Question

Two springs of spring constants $1500{\text{N}}{{\text{m}}^{ - 1}}$ and $3000{\text{N}}{{\text{m}}^{ - 1}}$ respectively are stretched with the same force. Find the ratio of their potential energies.
A) 4 : 1
B) 1 : 4
C) 2 : 1
D) 1 : 2

Answer 1

Hint:The two springs on stretching will constitute simple harmonic motion and so the potential energy of the two springs will be similar to the potential energy of a particle executing simple harmonic motion. . This potential energy is known to be proportional to the product of the spring constant and the square of the displacement of the spring. The force exerted on the springs is given to be same and this force is the product of the displacement of each spring and their respective spring constants

Formulas used:
-The force exerted on a spring is given by, $F = kx$ where $k$ is the spring constant of the spring and $x$ is the displacement of the spring.
-The potential energy of the spring is given by, $U = \dfrac{1}{2}k{x^2}$ where $k$ is the spring constant of the spring and $x$ is the displacement of the spring.

Complete step by step answer.
Step 1: List the parameters of the two springs.
Let the two springs be named as spring A and spring B.
Then the spring constant of spring A is given to be ${k_A} = 1500{\text{N}}{{\text{m}}^{ - 1}}$ .
The spring constant of spring B is given to be ${k_B} = 3000{\text{N}}{{\text{m}}^{ - 1}}$ .
Let ${x_A}$ and ${x_B}$ be the displacement of spring A and spring B respectively.
The force exerted on both the springs is mentioned to be the same.
i.e., ${F_A} = {F_B} = F$ .
Step 2: Express the force exerted on both springs.
The force exerted on spring A can be represented as $F = {k_A}{x_A}$ .
$ \Rightarrow {x_A} = \dfrac{F}{{{k_A}}}$ ------- (1)
The force exerted on spring B can be represented as $F = {k_B}{x_B}$ .
$ \Rightarrow {x_B} = \dfrac{F}{{{k_B}}}$ ------- (2)
Step 3: Express the potential energies of the two stretched springs.
The potential energy of spring A can be expressed as ${U_A} = \dfrac{1}{2}{k_A}{x_A}^2$ ------- (3)
Substituting equation (1) in (3) we get, ${U_A} = \dfrac{1}{2}{k_A}{\left( {\dfrac{F}{{{k_A}}}} \right)^2} = \dfrac{{{F^2}}}{{2{k_A}}}$
Thus the potential energy of stretched spring A is ${U_A} = \dfrac{{{F^2}}}{{2{k_A}}}$ ------- (4)
The potential energy of spring B can be expressed as ${U_B} = \dfrac{1}{2}{k_B}{x_B}^2$ ------- (5)
Substituting equation (2) in (5) we get, ${U_B} = \dfrac{1}{2}{k_B}{\left( {\dfrac{F}{{{k_B}}}} \right)^2} = \dfrac{{{F^2}}}{{2{k_B}}}$
Thus the potential energy of stretched spring B is ${U_B} = \dfrac{{{F^2}}}{{2{k_B}}}$ ------- (6)
Now dividing equation (4) by (6) we get, $\dfrac{{{U_A}}}{{{U_B}}} = \dfrac{{\left( {\dfrac{{{F^2}}}{{2{k_A}}}} \right)}}{{\left( {\dfrac{{{F^2}}}{{2{k_B}}}} \right)}} = \dfrac{{{k_B}}}{{{k_A}}}$ -------- (7)
Substituting for ${k_A} = 1500{\text{N}}{{\text{m}}^{ - 1}}$ and ${k_B} = 3000{\text{N}}{{\text{m}}^{ - 1}}$ in equation (7) we get, $\dfrac{{{U_A}}}{{{U_B}}} = \dfrac{{3000}}{{1500}} = \dfrac{2}{1}$
Thus the ratio of the potential energies of the two springs is obtained as 2 : 1.

So the correct option is C.

Note: Here it is mentioned that the two springs are stretched by exerting the same force. As the springs are stretched the displacement of each spring will be positive. The displacement corresponds to the length through which the spring gets stretched. Equation (7) suggests that the ratio of the potential energies of the two springs is inversely proportional to the ratio of their respective spring constants.