Vị trí, vận tốc khi biết quan hệ năng lượng

Biểu thức cơ năng với vật 1: ${{\text{W}}_{1}}=\dfrac{1}{2}k{{A}_{1}}^{2}$

Biểu thức cơ năng với vật 1: ${{\text{W}}_{2}}=\dfrac{1}{2}k{{A}_{2}}^{2}$

Mà ${{\text{W}}_{1}}={{\text{W}}_{2}}$ $\Rightarrow \dfrac{1}{2}k{{A}_{1}}^{2}=\dfrac{1}{2}k{{A}_{2}}^{2}$ $\Rightarrow {{A}_{1}}={{A}_{2}}$ $\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=1$

Cơ năng của con lắc 1: ${{\text{W}}_{1}}=\dfrac{mg{{\ell }_{1}}{{\alpha }_{1}}^{2}}{2}$

Cơ năng của con lắc 1: ${{\text{W}}_{2}}=\dfrac{mg{{\ell }_{2}}{{\alpha }_{2}}^{2}}{2}$

Mà ${{\text{W}}_{1}}={{\text{W}}_{2}}$ $\dfrac{mg{{\ell }_{1}}{{\alpha }_{1}}^{2}}{2}=\dfrac{mg{{\ell }_{2}}{{\alpha }_{2}}^{2}}{2}$ $\Rightarrow 2{{\ell }_{2}}{{\alpha }_{1}}^{2}={{\ell }_{2}}{{\alpha }_{2}}^{2}$ $\Rightarrow {{\alpha }_{1}}=\dfrac{{{\alpha }_{2}}}{\sqrt{2}}$

Ta có: ${{\text{W}}_{t}}\text{=3}{{\text{W}}_{d}}$ $\Rightarrow 3.\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{\text{x}}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}.{{x}^{2}}$ $\Rightarrow \omega .x=\sqrt{3}v$

Ta có: ${{\text{W}}_{t}}\text{=}{{\text{W}}_{d}}$ $\Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{\text{x}}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}.{{x}^{2}}$ $\Rightarrow \omega .x=v$

Li độ x: $x=\dfrac{40}{100}A=\dfrac{2}{5}A$

$\dfrac{{{\text{W}}_{d}}}{{{\text{W}}_{t}}}=\dfrac{\dfrac{1}{2}k{{\text{A}}^{2}}-\dfrac{1}{2}k{{x}^{2}}}{\dfrac{1}{2}k{{x}^{2}}}=\dfrac{{{A}^{2}}-{{x}^{2}}}{{{x}^{2}}}=\dfrac{{{A}^{2}}}{{{x}^{2}}}-1=\dfrac{{{A}^{2}}}{\dfrac{4}{25}{{A}^{2}}}-1=\dfrac{25}{4}-1=\dfrac{21}{4}$

Ta có: ${{\text{W}}_{d}}\text{=2}{{\text{W}}_{t}}$

$\text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=3{{\text{W}}_{t}}$

$\Rightarrow \dfrac{1}{2}k{{\text{A}}^{2}}=3.\dfrac{1}{2}k{{\text{x}}^{2}}$ $\Rightarrow x=\pm \dfrac{A}{\sqrt{3}}$

Ta có: ${{\text{W}}_{t}}\text{=3}{{\text{W}}_{d}}$ $\Rightarrow {{W}_{d}}+3{{W}_{d}}=\text{W}$ $\Rightarrow 4{{W}_{d}}=\text{W}$ $\Rightarrow {{W}_{d}}=\dfrac{\text{W}}{4}$ $\Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{8}m{{\omega }^{2}}.{{A}^{2}}$ $\Rightarrow v=\dfrac{A\omega }{2}$

Ta có: $\dfrac{{{\text{W}}_{t}}}{\text{W}}=\dfrac{\dfrac{1}{2}k{{x}^{2}}}{\dfrac{1}{2}k{{A}^{2}}}=\dfrac{{{x}^{2}}}{{{A}^{2}}}=\dfrac{\dfrac{1}{4}{{A}^{2}}}{{{A}^{2}}}=\dfrac{1}{4}$

Cơ năng của hệ: $\text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}$ mà ${{\text{W}}_{d}}\text{=3}{{\text{W}}_{t}}$ $\Rightarrow 4{{\text{W}}_{t}}=\text{W}$ $\Rightarrow 4.\dfrac{1}{2}m{{\omega }^{2}}{{x}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}$

$\Rightarrow {{x}^{2}}=\dfrac{{{A}^{2}}}{4}$ $\Rightarrow x=\pm \dfrac{A}{2}$

Vì con lắc lò xo nằm ngang nên $\Rightarrow x=\Delta \ell =\pm \dfrac{A}{2}$

Độ lớn lực đàn hồi: ${{F}_{dh}}=k.\Delta \ell =\dfrac{kA}{2}$

Mà: ${{F}_{dhm\text{ax}}}=kA$ $\Rightarrow {{F}_{dh}}=\dfrac{kA}{2}=\dfrac{{{F}_{\text{max}}}}{2}$

