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  • 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 <img a

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 {E_{A}/}{E_{B}} is  equal to :

Option: 1

\frac{4}{3}


Option: 2

\frac{16}{9}


Option: 3

\frac{3}{4}


Option: 4

\frac{9}{16}


Answers (1)

best_answer

Displacement \ of \ spring \ A \ is, d_a =kx=300x_1 ? Displacement \ of \ spring \ B \ is, d_b = 400x_b ?

Displacement \ of \ spring \ B \ is, d_b = 400x_2


\begin{aligned} &x_{1}+x_{2}=8.75 \ldots(i)\\ &300 x_{1}=400 x_{2}\\ &3 x_{1}=4 x_{2} \dots(i i)\\ &\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2}(300)\left(\frac{8.75 \times 4}{7}\right)^{2}}{\frac{1}{2}(400)\left(\frac{8.75 \times 4}{7}\right)^{2}}=\frac{4}{3} \end{aligned}

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shivangi.shekhar

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