Câu 25: B

Ta có:

\(\begin{array}{l}
\omega  = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{2} = \pi \left( {rad/s} \right)\\
 \Rightarrow \sqrt {\dfrac{g}{l}}  = \pi  \Rightarrow l = 1m
\end{array}\)

Câu 26: B

Ta có:

\(\begin{array}{l}
A = \dfrac{{{a_{\max }}}}{{{\omega ^2}}} = \dfrac{{100{\pi ^2}}}{{{{\left( {10\pi } \right)}^2}}} = 1\left( {cm} \right)\\
s = 4A = 4cm
\end{array}\)

Câu 27: B

Ta có: \(T = 1s\)

Câu 28: C

Biên độ tổng hợp là:

\(A = \sqrt {A_1^2 + A_2^2}  = \sqrt {{8^2} + {{15}^2}}  = 17cm\)

Câu 29: B

Ta có:

\(\begin{array}{l}
\omega  = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{2} = \pi \left( {rad/s} \right)\\
{x^2} + {\left( {\dfrac{v}{\omega }} \right)^2} = {A^2}\\
 \Rightarrow {6^2} + {\left( {\dfrac{v}{\pi }} \right)^2} = {10^2}\\
 \Rightarrow v = 8\pi \left( {cm/s} \right)
\end{array}\)

Câu 30: B

Li độ là:

\(x = 4\cos \left( {\pi t + \dfrac{\pi }{4}} \right) = 4\cos \left( {\pi .0,5 + \dfrac{\pi }{4}} \right) =  - 2\sqrt 2 \left( {cm} \right)\)

Vận tốc là:

\(v =  - 4\pi \sin \left( {\pi t + \dfrac{\pi }{4}} \right) =  - 4\pi \sin \left( {\pi .0,5 + \dfrac{\pi }{4}} \right) =  - 2\pi \sqrt 2 \left( {cm/s} \right)\)

Câu 31: B

Ta có:

\(\begin{array}{l}
T = \dfrac{{10}}{{10}} = 1\left( s \right)\\
\omega  = \dfrac{{2\pi }}{T} = 2\pi \left( {rad/s} \right)\\
 \Rightarrow \sqrt {\dfrac{g}{l}}  = 2\pi  \Rightarrow g = 9,87\left( {m/{s^2}} \right)
\end{array}\)

Câu 32: A

Ta có: 

\({W_d} = W - {W_t} = \dfrac{1}{2}k\left( {{A^2} - {x^2}} \right) = \dfrac{1}{2}.100.\left( {0,{1^2} - 0,{{06}^2}} \right) = 0,32J\)

Câu 33: C

Ta có:

\(\begin{array}{l}
\omega  = \sqrt {\dfrac{k}{m}}  = \sqrt {\dfrac{{80}}{{0,2}}}  = 20\left( {rad/s} \right)\\
A = \dfrac{{{v_{\max }}}}{\omega } = \dfrac{{60}}{{20}} = 3\left( {cm} \right)
\end{array}\)

Câu 34: B

Ta có:

\(\begin{array}{l}
{T_1} = 2\pi \sqrt {\dfrac{{{l_1}}}{{{g_1}}}} \\
{T_2} = 2\pi \sqrt {\dfrac{{{l_2}}}{{{g_2}}}} \\
 \Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{l_1}}}{{{l_2}}}} .\sqrt {\dfrac{{{g_2}}}{{{g_1}}}}  = \sqrt 2 .\sqrt {\dfrac{1}{{4,5}}}  = \dfrac{2}{3}\\
 \Rightarrow {T_2} = 1,5{T_1}
\end{array}\)