
A mass M is suspended by two springs of force constants \[{K_1}\] and \[{K_2}\] respectively as shown in the diagram. The total elongation (stretch) of the two springs is

A. \[\dfrac{{Mg}}{{{K_1} + {K_2}}} \\ \]
B. \[\dfrac{{Mg\left( {{K_1} + {K_2}} \right)}}{{{K_1}{K_2}}} \\ \]
C. \[\dfrac{{Mg\left( {{K_1}{K_2}} \right)}}{{{K_1} + {K_2}}} \\ \]
D. \[\dfrac{{{K_1} + {K_2}}}{{{K_1}{K_2}Mg}}\]
Answer
162.3k+ views
Hint: Force on the system will be a product of effective constant for the series combination and elongation in the spring.
Formula used:
The expression of restoring force is,
\[F = Kx\]
Where, F = Force, k= Spring constant and x = Elongation (stretch) of the spring.
Complete step by step solution:
Given here is a spring mass system of two springs having constant \[{K_1}\] and \[{K_2}\] respectively. Springs are combined together in series combination and a mass M is suspended by the springs, we have to find the elongation in springs.
First we need to find the effective constant for the given combination of springs and it will be,
\[{K_{eff}} = \dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}\,.........(1)\]
Let the elongation in spring be x then free body diagram of suspended mass M will be,

Image: Free body diagram of mass M
From free body diagram we have,
\[{K_{eff}}x = Mg \\
\Rightarrow x = \dfrac{{Mg}}{{{K_{eff}}}}\,.......(2)\]
Substituting value of \[{K_{eff}}\] form equation (1) in equation (2) we get,
\[x = \dfrac{{Mg\left( {{K_1} + {K_2}} \right)}}{{{K_1}{K_2}}}\,\]
Hence, elongation in the spring will be \[\dfrac{{Mg\left( {{K_1} + {K_2}} \right)}}{{{K_1}{K_2}}}\,\].
Therefore, option B is the correct answer.
Note: Even though in combination of springs we have multiple springs connected to each other either in series or parallel, they behave like a single spring and to solve numerical problems for combination of springs their effective constant is to be calculated first.
Formula used:
The expression of restoring force is,
\[F = Kx\]
Where, F = Force, k= Spring constant and x = Elongation (stretch) of the spring.
Complete step by step solution:
Given here is a spring mass system of two springs having constant \[{K_1}\] and \[{K_2}\] respectively. Springs are combined together in series combination and a mass M is suspended by the springs, we have to find the elongation in springs.
First we need to find the effective constant for the given combination of springs and it will be,
\[{K_{eff}} = \dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}\,.........(1)\]
Let the elongation in spring be x then free body diagram of suspended mass M will be,

Image: Free body diagram of mass M
From free body diagram we have,
\[{K_{eff}}x = Mg \\
\Rightarrow x = \dfrac{{Mg}}{{{K_{eff}}}}\,.......(2)\]
Substituting value of \[{K_{eff}}\] form equation (1) in equation (2) we get,
\[x = \dfrac{{Mg\left( {{K_1} + {K_2}} \right)}}{{{K_1}{K_2}}}\,\]
Hence, elongation in the spring will be \[\dfrac{{Mg\left( {{K_1} + {K_2}} \right)}}{{{K_1}{K_2}}}\,\].
Therefore, option B is the correct answer.
Note: Even though in combination of springs we have multiple springs connected to each other either in series or parallel, they behave like a single spring and to solve numerical problems for combination of springs their effective constant is to be calculated first.
Recently Updated Pages
How To Find Mean Deviation For Ungrouped Data

Difference Between Molecule and Compound: JEE Main 2024

Ammonium Hydroxide Formula - Chemical, Molecular Formula and Uses

Difference Between Area and Surface Area: JEE Main 2024

Difference Between Work and Power: JEE Main 2024

Difference Between Acetic Acid and Glacial Acetic Acid: JEE Main 2024

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
