Question

Two springs having force constants in the ratio of 16/25 Both the springs are stretched such that both have equal elastic potential energy. If elongation in first spring is x, then elongation in second spring will be​

Answer 1

$potential \: energy = \frac{1}{2} \times k \times {x}^{2} \\ \\ as \: (pe)1 = (pe)2 \\ hence \: k1 \times {(x1)}^{2} = k2 \times {(x2)}^{2} \\ \frac{k1}{k2} = \frac{ {(x2)}^{2} }{ {(x1)}^{2} } \\ \frac{16}{25} = \frac{ {(x2)}^{2} }{ {x}^{2} } \\ as \: x1 = x \\ \\ x2 = \frac{4}{5} \times x$

Hence this is the elongation in the spring.

Answer 2

Given:

Two springs having force constants in the ratio of 16:25. They have equal PE.

To find:

Elongation of 2nd spring ?

Calculation:

• Let the elongation of 2nd spring be 'y'.

$\dfrac{PE_{1}}{PE_{2}} = \dfrac{ \dfrac{1}{2} k_{1}( {x}^{2} ) }{\dfrac{1}{2} k_{2}( {y}^{2})}$

$\implies 1 = \dfrac{ 16( {x}^{2} ) }{25( {y}^{2})}$

$\implies {y}^{2} = \dfrac{ 16( {x}^{2} ) }{25}$

$\implies y= \sqrt{\dfrac{ 16( {x}^{2} ) }{25} }$

$\implies y= \dfrac{4x }{5}$

So, elongation in 2nd spring is 4x/5.