Answer:
K'=4K
Explanation:
The electric potential energy is given by :
[tex]E=\dfrac{1}{2}kx^2[/tex]
Where
k is spring constant
x is compression or extension in the spring
If the displacement of a horizontal mass-spring system is doubled, x'= 2x
New elastic potential energy :
[tex]E'=\dfrac{1}{2}kx'^2\\\\E'=\dfrac{1}{2}k(2x)^2\\\\=\dfrac{1}{2}k\times 4x^2\\\\K'=4\times \dfrac{1}{2}kx^2\\\\K'=4K[/tex]
So, new elastic potential energy 4 times the initial elastic potential energy.
