First slide

Applications of SHM

Question

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

Moderate

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

Given elastic energies are equal i.e., $\frac{1}{2} k_{1} x_{1}^{2} = \frac{1}{2} k_{2} x_{2}^{2}$
$\Rightarrow \frac{k_{1}}{k_{2}} = {(\frac{x_{2}}{x_{1}})}^{2}$ 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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