Câu hỏi 5

Ta có: \(v = \sqrt {2gl\left( {\cos \alpha  - \cos {\alpha _0}} \right)}  = \sqrt {{{{T^2}g} \over {4{\pi ^2}}}.2g\left( {\cos \alpha  - \cos {\alpha _0}} \right)}  = {{Tg} \over {2\pi }}\sqrt {2\left( {\cos \alpha  - \cos {\alpha _0}} \right)}  = 0,287m/s\)