Ai cứu iem đi ạk

nhma mấy đứa OCD nặng như e thì nhìn hơi nhức mắt =))

Cái ảnh mốt để chữ theo 1 kiểu, theo từng hàng thì dễ đọc, thuận mắt với đẹp hơn ấy a :vv

Ta có: \(\omega=\sqrt{\dfrac{g}{l}}=\dfrac{5\pi}{4}\left(\dfrac{rad}{s}\right)\)

\(\rightarrow A=\dfrac{v_{max}}{\omega}=\dfrac{0,4}{\dfrac{5\pi}{4}}=\dfrac{8\pi}{25}\left(m\right)\)

\(t:0\left\{{}\begin{matrix}x=0\\v>0\end{matrix}\right.\Rightarrow\varphi=-\dfrac{\pi}{2}\left(rad\right)\)

\(\rightarrow v=-4sin\left(\dfrac{5\pi}{4}t-\dfrac{\pi}{2}\right)\left(\dfrac{m}{s}\right)\)

Thay t = 2 vào \(\Rightarrow v=0\left(\dfrac{m}{s}\right)\)

Ở VTCB: \(\left|v\right|=\sqrt{2gl\left(cos\alpha-cos\alpha_0\right)}=\sqrt{2gl\left(1-cos\alpha_0\right)}\)

\(\Rightarrow l=2,5m\)

Ta có: \(\left\{{}\begin{matrix}T=2\pi\sqrt{\dfrac{l}{g}}=\dfrac{t}{12}\\T'=2\pi\sqrt{\dfrac{l-16}{g}}=\dfrac{t}{20}\end{matrix}\right.\) 

\(\Rightarrow\sqrt{\dfrac{l-16}{l}}=\dfrac{12}{20}\)

\(\Rightarrow l=25cm\)

Chọn B

Ta có: \(T=2\pi\sqrt{\dfrac{l}{g}}\Rightarrow l=\dfrac{T\sqrt{g}}{2\pi}\)

Theo đề: \(\left\{{}\begin{matrix}l_3=l_1+l_2\\l_4=l_1-l_2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}l_1=\dfrac{l_3+l_4}{2}\\l_2=\dfrac{l_3-l_4}{2}\end{matrix}\right.\)

Ta có: \(l_1=\dfrac{\sqrt{g}\left(T_3+T_4\right)}{4\pi}=0,8\)

\(l_2=\dfrac{g\left(T_3^2-T_4^2\right)}{8\pi^2}=0,64\)

Ta có: \(\dfrac{T'}{T}=\dfrac{2\pi\sqrt{\dfrac{l'}{g}}}{2\pi\sqrt{\dfrac{l}{g}}}=\sqrt{\dfrac{l'}{l}}=\dfrac{1,2T}{T}=1,2\Rightarrow l'=1,44l\)

\(\Rightarrow\dfrac{l'-l}{l}100\%=\dfrac{1,44l-l}{l}100\%=44\%\)

Chọn D

geee