Two springs of force constants $300 \mathrm{~N} / \mathrm{m}$ (Spring A) and $400 \mathrm{~N} / \mathrm{m}$…
Two springs of force constants $300 \mathrm{~N} / \mathrm{m}$ (Spring A) and $400 \mathrm{~N} / \mathrm{m}$ (Spring B) are joined together in series. The combination is compressed by $8.75 \mathrm{~cm}$. The ratio of energy stored in $\mathrm{A}$ and $\mathrm{B}$ is $\frac{E_A}{E_B}$. Then $\frac{E_A}{E_B}$ is equal to:
$\frac{4}{3}$
$\frac{16}{9}$
$\frac{3}{4}$
$\frac{9}{16}$
Solution
Given : $\mathrm{k}_{\mathrm{A}}=300 \mathrm{~N} / \mathrm{m}, \mathrm{k}_{\mathrm{B}}=400 \mathrm{~N} / \mathrm{m}$
Let when the combination of springs is compressed by force F. Spring A is compressed by $x$. Therefore compression in spring $\mathrm{B}$
$
\begin{aligned}
& x_B=(8.75-x) \mathrm{cm} \\
& \mathrm{F}=300 \times \mathrm{x}=400(8.75-\mathrm{x})
\end{aligned}
$
Solving we get, $x=5 \mathrm{~cm}$
$
\mathrm{x}_{\mathrm{B}}=8.75-5=3.75 \mathrm{~cm}
$
$\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{E}_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{k}_{\mathrm{A}}\left(\mathrm{x}_{\mathrm{A}}\right)^2}{\frac{1}{2} \mathrm{k}_{\mathrm{B}}\left(\mathrm{x}_{\mathrm{B}}\right)^2}=\frac{300 \times(5)^2}{400 \times(3.75)^2}=\frac{4}{3}$