Question

A uniform spring has an unstretched length l and a force constant k. The spring is cut into two parts of unstretched length ${l_1}$ and ${l_2}$ such that ${l_1} = \eta {l_2}$, where $\eta$ is an integer. The corresponding force constants ${k_1}$ and ${k_2}$ are
A. $k\eta$ and $k\left( {\eta + 1} \right)$
B. $\dfrac{{k\left( {\eta + 1} \right)}}{\eta }$ and $k\left( {\eta - 1} \right)$
C. $\dfrac{{k\left( {\eta - 1} \right)}}{\eta }$ and $k\left( {\eta + 1} \right)$
D. $\dfrac{{k\left( {\eta + 1} \right)}}{\eta }$ and $k\left( {\eta + 1} \right)$

Answer 1

The spring force balances the weight of the mass attached to spring. We can use this concept to determine the force constant of spring of length l. The extension produced in the spring is proportional to the length of the spring. Determine the length ${l_1}$ and ${l_2}$ in terms of total length l and use the above discussed concept to solve this question.

Formula used:

$k = \dfrac{{mg}}{x}$

Here, k is force constant, m is the mass, g is the acceleration due to gravity and x is the elongation in the spring.

Complete step by step answer:
If the mass m is attached to the lower end of the spring of force constant k, the spring force balances the weight of mass m. therefore, we can write,
$kx = mg$
$k = \dfrac{{mg}}{x}$ …… (1)

Here, g is the acceleration due to gravity.

The extension produced in the spring is proportional to its length, $x \propto l$. Therefore,
$x = cl$

Here, c is the proportionality constant.

Substitute the value of x in equation (1).
$k = \dfrac{{mg}}{{cl}}$ ……. (2)
Now, since the spring is cut into two pieces, ${l_1}$ and ${l_2}$, we can write,
$\;{l_1} + {l_2} = l$ …… (3)

We have given,
${l_1} = \eta {l_2}$ …… (4)

Substitute ${l_1} = \eta {l_2}$ in equation (3).
$\eta {l_2} + {l_2} = l$
$\Rightarrow {l_2}\left( {\eta + 1} \right) = l$
$\Rightarrow {l_2} = \dfrac{l}{{\eta + 1}}$

Substitute the above equation in equation (4).
${l_1} = \eta \left( {\dfrac{l}{{\eta + 1}}} \right)$
$\Rightarrow {l_1} = \dfrac{{\eta l}}{{\eta + 1}}$

We can express the force constant of length ${l_1}$ using equation (2) as follows,
${k_1} = \dfrac{{mg}}{{c{l_1}}}$
$\Rightarrow {k_1} = \dfrac{{mg}}{{c\left( {\dfrac{{\eta l}}{{\eta + 1}}} \right)}}$
$\Rightarrow {k_1} = \dfrac{{mg}}{{cl}}\left( {\dfrac{{\eta + 1}}{\eta }} \right)$

Substitute k for $\dfrac{{mg}}{{cl}}$ in the above equation.
$\Rightarrow {k_1} = \dfrac{{k\left( {\eta + 1} \right)}}{\eta }$

Also, we can express the force constant of length ${l_2}$ using equation (2) as follows,
${k_2} = \dfrac{{mg}}{{c{l_2}}}$
$\Rightarrow {k_2} = \dfrac{{mg}}{{c\left( {\dfrac{l}{{\eta + l}}} \right)}}$
$\Rightarrow {k_2} = \dfrac{{mg}}{{cl}}\left( {\eta + l} \right)$
$\Rightarrow {k_2} = k\left( {\eta + l} \right)$

Therefore, ${k_1} = \dfrac{{k\left( {\eta + 1} \right)}}{\eta }$ and ${k_2} = k\left( {\eta + l} \right)$.

So, the correct answer is option (D).

Note: The force constant is inversely proportional to the length of the spring. Therefore, when we cut the spring into two pieces, the force constant is different for different lengths of the spring.