Đáp án:
\({x^2} + \dfrac{{{v^2}}}{{{\omega ^2}}} = {A^2}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
x = A\cos \left( {\omega t + \varphi } \right) \Rightarrow \cos \left( {\omega t + \varphi } \right) = \dfrac{x}{A}\\
v = - \omega A\sin \left( {\omega t + \varphi } \right) \Rightarrow \sin \left( {\omega t + \varphi } \right) = - \dfrac{v}{{\omega A}}
\end{array}\)
Mà:
\(\begin{array}{l}
{\cos ^2}\left( {\omega t + \varphi } \right) + {\sin ^2}\left( {\omega t + \varphi } \right) = 1\\
\Rightarrow {\left( {\dfrac{x}{A}} \right)^2} + {\left( { - \dfrac{v}{{\omega A}}} \right)^2} = 1\\
\Rightarrow {x^2} + \dfrac{{{v^2}}}{{{\omega ^2}}} = {A^2}
\end{array}\)
