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13.Oscillations
medium
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 ${F_1}$ and ${F_2}$, then ${F_1}:{F_2}$ is
A
${K_1}:{K_2}$
B
${K_2}:{K_1}$
C
$\sqrt {{K_1}} :\sqrt {{K_2}} $
D
$K_1^2:K_2^2$
Solution
(c) 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}}}} $
Standard 11
Physics
