Question

Two springs have force constant $K_1$ and $(K_1>K_2)$. Each spring is extended by same force. It their elastic potential energy are $E_1$ and $E_2$ then $\frac{E_1}{E_2}$ is

• A. $\frac{K_1}{K_2}$
• B. $\frac{K_2}{K_1}$
• C. $\sqrt{\frac{K_1}{K_2}}$
• D. $\sqrt{\frac{K_2}{K_1}}$

Answer 1

The correct option is B $\frac{K_2}{K_1}$

Let the two spring constants are $K_1$ and $K_2$. The force applied on the spring be $f$.

So the expansion occurred in first spring is given by,

$\Delta x=\frac{f}{K_1}$

The expansion occurred in the second spring is given as,

$\Delta x_1=\frac{f}{K_2}$

Thereby,

$E_1=\frac{1}{2}{K}_1\Delta x^2$

Similarly,

$E_2=\frac{1}{2}{K}_2\Delta x_1^2$

Hence,

$\frac{E_1}{E_2}=\frac{\frac{1}{2}{K}_2\Delta x^2}{\frac{1}{2}{K}_2\Delta x_1^2}$

$\frac{E_1}{E_1}=\frac{K_1(\frac{f}{K_1}{)}^2}{K_2(\frac{f}{K_2}{)}^2}$

$\frac{E_1}{E_2}=\frac{\frac{1}{K_1}}{\frac{1}{K_2}}$

$⟹\frac{E_1}{E_2}=\frac{K_2}{K_1}$