Potential Energy of Spring

Elastic potential energy stored in the spring

$U=\frac{1}{2}k{x}^2$

$=\frac{1}{2}Fx=\frac{F^2}{2k}$

$Hence\frac{U_1}{U_2}=\frac{k_2}{k_1}\left. [Fissameforbothsprings\right]$

$=\frac{500}{1200}=\frac{5}{12}$

Potential energy stored in the spring $=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2$

$\begin{array}{rr}& \text{Now}\frac{1}{2}\mathrm{k}(2{)}^2=\mathrm{U}\\ & \&\frac{1}{2}\mathrm{k}(8{)}^2=U^{\mathrm{\text{'}}}\mathrm{}\text{(say)}\\ & \Rightarrow U^{\mathrm{\text{'}}}=\frac{64}{4}U=16U\end{array}$

ME = PE + KE = 8J
At x? , PE=4J, so KE = 8-4=4J. (A is correct)
At x>x? , PE=6J, so KE=2J, which is constant. (B is correct)
At x8J. This is not possible. Particle cannot reach here. So it is not a correct statement. (C is incorrect)
At x? , PE=0J, so KE=8J, which is maximum. (D is correct)