cos^{2} alpha=1} o sin^{2} alpha = 1-cos^{2} alpha} ight \ 1-cos^{2} alpha - cos^{2} alpha=0 \ [/latex]   [latex]cos^{2} alpha=frac{1}{2} \ (cos alpha - frac{sqrt{2}}{2})(cos alpha + frac{sqrt{2}}{2})=0 \ cos alpha - frac{sqrt{2}}{2}=0 \ [/latex]   [latex]cos alpha = frac{sqrt{2}}{2} \ alpha _{1}=frac{ pi }{4} \ alpha_{2}=frac{7 pi }{4} \ alpha= omega t_{1} \ frac{ pi }{4}=omega t_{1} \ [/latex]   [latex]t_{1}=frac{ pi }{4 omega } \ frac{7 pi }{4}=omega t_{2} \ \ t_{2}=frac{7 pi }{4 omega } \ cos alpha+frac{sqrt{2}}{2}=0 \ [/latex]   [latex]cos alpha=-frac{sqrt{2}}{2} \ alpha_{3}=frac{3 pi }{4} \ alpha_{4}=frac{4 pi }{3} \ alpha= omega t \ [/latex]   [latex]frac{3 pi }{4}=omega t_{3} \ t_{3}=frac{3 pi }{4 omega } \ frac{4 pi }{3}=omega t_{4} \ t_{4}=frac{4 pi }{3 omega}[/latex]   [latex]x(t)=x_{max} sin (omega t) \ egin{cases} t_{1}=frac{ pi }{4 omega } \ t_{2}=frac{7 pi }{4 omega } \t_{3}=frac{3 pi }{4 omega } \ t_{4}=frac{4 pi }{3 omega } end{cases} \ [/latex]   [latex]x(t_{1})=x_{max} sin (omega frac{ pi }{4 omega }) =x_{max}sinfrac{ pi }{4}=frac{sqrt{2}}{2}x_{max} \ [/latex]   [latex]x(t_{2})=x_{max} sin (omega frac{7 pi }{4 omega })=x_{max}sinfrac{7 pi }{4}=-frac{1}{sqrt{2}}x_{max}=frac{-x_{max}}{sqrt{2}}=-frac{sqrt{2}}{2}}x_{max}[/latex]   [latex]x(t_{3})=x_{max}sin omega frac{3 pi }{4 omega}=x_{max}sinfrac{3 pi }{4}=frac{sqrt{2}}{2} x_{max} \ [/latex]   [latex]x(t_{4})=x_{max}sin omega frac{4 pi }{3 omega }=x_{max}sinfrac{4 pi }{3}=-frac{sqrt{3}}{2}x_{max}[/latex]   [latex]egin{cases} x_{1}=frac{sqrt{2}x_{max}}{2} \ x_{2}=frac{sqrt{3}x_{max}}{2} \x_{3}=-frac{sqrt{2}x_{max}}{2} \ x_{4} -frac{sqrt{3}x_{max}}{2} end{cases}[/latex]     pozostałe przykłady analogicznie.

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