Two springs A and B of spring constants $1500 N / m$ and $3000 N / m$ respectively are stretched with the same force. The potential energy of B and A are in the ratio-

4:1

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1:4

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2:1

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1:2

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Solution

The correct option is C 2:1
$\frac{U_{1}}{U_{2}} = \frac{\frac{1}{2} k_{1} x_{1}^{2}}{\frac{1}{2} k_{2} x_{2}^{2}} = (\frac{k_{1} x_{1}}{k_{2} x_{2}})^{2} \frac{k_{2}}{k_{1}} = \frac{3000}{1500} = 2 : 1$ since $F = k_{1} x_{1} = k_{2} x_{2}$

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