Question

Two springs $A$ and $B$ are stretched by applying forces of equal magnitudes. If the energy stored in $A$ is E, then the energy stored in B will be-
(Given relation between spring constants of $A$ and $B$ is $(k_A=2k_B)$)

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

Answer 1

The correct option is B $2E$

Given that
$F_A=F_B$
$k_A{x}_A=k_B{x}_B$
$\frac{k_A}{k_B}=\frac{x_B}{x_A}....\left(1\right)$
$\frac{E_A}{E_B}=\frac{\frac{1}{2}{k}_A{x}_A^2}{\frac{1}{2}{k}_B{x}_B^2}=\left(\frac{k_A}{k_B}\right){\left(\frac{x_A}{x_B}\right)}^2$
From eqution (1),
$\frac{E_A}{E_B}=\left(\frac{k_A}{k_B}\right)\times {\left(\frac{k_B}{k_A}\right)}^2=\frac{k_B}{k_A}\left\{\begin{array}{c}\because \text{Given that}\\ \frac{k_B}{k_A}=\frac{1}{2}\end{array}\right\}$
$\frac{E_A}{E_B}=\frac{1}{2}$
$E_B=2E_A=2E$
Hence, the correct option is (B).