Two, spring $$P$$ and $$Q$$ of force constants $${k_p}$$ and $${k_Q}\left( {{k_Q} = \frac{{{k_p}}}{2}} \right)$$   are stretched by applying forces of equal magnitude. If the energy stored in $$Q$$ is $$E,$$ then the energy stored in $$P$$ is

Solution :
Here, $${k_Q} = \frac{{{k_p}}}{2}$$
According to Hooke's law
$$\eqalign{ & \therefore {F_p} = - {k_p}{x_p} \cr & {F_Q} = - {k_Q}{x_Q} \Rightarrow \frac{{{F_p}}}{{{F_Q}}} = \frac{{{k_p}}}{{{k_Q}}}\frac{{{x_p}}}{{{x_Q}}} \cr & {F_p} = {F_Q}\,\left[ {{\text{Given}}} \right] \cr & \therefore \frac{{{x_p}}}{{{x_Q}}} = \frac{{{k_Q}}}{{{k_p}}}\,......\left( {\text{i}} \right) \cr} $$
Energy stored in a spring is $$U = \frac{1}{2}k{x^2}$$
$$\eqalign{ & \therefore \frac{{{U_p}}}{{{U_Q}}} = \frac{{{k_p}x_p^2}}{{{k_Q}x_Q^2}} = \frac{{{k_p}}}{{{k_Q}}} \times \frac{{k_Q^2}}{{k_p^2}} = \frac{1}{2}\,\,\left[ {\because {k_Q} = \frac{{{k_p}}}{2}} \right] \cr & \Rightarrow {U_p} = \frac{{{U_Q}}}{2} = \frac{E}{2}\,\,\left[ {\therefore {U_Q} = E} \right] \cr} $$