Cho em hỏi mấy câu này ạ

Giúp thích đáp án giúp em luôn ạ

Xin cảm ơn

Câu 61: B

Ta có:

\(\begin{array}{l}
W = {W_t} + {W_d} = {W_t} + \dfrac{{{W_t}}}{3} = \dfrac{4}{3}{W_t}\\
 \Rightarrow \dfrac{1}{2}k{A^2} = \dfrac{4}{3}.\dfrac{1}{2}k{x^2}\\
 \Rightarrow x =  \pm \dfrac{{A\sqrt 3 }}{2}\\
\cos \alpha  = \dfrac{x}{A} = \dfrac{{\dfrac{{A\sqrt 3 }}{2}}}{A} = \dfrac{{\sqrt 3 }}{2} \Rightarrow \alpha  = \dfrac{\pi }{6}
\end{array}\)

Thời gian ngắn nhất là:

\({t_1} = \dfrac{{2\alpha }}{\omega } = \dfrac{{2.\dfrac{\pi }{6}}}{{\dfrac{{2\pi }}{T}}} = \dfrac{T}{6}\)

Câu 62: A

Ta có:

\(\begin{array}{l}
{W_t} = \dfrac{W}{4} \Rightarrow \dfrac{1}{2}k{x^2} = \dfrac{1}{4}.\dfrac{1}{2}k{A^2} \Rightarrow x =  \pm \dfrac{A}{2}\\
\cos \alpha  = \dfrac{x}{A} = \dfrac{{\dfrac{A}{2}}}{A} = \dfrac{1}{2} \Rightarrow \alpha  = \dfrac{\pi }{3}
\end{array}\)

Thời gian lớn nhất là:

\({t_1} = \dfrac{{2\alpha }}{\omega } = \dfrac{{2.\dfrac{\pi }{3}}}{{\dfrac{{2\pi }}{T}}} = \dfrac{T}{3}\)

Câu 63: C

Ta có:

\(\begin{array}{l}
\dfrac{{{W_t}}}{{{W_d}}} = \dfrac{{W - {W_d}}}{{{W_d}}} = \dfrac{{\dfrac{1}{2}mv_{\max }^2}}{{\dfrac{1}{2}m{v^2}}} - 1 = {\left( {\dfrac{{{v_{\max }}}}{v}} \right)^2} = 1\\
 \Rightarrow \dfrac{{{W_t}}}{{{W_d}}} = {\left( {\dfrac{{0,6}}{{0,2}}} \right)^2} - 1 = 8
\end{array}\)

Câu 64: B

Ta có: 

\(\dfrac{{{W_d}}}{W} = \dfrac{{\dfrac{1}{2}m{v^2}}}{{\dfrac{1}{2}mv_{\max }^2}} = {\left( {\dfrac{v}{{{v_{\max }}}}} \right)^2} = 0,{5^2} = 0,25\)

Câu 65: C

Ta có:

\(\begin{array}{l}
W = {W_d} + {W_t} = 8{W_t} + {W_t} = 9{W_t}\\
 \Rightarrow \dfrac{1}{2}k{A^2} = 9.\dfrac{1}{2}k{x^2}\\
 \Rightarrow x = \dfrac{A}{3}
\end{array}\)

Câu 66: D

Ta có:

\(\dfrac{{{W_d}}}{W} = \dfrac{{W - {W_t}}}{W} = 1 - \dfrac{{\dfrac{1}{2}k{x^2}}}{{\dfrac{1}{2}k{A^2}}} = 1 - {\left( {\dfrac{x}{A}} \right)^2} = 1 - {\left( {\dfrac{1}{4}} \right)^2} = \dfrac{{15}}{{16}}\)

Đáp án + Giải thích các bước giải: