Question

Two springs A and B having spring constants $k_A$ and $k_B(k_A=4k_B)$ are stretched by applying force of equal magnitude. If energy stored in spring A is $E_A$, then energy stored in B will be:

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

Answer 1

The correct option is D $4E_A$

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),
$\frac{E_A}{E_B}=\left(\frac{k_A}{k_B}\right)\times {\left(\frac{k_B}{k_A}\right)}^2=\frac{k_B}{k_A}\left\{\begin{array}{c}\because \text{Given that}\\ \frac{k_B}{k_A}=\frac{1}{4}\end{array}\right\}$
$\frac{E_A}{E_B}=\frac{1}{4}$
$E_B=4E_A=4E$
Hence, the correct option is (d).