- 53

- 0

We now need to find an equation that satisfies: [tex]\frac{d^2x}{d^2t}=\frac{kx}{m}[/tex]

We get (letting w^2=k/m): [tex] x(t)=A*Cos(wt+\phi)[/tex]

The time it takes for one cycle is:[tex]T_1=2 \pi\sqrt{\frac{m}{k}[/tex]

Now, lets solve for T again using a different method.

[tex] F_{ave}=\frac{1}{x}\int{(kx)}dx=\frac{kx}{2}[/tex]

[tex]a_{ave}=\frac{kx}{2m}[/tex]

So, we get: [tex]x_f=x_i+\frac{at^2}{2}--->T^2=\frac{4m}{k}--->T=2\sqrt{\frac{2m}{k}}[/tex]

We multiply our new T by 4 to get the time over a complete cycle: [tex]T_2=4*2\sqrt{\frac{2m}{k}}[/tex]

[tex]T_1/=T_2[/tex]

? ? ? ? ? ? ? ? ?