Question

Two springs with spring constants \[{K_1} = 1500\,N/m\] and \[{K_1} = 3000\,N/m\] are stretched by the same force. The ratio of potential energy stored in spring will be
A. 2:1
B. 1:2
C. 4:1
D. 1:4

Answer 1

Hint:When a spring is stretched or compressed it undergoes some displacement then comes back to its equilibrium after some time, therefore the spring exerts an equal and opposite force on a body that compresses or stretches the spring. This energy stored in the spring is called potential energy of spring and is equal to the product of force with displacement.

Formula used :
\[F = kx\,\,\]and \[U = \dfrac{1}{2}k{x^2}\]
Here, F= spring force K = Spring constant, x = Elongation in spring and U = Spring potential energy.

Complete step by step solution:
Two springs of constant \[{K_1} = 1500\,N/m\] and \[{K_1} = 3000\,N/m\] are given here and we have to find the ratio of potential energy for the spring when they are stretched by same force. Form spring force, \[F = kx\,\,\] we have elongation in spring as,
\[x = \dfrac{F}{k}\,\,\]
As the force on both the springs is the same, let the elongation be \[{x_1}\] and \[{x_2}\].
Then, \[{x_1} = \dfrac{F}{{{K_1}}}\] and \[{x_2} = \dfrac{F}{{{K_2}}}\,..........(1)\].

Potential energy of spring is defined by the amount of energy stored in it, which the spring applies to come to equilibrium when it is either stretched or compressed. Potential energy U for spring constant K and elongation in spring x is given by,
\[U = \dfrac{1}{2}k{x^2}\]
Using above equation potential energy \[{U_1}\] and \[{U_2}\] for springs of constant \[{K_1}\] and \[{K_2}\] and elongation \[{x_1}\] and \[{x_2}\] respectively will be,
\[{U_1} = \dfrac{1}{2}{K_1}x_1^2\,.......(2)\]
\[\Rightarrow {U_2} = \dfrac{1}{2}{K_2}x_2^2\,.......(3)\]

Dividing equation (2) by (3)
\[\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{K_1}x_1^2}}{2} \times \dfrac{2}{{{K_2}x_2^2}} \Rightarrow \dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{K_1}x_1^2}}{{{K_2}x_2^2}}\]
Substituting values of $x_1$ and $x_2$ from equation (1) in above equation we get,
$\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{K_1\,(F/K_1)^2}{K_2\,(F/K_2)^2} \\
\Rightarrow \dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{K_1}{F^2}K_2^2}}{{{K_2}K_1^2{F^2}}}\,.......(3)$

Further solving equation (3) we get,
\[\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{K_2}}}{{{K_1}}}\,........(4)\]
Substituting \[{K_1} = 1500\,N/m\] and \[{K_1} = 3000\,N/m\] in equation (4) we get,
\[\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{3000}}{{1500}} = \dfrac{2}{1}\]
Hence, the ratio of potential energy stored in the springs is 2:1.

Therefore, option A is the correct answer.

Note: When the spring is at equilibrium position the potential energy of the spring is minimum i. e. zero and when the spring is stretched to length x where kinetic energy is zero, at that point potential energy is equal to the total external work done on the system.