Question

Two springs of constant\[{k_1}\] and \[{k_2}\] are joined in series. The effective spring constant of the combination is given by
A. \[\sqrt {{k_1}{k_2}} \\ \]
B. \[\dfrac{{{k_1} + {k_2}}}{2} \\ \]
C. \[{k_1} + {k_2} \\ \]
D. \[\dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}\]

Answer 1

Hint: According to Hooke’ law spring force is given by \[F = - kx\], where x is deformation of spring which can be mathematically calculated from equation \[x = - \left( {\dfrac{F}{k}} \right)\].

Formula used :
\[F = - kx\]
Here, F = Spring force, K = Spring constant and x = Deformation in spring.

Complete step by step solution:
Two springs of constant \[{k_1}\] and \[{k_2}\] are joined in series, we have to calculate the effective constant of the combination. From Hooke’s law spring force for constant k and deformation x is given by,
\[F = - kx\,......(1)\]

Let the deformation in springs in series combination be \[{x_1}\] and \[{x_2}\] then spring force \[{F_1}\] and \[{F_2}\] will be,
\[{F_1} = - {k_1}{x_1}\] and \[{F_1} = - {k_1}{x_2}\]
Then we have, \[{x_1} = - \dfrac{{{F_1}}}{{{k_1}}}\] and \[{x_2} = - \dfrac{{{F_1}}}{{{k_1}}}\].
In a series combination of two or more springs deformation of each spring may be different but total deformation of the combination is the sum of individual deformations of springs.

Therefore for given series combination of springs, deformation of combination(x) will be,
\[x = {x_1} + {x_2}\]
Substituting the values of \[{x_1}\]and\[{x_2}\]in above equation we get,
\[x = \left( { - \dfrac{{{F_1}}}{{{k_1}}}} \right) + \left( { - \dfrac{{{F_2}}}{{{k_2}}}} \right)\,.....(2)\]

As in series combination of springs force on each spring is same therefore, \[{F_1} = {F_2}\]
Let \[{F_1} = {F_2} = F\]
Then, equation (2) can be rewritten as,
\[x = - F\left( {\dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}}} \right)..........(3)\]
From Hooke’s law, force on combination of spring of constant will be,
\[F = - kx\]
Then \[x = - \dfrac{F}{{{k_{eff}}}}\,.....(4)\]
On comparing equation (3) and (4) we get,
\[\dfrac{1}{{{k_{eff}}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}}\]
Hence, \[{k_{eff}} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}\]

Therefore, option D is the correct answer.

Note: In series or parallel combination of two or more springs even when the spring constants are different for each individual but they act as one.