Answer:0.4 m
Explanation:
Given
Maximum displacement A=0.49
The sum of kinetic and elastic potential energy is [tex]\frac{1}{2}kA^2[/tex]
where k=spring constant
U+K.E.=[tex]\frac{1}{2}kA^2[/tex]
when K.E.=U/2
K.E.=kinetic energy
U=Elastic potential Energy
[tex]\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m[/tex]
