Question

Two springs of force constants $300\, N/m$ (Spring A) and $400\, N/m$ (Spring B) are joined together in series. The combination is compressed by $8.75\, cm$. The ratio of energy stored in $A$ and $B$ is $\frac{E_{A}}{E_{B}}. $ Then $\frac{E_{A}}{E_{B}}$ is equal to :

• A. $\frac{4}{3}$
• B. $\frac{16}{9}$
• C. $\frac{3}{4}$
• D. $\frac{9}{16}$

Answer 1

Answer: (A)

Given: $k_A = 300N/m, \,kB = 400N/m$ Let when the combination of springs is compressed by force $F$. Spring $A$ is compressed by $x$. Therefore compression in spring $B$.
$B_x = (8.75-x)\, cm$
$F = 300 \times x = 400(8.75 - x)$
Solving we get, $x = 5 \,cm$
$x_B = 8.75-5 = 3.75 \,cm$
$\frac{E_{A}}{E_{B}} = \frac{\frac{1}{2}k_{A}\left(X_{A}\right)^{2}}{\frac{1}{2}k_{B}\left(X_{B}\right)^{2}} = \frac{300\times\left(5\right)^{2}}{400\times\left(3.75\right)^{2}} = \frac{4}{3}$