Answer:
potential energy is maximum when streach distance of string is doubled
Explanation:
we know that elastic potential energy of a spring is given as
[tex]E_{p}=\frac{1}{2}kx^2[/tex]..................(1)
here k is spring constant
and x is stretched distance of spring
if we double the spring constant then potential energy will be given as
[tex]E_{1}=\frac{1}{2}\times 2kx^2[/tex]
[tex]E_{1}=kx^2[/tex]...............(2)
similarly if we double the stretch distance then potential energy in spring will be given as
[tex]E_{2}=\frac{1}{2}\times 4kx^2[/tex]
[tex]E_{2}=2kx^2[/tex].............(3)
from equation 2 and equation 3 it is clear that
[tex]E_{2}>E_{1}[/tex]
it means potential energy is maximum when stretch distance of string is doubled
