The force constants of two springs are $K_{1}$ and $K_2$ . …

Physics Simple Harmonic Motion (Oscillations) Spring Pendulum Single Correct MCQ

Published on: September 12, 2026

The force constants of two springs are $K_{1}$ and $K_2$ . Both are stretched till their elastic energies are equal. If the stretching forces are $\mathbf{F}_1$ and $\mathbf{F}_{2}$ , then $F_1 : F_2$ is

$K_1 : K_2$

$K_{2} : K_{1}$

$\sqrt{K_1} : \sqrt{K_2}$

$K_1^2 : K_2^2$

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The correct answer is:

Given elastic energies are equal i.e., $\frac{1}{2} k_1 x_1^2 = \frac{1}{2} k_2 x_2^2$ $\Rightarrow \frac{k_1}{k_2} = \left(\frac{x_2}{x_1}\right)^2$ and using F = kx $\Rightarrow \frac{F_1}{F_2} = \frac{k_1 x_1}{k_2 x_2} = \frac{k_1}{k_2} \times \sqrt{\frac{k_2}{k_1}} = \sqrt{\frac{k_1}{k_2}}$

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