Answer:[tex]x=\frac{x_m}{\sqrt{2}}[/tex]
Explanation:
Given
initially mass is stretched to [tex]x_m[/tex]
Let k be the spring Constant of spring
Therefore Total Mechanical Energy is [tex]\frac{kx_m^2}{2}[/tex]
Position at which kinetic Energy is equal to Elastic Potential Energy
[tex]K=\frac{mv^2}{2}[/tex]
[tex]U=\frac{kx^2}{2}[/tex]
it is given
[tex]k=U[/tex]
thus [tex]2U=\frac{kx_m^2}{2}[/tex]
[tex]2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}[/tex]
[tex]2x^2=x_m^2[/tex]
[tex]x=\frac{x_m}{\sqrt{2}}[/tex]
