Question

Two springs have force constant K1 and K2 (k1k2) . EACH SPRING IS EXTENDED BY SAME FORCE f , IF THEIR elastic potential energy are E1 and E2 then, E1/E2 IS = ?

Answer 1

expansion of 1st string
change in x= f/k1

expansion in 2nd string
change in x= f/k2

sso,
ΔE1/ΔE2= 1/2 k1(Δx1)² divided by 1/2 k2(Δx2)²
E1/E2= k2/k1

Answer 2

$\frac{E 1}{E 2} \ \text{is} \ \frac{K 2}{K 1}$

Given:

Spring force constants = K1 and K2

Solution:  

Force applied on spring = f

So, the expansion occurred in first spring is given as:

$\Delta x=\frac{f}{K 1}$

The expansion occurred in the second spring is given as:

$\Delta x^{\prime}=\frac{f}{K 2}$

Thereby,

$E 1=\frac{1}{2}(K 1)(\Delta x)^{2}$

Similarly,  

$E 2=\frac{1}{2}(K 2)\left(\Delta x^{\prime}\right)^{2}$

Hence,  

$\frac{E 1}{E 2}=\frac{\left(\frac{1}{2}(K 1)(\Delta x)^{2}\right)}{\left(\frac{1}{2}(K 2)\left(\Delta x^{\prime}\right)^{2}\right)}$

$\frac{E 1}{E 2}=\frac{K 1\left(\frac{f}{K 1}\right)^{2}}{K 2\left(\frac{f}{K 2}\right)^{2}}$

$\frac{E 1}{E 2}=\frac{K 1\left(\frac{1}{K 1}\right)^{2}}{K 2\left(\frac{1}{K 2}\right)^{2}}$

$\frac{E 1}{E 2}=\frac{\frac{1}{K 1}}{\frac{1}{K 2}}$

$\Rightarrow \frac{E 1}{E 2}=\frac{K 2}{K 1}$