
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
164.1k+ views
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.
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.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Charging and Discharging of Capacitor

A body of mass 3Kg moving with a velocity of 4ms towards class 11 physics JEE_Main

Class 11 JEE Main Physics Mock Test 2025

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
JEE Advanced 2025 Notes

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
