Lm giúp mk câu này mk đag cần gấp

Đáp án:

a) \(0,128J\)

b) \(\dfrac{\pi }{5}\left( s \right)\)

c) \(\left\{ \begin{array}{l}
{W_t} = \dfrac{8}{{75}}J\\
{W_d} = \dfrac{8}{{375}}J
\end{array} \right.\)

Giải thích các bước giải:

a) Cơ năng là:

\(W = \dfrac{1}{2}k{A^2} = \dfrac{1}{2}.40.0,{08^2} = 0,128J\)

b) Tần số góc là:

\(\omega  = \sqrt {\dfrac{k}{m}}  = \sqrt {\dfrac{{40}}{{0,4}}}  = 10\left( {rad/s} \right)\)

Tần số là:

\(f = \dfrac{\omega }{{2\pi }} = \dfrac{{10}}{{2\pi }} = \dfrac{5}{\pi }\left( {Hz} \right)\)

Chu kì là:

\(T = \dfrac{1}{f} = \dfrac{\pi }{5}\left( s \right)\)

c) Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
W = {W_d} + {W_t} = 0,128J\\
{W_t} = 5{W_d}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{W_t} = \dfrac{8}{{75}}J\\
{W_d} = \dfrac{8}{{375}}J
\end{array} \right.
\end{array}\)

\(a)\\\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{40}{0,4}}=10(rad/s)\\W=\dfrac{1}{2}m\omega^2A^2=\dfrac{1}{2}.0,4.10^2.0,08^2=0,128(J)\\b)\\\omega=10(rad/s)\\T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{10}=\dfrac{\pi}{5}(s)\\f=\dfrac{1}{T}=\dfrac{5}{\pi}(Hz)\\c)\\\begin{cases}W_t=5W_d\\W_d+W_t=W\end{cases}\Rightarrow \dfrac{1}{5}W_t+W_t=W\\\Rightarrow \dfrac{6}{5}.\dfrac{1}{2}m\omega^2 x^2=\dfrac{1}{2}m\omega^2 A^2\\\Rightarrow x=\pm\dfrac{A\sqrt{30}}{6}=\pm\dfrac{4\sqrt{30}}{3}\\\Rightarrow W_t=\dfrac{1}{2}m\omega^2x^2=\dfrac{1}{2}.0,4.10^2.\left(\dfrac{4\sqrt{30}}{3}.10^{-2}\right)^2\approx 0,1(J)\\\Rightarrow W_d=W-W_t=0,128-0,1=0,028(J)\)