Hai con lắc đơn có cùng chiều dài\(\ell \), cùng khối lượng m

* Hướng dẫn giải

Có:\(T=2\pi \sqrt{\frac{\ell }{g}}\Rightarrow T\sim \frac{1}{\sqrt{g}}\)

\(\frac{{{T}_{1}}}{{{T}_{0}}}=5\Rightarrow \sqrt{\frac{g}{{{g}_{1}}}}=5\Rightarrow {{g}_{1}}=\frac{g}{25}\)mà \(\overrightarrow{E}\)hướng thẳng đứng xuống dưới\(\Rightarrow {{g}_{1}}=g-{{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 1}}}\left( \overrightarrow{{{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 1}}}}\uparrow \downarrow \overrightarrow{g} \right)\)

\(\Rightarrow {{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 1}}}=\frac{24g}{25}=\frac{\left| {{q}_{1}}E \right|}{m};{{q}_{1}}<0\) do \(\overrightarrow{E}\uparrow \downarrow \overrightarrow{{{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 1}}}}\)                (1)

\(\frac{{{T}_{2}}}{{{T}_{0}}}=\frac{5}{7}\Rightarrow \sqrt{\frac{g}{{{g}_{2}}}}=\frac{5}{7}\Rightarrow {{g}_{2}}=\frac{49}{25}g\)

mà \(\overrightarrow{E}\)hướng thẳng đứng xuống dưới\(\Rightarrow {{g}_{2}}=g+{{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 2}}}\left( \overrightarrow{{{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 2}}}}\uparrow \uparrow \overrightarrow{g} \right)\)

\(\Rightarrow {{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 2}}}=\frac{24g}{25}=\frac{\left| {{q}_{2}}E \right|}{m};{{q}_{2}}>0\) do \(\overrightarrow{E}\uparrow \uparrow \overrightarrow{{{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 2}}}}\)                (2)

\(\left( 1 \right),\left( 2 \right)\Rightarrow {{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 1}}}={{g}_{\text{l }\!\!{}^\text{1}\!\!\text{ 2}}}\Rightarrow \left| {{q}_{1}} \right|=\left| {{q}_{2}} \right|\) mà \({{q}_{1}},{{q}_{2}}\)trái dấu \(\Rightarrow \frac{{{q}_{1}}}{{{q}_{2}}}=-1\)