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

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

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

\(\Rightarrow l=25cm\)

Chọn B

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

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

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

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

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

\(T=2\pi\sqrt{\dfrac{l}{g}}\Rightarrow T^2\) tỉ lệ thuận với l

Theo bài ra 

\(\left\{{}\begin{matrix}T_1^2+T_2^2=\left(2,7\right)^2=7,29\\T_1^2-T_2^2=\left(0,9\right)^2=0,81\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}T_1^2=4,05\\T_2^2=3,24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}T_1=2,01s\\T_2=1,8s\end{matrix}\right.\)

Vẽ đường tròn lượng giác:

1. \(7=10cos\left(\dfrac{4\pi}{3}t-\dfrac{2\pi}{3}\right)\)

Thời gian min từ VTCB đến \(x=7:\)

\(t=\dfrac{\alpha}{\omega}=\dfrac{\dfrac{\pi}{2}-acrcos\dfrac{1}{10}}{\dfrac{4\pi}{3}}=0,024s\)

2. \(3=10cos\left(\dfrac{4\pi}{3}t-\dfrac{2\pi}{3}\right)\)

Thời gian min từ biên âm đến \(x=3:\)

\(t=\dfrac{\beta}{\omega}=\dfrac{\pi-acrcos\dfrac{3}{10}}{\dfrac{4\pi}{3}}=0,45s\)

\(T=2\pi\sqrt{\dfrac{l}{g}}\Rightarrow T\) tỉ lệ thuận với \(\sqrt{l}\) => \(T^2\) tỉ lệ thuận với l

=> Con lắc đơn có chiều dài l=l1+l2 thì có chu kì \(T=\sqrt{T_1^2+T_2^2}=\sqrt{2^2+\left(1,5\right)^{^2}}=2,5s\)

=> Chọn B