Đáp án:

 \(A = 0,015\left( m \right)\)

Giải thích các bước giải:

Tại \(t = 8ms\):

Ta có: 

\(\begin{array}{l}
{E_{{d_1}}} = \dfrac{6}{8}E = \dfrac{3}{4}E \Rightarrow {E_{{t_1}}} = \dfrac{E}{4} \Rightarrow x_1^2 = \dfrac{{{A^2}}}{4}\\
 \Rightarrow {x_1} = \dfrac{A}{2}
\end{array}\)

\(\sin \alpha  = \dfrac{{{x_1}}}{A} = \dfrac{1}{2} \Rightarrow \alpha  = \dfrac{\pi }{6}\left( {rad} \right)\)

Tại \(t = 26ms\):

Ta có: 

\(\begin{array}{l}
{E_{{d_2}}} = \dfrac{4}{8}E = \dfrac{1}{2}E \Rightarrow {E_{{t_2}}} = \dfrac{E}{2} \Rightarrow x_2^2 = \dfrac{{{A^2}}}{2}\\
 \Rightarrow {x_2} =  - \dfrac{A}{{\sqrt 2 }}\\
\sin \beta  = \dfrac{{\left| {{x_2}} \right|}}{A} = \dfrac{1}{{\sqrt 2 }} \Rightarrow \beta  = \dfrac{\pi }{4}\left( {rad} \right)
\end{array}\)

Góc quét là:

\(\begin{array}{l}
\gamma  = \alpha  + \beta  = \dfrac{\pi }{6} + \dfrac{\pi }{4} = \dfrac{{5\pi }}{{12}}\left( {rad} \right)\\
 \Rightarrow t = \dfrac{\gamma }{\omega } = \dfrac{{\dfrac{{5\pi }}{{12}}}}{{\dfrac{{2\pi }}{T}}} = \dfrac{5}{{24}}T\\
 \Rightarrow \left( {26 - 8} \right){.10^{ - 3}} = \dfrac{5}{{24}}T\\
 \Rightarrow T = 0,0864s\\
 \Rightarrow \omega  = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{{0,0864}} = 72,72\left( {rad/s} \right)
\end{array}\)

Ta có:

\(\begin{array}{l}
E = \dfrac{1}{2}m{\omega ^2}{A^2} \Rightarrow {30.10^{ - 3}} = \dfrac{1}{2}.\dfrac{{50}}{{1000}}.72,{72^2}.{A^2}\\
 \Rightarrow A = 0,015\left( m \right)
\end{array}\)