Question

A rubber band (two parallel strands of elastic material) has a spring constant of $10{\text{ N}}{{\text{m}}^{ - 1}}$, if the band cut in one place such that it now forms a single long strand of elastic material, then its new spring constant is
A. $20{\text{ N}}{{\text{m}}^{ - 1}}$
B. $40{\text{ N}}{{\text{m}}^{ - 1}}$
C. $5{\text{ N}}{{\text{m}}^{ - 1}}$
D. $2.5{\text{ N}}{{\text{m}}^{ - 1}}$

Answer 1

In the question, spring constant of combination of two springs (strands can be considered as springs) connected in parallel is given as $10{\text{ N}}{{\text{m}}^{ - 1}}$ . So, find the spring constant of individual spring. If two springs having spring constants ${k_1}$ and ${k_2}$ are connected in parallel, then the spring constant for the combination is given by ${k_p} = {k_1} + {k_2}$ .
If two springs having spring constants ${k_1}$ and ${k_2}$ are connected in series, then the spring constant for the combination is given by $\dfrac{1}{{{k_s}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}}$ .

Complete step by step answer:
As given in the question, a rubber band (two parallel strands of elastic material) has a spring constant of $10{\text{ N}}{{\text{m}}^{ - 1}}$ . We can consider a strand to be a spring. We have to find the spring constant of individual spring. It is given that initially the two springs are connected in parallel.
Let two springs having spring constants ${k_1}$ and ${k_2}$ be connected in parallel. So, the displacement $\left( x \right)$ of both the springs will be the same but the restoring force will be different. Total force will be the sum of individual forces i.e.
$F = {F_1} + {F_2}$
The restoring force is given by $F = - kx$ , so substituting this in the above equation we have
$- {k_p}x = - {k_1}x - {k_2}x$
On simplifying we get
${k_p} = {k_1} + {k_2}$
According to the question, the springs are identical so they have same spring constant (say $k$ )
So, $10 = k + k = 2k$
Therefore, the spring constant of an individual string is $k = 5{\text{ N}}{{\text{m}}^{ - 1}}$ .
Now, if the band cut in one place such that it now forms a single long strand of elastic material. So, the two springs are now connected in series.
In this case, the restoring force will be the same but the displacement will be different for the springs.
The total displacement will be $x = {x_1} + {x_2}$
Substituting the value of the forces we have
$- \dfrac{F}{{{k_s}}} = - \dfrac{F}{{{k_1}}} - \dfrac{F}{{{k_2}}}$ (as $x = - \dfrac{F}{k}$ )
On simplifying we have
$\dfrac{1}{{{k_s}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}}$
As, ${k_1} = {k_2} = k$ then
$\dfrac{1}{{{k_s}}} = \dfrac{1}{k} + \dfrac{1}{k} = \dfrac{2}{k}$
So, the new spring constant is ${k_s} = \dfrac{k}{2} = \dfrac{5}{2} = 2.5{\text{ N}}{{\text{m}}^{ - 1}}$
Hence, option D is correct.

Note: The force applied if the displacement in the spring is one is known as spring constant. If a force $F$ is considered that elongates the spring so that it displaces the equilibrium position by $x$. That’s why the restoring force (the force which tries to retain its original position) is given by $F = - kx$. Here, the negative sign denotes that the restoring force is applied in the opposition direction of the displacement of the spring.