Question

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

• A. $E$
• B. $\frac{E}{8}$
• C. $\frac{E}{4}$
• D. $\frac{E}{2}$

Answer 1

The correct option is D $\frac{E}{2}$

$\begin{array}{cc}U& =\frac{1}{2}k{x}^2\\ U_p& =\frac{1}{2}{k}_p{x}_p^2\\ U_Q& =\frac{1}{2}{k}_Q{x}_Q^2\end{array}$

$F_p=F_q$

So, $\frac{U_p}{U_Q}=\frac{x_p}{x_Q}$

$\because F_p=-k_p{x}_p$
and $F_Q=-k_Q{x}_Q$

$⟹\frac{k_p}{k_Q}=\frac{x_Q}{x_p}=2$

$⟹U_p=\frac{E}{2}$