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Q. The force constants of two springs are $ k_{1} $ and $ k_{2} $ respectively. Both are stretched till their potential energies are equal. The forces $ f_{1} $ and $ f_{2} $ applied on them are in the ratio

Chhattisgarh PMTChhattisgarh PMT 2004

Solution:

Energy of spring $U=\frac{1}{2} k x^{2}$
and force $f=k x$
From these two, we get
$U=\frac{1}{2} k \frac{f^{2}}{k^{2}}=\frac{1}{2} \frac{f^{2}}{k}$
$ \because$ Energies are equal,
therefore $\frac{f_{1}^{2}}{k_{1}}=\frac{f_{2}^{2}}{k_{2}}$
$ \Rightarrow \frac{f_{1}}{f_{2}}=\frac{\sqrt{k_{1}}}{\sqrt{k_{2}}}$