Answer:
[tex]d=0.165m[/tex]
Explanation:
Given
[tex]m=0.25kg[/tex],[tex]x_{1}=5cm*\frac{1m}{100cm}=0.05m[/tex],[tex]x_{2}=4cm*\frac{1m}{100cm}=0.04m[/tex],[tex]v=2\frac{rev}{s}[/tex]
The tension of the spring is
[tex]F_{k}=K*x_{1}=m*g[/tex]
[tex]K=\frac{m*g}{x_{1}}[/tex]
[tex]K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m[/tex]
The force in the spring is equal to centripetal force so
[tex]F_{c}=\frac{m*v^2}{r}[/tex]
[tex]v=w*r=2\pi*r[/tex]
But Fc is also
Fc=KxΔr
[tex]F_{c}=K*(r-x_{2})[/tex]
Replacing
[tex]m*4\pi^2*r=K*(r-x_{2})[/tex]
[tex]0.25kg*4\pi^2*r=49*(r-0.04m)[/tex]
[tex]r=0.205m[/tex]
total distance is
[tex]d=0.205-0.04=0.165m[/tex]
