Two springs of force constants 300, N / m (Spring A ) and 400 N / m (Spring B ) are joined together

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13.Oscillations

hard

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

$4 / 3$

B

$16 / 9$

C

$3 / 4$

D

$9 / 6$

(AIIMS-2019)

Solution

Consider the spring $A$ is compressed by $x$ and spring $B$ is

compressed by $(8.75-x)$

The compressed force is,

$F=k x$

$F=300 x=400(8.75-x)$

On solving the above equation,

$x=5 \,cm$

Spring $B$ is compressed by $(8.75-5)=3.75 cm$

Now,

$\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A} x_{A}^{2}}{\frac{1}{2} k_{B} x_{B}^{2}}$

$=\frac{\frac{1}{2} \times 300 \times 5^{2}}{\frac{1}{2} \times 400 \times 3.75^{2}}$

$=\frac{4}{3}$

Standard 11

Physics

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