Ответ:

В контуре возникнут колебания частотой приблизительно 111,8 Гц

Объяснение:

Дано:

[tex]L[/tex]

[tex]C_{1}[/tex]

[tex]\nu_{1} =[/tex] 50 Гц

[tex]C_{2}[/tex]

[tex]\nu_{2} =[/tex] 100 Гц

Найти:

[tex]\nu_{12} \ - \ ?[/tex]

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Решение:

По формуле Томсона:

[tex]\boxed{T = 2\pi \sqrt{LC} }[/tex] - период колебаний

[tex]\nu = \dfrac{1}{T} = \dfrac{1}{2\pi \sqrt{LC} }[/tex]

[tex]\bigg(\nu \bigg)^{2} = \bigg( \dfrac{1}{2\pi \sqrt{LC} } \bigg)^{2}[/tex]

[tex]\nu^{2} = \dfrac{1}{4\pi ^{2}LC} \Longrightarrow \boxed{ C = \dfrac{1}{4\pi ^{2}\nu^{2}L} }[/tex]

При последовательном соединении конденсаторов:

[tex]\dfrac{1}{C_{12}} = \dfrac{1}{C_{1}} +\dfrac{1}{C_{2}}[/tex]

[tex]\dfrac{1}{C_{12}} = \dfrac{C_{1} +C_{2}}{C_{1}C_{2}} \Longrightarrow C_{12} = \dfrac{C_{1}C_{2}}{C_{1} +C_{2}}[/tex]

[tex]C_{12} = \dfrac{C_{1}C_{2}}{C_{1} +C_{2}} = \dfrac{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} \cdot \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}}{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} + \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}} = \dfrac{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} \cdot \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}}{\dfrac{1}{4\pi ^{2}L} \bigg(\dfrac{1}{\nu_{1}^{2}} + \dfrac{1}{\nu_{2}^{2}} \bigg)} =[/tex][tex]= \dfrac{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} \cdot \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}}{\dfrac{1}{4\pi ^{2}L} \bigg(\dfrac{\nu_{2}^{2} + \nu_{1}^{2}}{\nu_{1}^{2}\nu_{2}^{2}} \bigg)} = \dfrac{4\pi ^{2}L\nu_{1}^{2}\nu_{2}^{2}}{4\pi ^{2}\nu_{1}^{2}L \cdot 4\pi ^{2}\nu_{2}^{2}L(\nu_{1}^{2} + \nu_{2}^{2})} =\dfrac{1}{4\pi ^{2}L(\nu_{1}^{2} + \nu_{2}^{2})}[/tex]

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Частота контура состоящего из конденсаторов ёмкостью [tex]C_{1}[/tex] и [tex]C_{2}[/tex].

[tex]\nu_{12} = \dfrac{1}{2\pi \sqrt{LC_{12}} } = \dfrac{1}{2\pi \sqrt{L \cdot\dfrac{1}{4\pi ^{2}L(\nu_{1}^{2} + \nu_{2}^{2})}} } = \dfrac{1}{\dfrac{2 \pi}{2 \pi}\sqrt{\dfrac{1}{\nu_{1}^{2} + \nu_{2}^{2}} } } =[/tex]

[tex]= \dfrac{\dfrac{1}{1} }{\dfrac{1}{\sqrt{\nu_{1}^{2} + \nu_{2}^{2}} } } =\sqrt{\nu_{1}^{2} + \nu_{2}^{2}}[/tex].

[tex]\boxed{ \boldsymbol{ \nu_{12}= \sqrt{\nu_{1}^{2} + \nu_{2}^{2}} } }[/tex]

[tex]\nu_{12} =[/tex] √(2500 Гц² + 10 000 Гц²) [tex]\approx[/tex] 111,8 Гц

Ответ: [tex]\nu_{12} \approx[/tex] 111,8 Гц.