Dạ mọi người giúp em 3 câu này với ạ

\(\textbf{Bai 4:}\\A=40(cm)=0,4(m)\\\dfrac{T}{4}=1\Rightarrow T=4\Rightarrow \omega=\dfrac{\pi}{2}\\\varphi=0\\\Rightarrow x=0,4\cos\dfrac{\pi}{2}t\\a)\\t=0\Rightarrow x=A\Rightarrow v-{min}=0\\b)\\v_{max}=A\omega=\dfrac{\pi}{5}\\c)\\\alpha=90+\dfrac{0,5}{4}.360=90^o+45^o\\\Rightarrow x=-0,4.\dfrac{\sqrt{2}}{2}=-\dfrac{\sqrt{2}}{5}\\v=-\omega\sqrt{A^2-x^2}=-\dfrac{\pi\sqrt{2}}{10}\\\textbf{Bai 5:}T=\dfrac{2\pi}{\omega}=\dfrac{2}{5}\\a)\\\alpha =\cos\dfrac{-3\sqrt{3}}{-6}-\varphi=150^o-60^o=90^o\\t=\dfrac{90}{360}.T=\dfrac{1}{10}(s)\\b)\\\alpha =\cos\dfrac{-3}{-6}-\varphi=120-60^o=60^o\\t=\dfrac{60}{360}.T=\dfrac{1}{15}(s)\\c)\\\alpha =270^o-\varphi=210^o\\t=\dfrac{210}{360}.T=\dfrac{7}{30}(s)\\d)\\\alpha =360^o-\varphi=300^o\\t=\dfrac{300}{360}.T=\dfrac{1}{3}(s)\\\textbf{Bai 6:}T=\dfrac{2\pi}{\omega}=\dfrac{1}{2}\\t_1=\dfrac{13}{6}=2+\dfrac{1}{6}\\\Rightarrow \alpha_1=4.360^o+\alpha'=4.360^o+\dfrac{\dfrac{1}{6}}{\dfrac{1}{2}}.360^o=4.360^o+120^o\\S_1=4.4A+S_1'=4A+\dfrac{3}{2}+\dfrac{3}{2}=51(cm)\\t_2=\dfrac{23}{6}=3+\dfrac{5}{6}\\\alpha_2=6.360^o+\alpha'=6.360^o+\dfrac{\dfrac{5}{6}}{\dfrac{1}{2}}.360^o=6.360^o+1.360^o+\dfrac{2}{3}.360\\S_2=6.4A+4A+3+2A=93(cm)\\\Rightarrow \Delta S=S_1-S_2=93-51=42(cm)\)