The elastic potential energy of a spring is given by:
[tex]PE=\frac{1}{2}kx^2[/tex]
Where:
[tex]\begin{gathered} PE=\text{ potential energy} \\ k=\text{ spring constant} \\ x=\text{ distance of the spring is compressed} \end{gathered}[/tex]
Now, the total energy of the spring is given when the value of "x" is equal to the amplitude "A":
[tex]E_{tot}=\frac{1}{2}kA^2[/tex]
The total energy is converted into kinetic energy and elastic energy, therefore, we have:
[tex]\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2[/tex]
Since the kinetic energy is twice the potential energy we have:
[tex]\frac{1}{2}mv^2=2(\frac{1}{2}kx^2)=kx^2[/tex]
Now, we substitute the values in the equation:
[tex]\frac{1}{2}kA^2=\frac{1}{2}kx^2+kx^2[/tex]
Now, we can cancel out the spring constant "k":
[tex]\frac{1}{2}A^2=\frac{1}{2}x^2+x^2[/tex]
Now, we add like terms in the right side:
[tex]\frac{1}{2}A^2=\frac{3}{2}x^2[/tex]
Now, we multiply both sides by 2:
[tex]A^2=3x^2[/tex]
Now, we divide both sides by 3:
[tex]\frac{A^2}{3}=x^2[/tex]
Now, we take the square root to both sides:
[tex]\sqrt{\frac{A^2}{3}}=x[/tex]
Solving the operations:
[tex]\pm0.5A=x[/tex]
Therefore, the distance is 0.5A.