Ta có: ${{\text{W}}_{t}}\text{=n}{{\text{W}}_{d}}$

Mà $\text{W}=\dfrac{1}{n}{{\text{W}}_{t}}+{{\text{W}}_{t}}=\dfrac{n+1}{n}{{\text{W}}_{t}}$

$\Rightarrow \dfrac{1}{2}k{{\text{A}}^{2}}=\dfrac{n+1}{n}.\dfrac{1}{2}k{{\text{x}}^{2}}$ $\Rightarrow x=\pm \dfrac{A\sqrt{n}}{\sqrt{n+1}}$

Vận tốc v : $v=\dfrac{20}{100}A\omega =\dfrac{1}{5}A\omega$

$\dfrac{\text{W}}{{{\text{W}}_{d}}}=\dfrac{\dfrac{1}{2}\text{m}{{\omega }^{2}}{{\text{A}}^{2}}}{\dfrac{1}{2}m{{v}^{2}}}=\dfrac{{{A}^{2}}{{\omega }^{2}}}{\dfrac{1}{25}{{A}^{2}}{{\omega }^{2}}}=25$

Theo bài ra: $v=\dfrac{1}{2}{{v}_{\text{max}}}$ $\Rightarrow A\omega =2v$

Ta có: $\dfrac{{{\text{W}}_{t}}}{{{\text{W}}_{d}}}=\dfrac{\dfrac{1}{2}\text{m}{{\omega }^{2}}{{\text{A}}^{2}}-\dfrac{1}{2}m{{v}^{2}}}{\dfrac{1}{2}m{{v}^{2}}}=\dfrac{\omega {{A}^{2}}-{{v}^{2}}}{{{v}^{2}}}=\dfrac{{{\omega }^{2}}{{A}^{2}}}{{{v}^{2}}}-1=4-1=3$

Ta có: ${{\text{W}}_{d}}\text{=n}{{\text{W}}_{t}}$

Mà $\text{W}={{\text{W}}_{d}}+\dfrac{1}{n}{{\text{W}}_{d}}=\dfrac{n+1}{n}{{\text{W}}_{d}}$

$\Rightarrow \dfrac{1}{2}\text{m}{{\omega }^{2}}{{\text{A}}^{2}}=\dfrac{n+1}{n}.\dfrac{1}{2}\text{m}{{\text{v}}^{2}}$ $\Rightarrow v=\pm \dfrac{A\omega \sqrt{n}}{\sqrt{n+1}}$

Ta có: ${{\text{W}}_{d}}\text{=}\dfrac{1}{2}m{{v}^{2}},{{\text{W}}_{t}}\text{=}\dfrac{1}{2}k{{x}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}{{x}^{2}}$

Mà $\dfrac{{{\text{W}}_{d}}}{{{\text{W}}_{t}}}=\dfrac{{{v}^{2}}}{{{x}^{2}}{{\omega }^{2}}}$

Tại li độ x: ${{x}^{2}}+\dfrac{{{v}^{2}}}{{{\omega }^{2}}}={{A}^{2}}\Rightarrow {{v}^{2}}=({{A}^{2}}-{{x}^{2}}){{\omega }^{2}}$ $\Rightarrow \dfrac{{{v}^{2}}}{{{x}^{2}}{{\omega }^{2}}}=\dfrac{({{A}^{2}}-{{x}^{2}}){{\omega }^{2}}}{{{x}^{2}}{{\omega }^{2}}}=\dfrac{{{A}^{2}}}{{{x}^{2}}}-1$

Li độ x: $x=\dfrac{60}{100}A=\dfrac{3}{5}A$

$\dfrac{\text{W}}{{{\text{W}}_{t}}}=\dfrac{\dfrac{1}{2}k{{\text{A}}^{2}}}{\dfrac{1}{2}k{{x}^{2}}}=\dfrac{{{A}^{2}}}{{{x}^{2}}}=\dfrac{{{A}^{2}}}{\dfrac{9}{25}{{A}^{2}}}=\dfrac{25}{9}$

Cơ năng của con lắc 1: ${{\text{W}}_{1}}=\dfrac{mg{{\ell }_{1}}{{\alpha }_{1}}^{2}}{2}$

Cơ năng của con lắc 1: ${{\text{W}}_{2}}=\dfrac{mg{{\ell }_{2}}{{\alpha }_{2}}^{2}}{2}$

Mà ${{\text{W}}_{1}}={{\text{W}}_{2}}$ $\Rightarrow \dfrac{mg{{\ell }_{1}}{{\alpha }_{1}}^{2}}{2}=\dfrac{mg{{\ell }_{2}}{{\alpha }_{2}}^{2}}{2}$ $\Rightarrow 1.{{\alpha }_{1}}^{2}=1,44{{\alpha }_{2}}^{2}$ $\Rightarrow {{\alpha }_{1}}=1,2{{\alpha }_{2}}$