Câu 56:
Ta có:
\(\begin{array}{l}
{T_1} = 2\pi \sqrt {\dfrac{{{m_1}}}{k}} \Rightarrow T_1^2 = 4{\pi ^2}.\dfrac{{{m_1}}}{k}\\
{T_2} = 2\pi \sqrt {\dfrac{{{m_2}}}{k}} \Rightarrow T_2^2 = 4{\pi ^2}.\dfrac{{{m_2}}}{k}\\
T = 2\pi \sqrt {\dfrac{{{m_1} + {m_2}}}{k}} \Rightarrow {T^2} = 4{\pi ^2}.\dfrac{{{m_1} + {m_2}}}{k}\\
\Rightarrow {T^2} = T_1^2 + T_2^2 = 1,{2^2} + 1,{6^2} = 4\\
\Rightarrow T = 2s
\end{array}\)
Câu 72:
Ta có:
\(\omega = \sqrt {\dfrac{k}{m}} = \sqrt {\dfrac{{80}}{{0,2}}} = 20rad/s\)
\(\begin{array}{l}
{x^2} + \dfrac{{{v^2}}}{{{\omega ^2}}} = {A^2} \Rightarrow A = 5cm\\
\cos \varphi = \dfrac{{2,5}}{5} = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi }{3}\\
\Rightarrow x = 5\cos \left( {20t + \dfrac{\pi }{3}} \right)
\end{array}\)