Fizică, întrebare adresată de MFM, 9 ani în urmă

1.7 si 1.9.Multumesc.

Anexe:

Răspunsuri la întrebare

Răspuns de stassahul
13
[tex]\displaystyle \text{1.7 Se da:}\\ \\ m=200g=0,2kg\\ \\ x_1=10cm=0,1m\\ \\ v_1=20\frac {cm}s=0,2\frac ms\\ \\ x_2=20cm=0,2m\\ \\ v_2=8\frac {cm}s=0,08\frac ms\\ \\ a=6\frac {cm}s^2=0,06\frac m{s^2}\\ \\ \\ a)\text{Ecuatia}\\ \\ b)k=?\frac Nm\\ \\ c)x=?m\\ \\ d)E_c'=E_p',\text{ }x'=?m\\ \\ \\[/tex]


[tex]Formule:\\ \\ a)\text{Ecuatia uzuala ce descrie miscarea unui oscilator armonic:}\\ \\ x(t)=A\times\sin{(w\times t+\varphi_0)}\\ \\ \text{Nu e nici o vorba despre faza initiala, deci reducem la:}\\ \\ x(t)=A\times\sin{(w\times t)},\text{ unde trebuie sa aflam A si w}\\ \\ \\ \text{Sa formam un sistem de ecuatii din cele 2 cazuri propuse in problema:}\\ \\ x_1=A\times\sin{(w\times t_1)}\\ \\ v_1=w\times A\times\cos{(w\times t_2)}(\text{formula vitezei})\\ \\ \text{Sistem:}\\ \\[/tex]

[tex]\displaystyle \begin{cases}x_1=A\times\sin{(w\times t_1)}\\v_1=w\times A\times\cos{(w\times t_1)}\end{cases}\Leftrightarrow \begin{cases}\frac {x_1}{A}=\sin(w\times t_1)\\\frac{v_1}{w\times A}=\cos(w\times t_1)\end{cases}\\ \\ \begin{cases}\frac {x_1^2}{A^2}=\sin^2(w\times t_1)\\\frac{v_1^2}{w^2\times A^2}=\cos^2(w\times t_1)\end{cases}\\ \\ \text{Adunam ecuatiile:}\\ \\ \frac {x_1^2}{A^2}+\frac{v_1^2}{w^2\times A^2}=\sin^2(w\times t_1)+\cos^2(w\times t_1)\\ \\[/tex]

[tex]\displaystyle \text{Iar din trigonometrie stim ca: } \sin^2(x)+\cos^2(x)=1\\ \\ \frac {x_1^2}{A^2}+\frac{v_1^2}{w^2\times A^2}=1\text{ (1)}\\ \\ \\ \text{Analogic cu } x_2 \text{ si } v_2!\\ \\ \frac {x_2^2}{A^2}+\frac{v_2^2}{w^2\times A^2}=1\\ \\ \text{Egalam ecuatiile:}\\ \\[/tex]

[tex]\displaystyle \frac {x_1^2}{A^2}+\frac{v_1^2}{w^2\times A^2}=\frac {x_2^2}{A^2}+\frac{v_2^2}{w^2\times A^2},\text{Scapam de } A^2\\ \\ x_1^2+\frac{v_1^2}{w^2}=x_2^2+\frac{v_2^2}{w^2}\\ \\ w^2\times x_1^2+v_1^2=w^2\times x_2^2+v_2^2\\ \\ w^2\times x_1^2-w^2\times x_2^2=v_2^2-v_1^2\\ \\ w^2\times(x_1^2-x_2^2)=v_2^2-v_1^2\\ \\ w=\sqrt\frac{v_2^2-v_1^2}{x_1^2-x_2^2}\text{ Formula finala}\\ \\ \\ [/tex]


[tex]\displaystyle \text{Pentru A, lucarm cu formula (1):}\\ \\ \frac {x_1^2}{A^2}+\frac{v_1^2}{w^2\times A^2}=1\\ \\ A^2=x_1^2+\frac{v_1^2}{w^2}\\ \\ A=\sqrt{x_1^2+\frac{v_1^2}{w^2}}\\ \\ \\ \text{Calculam si scriem ecuatia:}\\ \\ w=\sqrt\frac{0,08^2-0,2^2}{0,1^2-0,2^2}=1,0583\approx 1,06\frac{rad}s\\ \\ A=\sqrt{0,1^2+\frac{0,2^2}{1,06^2}}=0,21354m\approx 0,214m\\ \\ \\ \boxed{x(t)=0,214\times\sin(1,06\times t)}[/tex]



[tex]\displaystyle b)\text{Era important sa facem a) si mai departe usor}\\ \\ w^2=\frac km\\ \\ k=m\times w^2\\ \\ \\[/tex]


[tex]\displaystyle c)\text{Ecuatia pentru acceleratie:}\\ \\ a=w^2\times A\times\sin(w\times t)\\ \\ x=A\times\sin(w\times t),\text{ inlocuim:}\\ \\ a=w^2\times x\\ \\ x=\frac{a}{w^2}\\ \\ \\[/tex]


\displaystyle d)E_c'+E_p'=E_c\\ \\2\times E_p'=E_c\\ \\2\times \frac{k\times x'^2}2=\frac{m\times v^2}2\\ \\2\times k\times x'^2=m\times v^2\\ \\x'=\sqrt\frac{m\times v^2}{2\times k}\\ \\\text{Unde: }\frac mk=\frac 1{w^2} \text{ ,iar } \frac {v^2}{w^2}=A^2\\ \\x'=\sqrt\frac{A^2}{2}\\ \\ \\


[tex]\displaystyle \text{Calcule:}\\ \\ b)k=0,2\times 1,06^2=0,22472\approx 0,224\frac Nm\\ \\ c)x=\frac{0,06}{1,06^2}=0,053399m\approx 0,054m\\ \\ d)x'=\sqrt\frac{0,214^2}{2}=0,15132\approx 0,151m[/tex]




[tex]\displaystyle \text{1.9 Se da:}\\ \\ x_1=2cm=0,02m\\ \\ v_1=5\frac ms\\ \\ x_2=3cm=0,03m\\ \\ v_2=4\frac ms\\ \\ a)A=?m\\ \\ b)T=?s\\ \\ c)V=?Hz\\ \\ \\ [/tex]


[tex]\displaystyle \text{Formule:}\\ \\ \text{Din problema precedenta:}\\ \\ w=\sqrt\frac{v_2^2-v_1^2}{x_1^2-x_2^2}\\ \\ \\ a)\text{Tot din problema precedenta:}\\ \\ A=\sqrt{x_1^2+\frac{v_1^2}{w^2}}\\ \\ \\ b)T=\frac{2\times \pi}{w}\\ \\ \\ c)V=\frac 1T[/tex]

stassahul: La 1.9 la frecventa nam prea inteles de ce se masoara in metri? sau la voi frecventa e alt ceva?!
Alte întrebări interesante