+ Tại thời điểm t:

$x_1=A\cos \phi \Rightarrow {\text{W}}_{t1}=\frac{k{x}_1^2}{2}=\frac{k{A}^2}{2}c{\text{os}}^2\phi =\mathrm{0,256}J\iff \frac{k{A}^2}{2}.\frac{1+c\text{os}2\phi }{2}=\mathrm{0,256}J(\ast ).$

+ Tại thời điểm t + 0,05:

$x_2=A\cos \left(\phi +\frac{\pi }{4}\right)\Rightarrow {\text{W}}_{t2}=\frac{k{A}^2}{2}-\frac{m{v}_2^2}{2}=\frac{k{A}^2}{2}c{\text{os}}^2\left(\phi +\frac{\pi }{4}\right).$

$\iff \frac{k{A}^2}{2}-\mathrm{0,288}=\frac{k{A}^2}{2}{\left(c\text{os}\phi .\text{cos}\frac{\pi }{4}-\sin \phi .\sin \frac{\pi }{4}\right)}^2\iff \frac{k{A}^2}{2}-\mathrm{0,288}=\frac{k{A}^2}{2}.\frac{1}{2}{\left(\cos \phi -\sin \phi \right)}^2.$

$\iff \frac{k{A}^2}{2}-\mathrm{0,288}=\frac{k{A}^2}{2}.\frac{1}{4}{\left(\cos \phi -\sin \phi \right)}^2\iff \frac{k{A}^2}{2}-\mathrm{0,288}=\frac{k{A}^2}{4}\left(1-\sin 2\phi \right)(**).$

Từ (*) và (**) ta có hệ phương trình:

 $\left\{\begin{array}{l}\frac{1}{4}k{A}^2\left(1+\sin 2\phi \right)=\mathrm{0,288}\\ \frac{1}{4}k{A}^2\left(1+\cos 2\phi \right)=\mathrm{0,256}\end{array}\right.\Rightarrow \frac{1+\sin 2\phi }{1+\cos 2\phi }=\frac{9}{8}\Rightarrow 8+8\sin 2\phi =9+9c\text{os}2\phi$

$\Rightarrow 1+9c\text{os}2\phi =8\sin 2\phi$

$\iff {\left(1+9c\text{os}2\phi \right)}^2=8^2\left(1-c{\text{os}}^{2}2\phi \right)\iff 145c{\text{os}}^{2}2\phi +18c\text{os}2\phi -63=0.$

$\iff \left[\begin{array}{l}\text{os}2\phi =\frac{3}{5}\Rightarrow \text{W}=\mathrm{0,32}J\\ c\text{os}2\phi =-\frac{21}{29}\Rightarrow \text{W}=\mathrm{1,856}J\left(loai\right)\end{array}\right..$

+ Với $\text{W}=\mathrm{0,32}J=\frac{k{A}^2}{2}\Rightarrow A=\mathrm{0,08}m.$

+ Độ biến dạng của lò xo ở VTCB: $\Delta l_0=\frac{mg}{k}=\frac{\mathrm{0,4.10}}{100}=\mathrm{0,04}m.$