A mass m performs oscillations of period T when hanged by spring of force constant K. If spring is cut in two parts and arranged in parallel and same mass is oscillated by them, then the new time period will be

Figure: The spring-mass system
A. \[2T\]
B. \[T\]
C. \[\dfrac{T}{{\sqrt 2 }}\]
D. \[\dfrac{T}{2}\]
Answer
248.1k+ views
Hint: The spring constant of the spring is inversely proportional to the length of the spring. If the same spring is divided into smaller spring lengths then the resulting spring will have more stiffness than the original one. We find the equivalent stiffness of the combination then use it in the formula of the period of spring-mass system.
Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Where T is the period, k is the spring constant and m is the mass of the block.
Complete step by step solution:

Figure: The spring-mass system
For the initial case, using period formula,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
As we know that the spring constant of the spring is inversely proportional to the length of the spring, so when the spring is divided into two equal parts, then the spring constant of each equal parts will be twice of the original spring.
It is given that the spring constant of original spring is k, so the spring constant of each part will be 2k. Then, the equivalent spring constant in parallel is,
\[{K_{eq}} = 2k + 2k = 4k\]
So, the final period of the system will be,
\[{T_2} = 2\pi \sqrt {\dfrac{m}{{4K}}} \]
\[{T_2} = \dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)\]
On dividing the expression for the periods, we get
\[\dfrac{T}{{{T_2}}} = \dfrac{{\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}{{\dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}\]
\[\dfrac{T}{{{T_2}}} = 2\]
\[{T_2} = \dfrac{T}{2}\]
So, the period of the new spring-mass system is half of the initial period.
Therefore, the correct option is (D).
Note: We should assume that the spring’s mass is negligible in relative to the mass suspended through the spring. Otherwise the motion of the block will not be perfect simple harmonic.
Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Where T is the period, k is the spring constant and m is the mass of the block.
Complete step by step solution:

Figure: The spring-mass system
For the initial case, using period formula,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
As we know that the spring constant of the spring is inversely proportional to the length of the spring, so when the spring is divided into two equal parts, then the spring constant of each equal parts will be twice of the original spring.
It is given that the spring constant of original spring is k, so the spring constant of each part will be 2k. Then, the equivalent spring constant in parallel is,
\[{K_{eq}} = 2k + 2k = 4k\]
So, the final period of the system will be,
\[{T_2} = 2\pi \sqrt {\dfrac{m}{{4K}}} \]
\[{T_2} = \dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)\]
On dividing the expression for the periods, we get
\[\dfrac{T}{{{T_2}}} = \dfrac{{\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}{{\dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}\]
\[\dfrac{T}{{{T_2}}} = 2\]
\[{T_2} = \dfrac{T}{2}\]
So, the period of the new spring-mass system is half of the initial period.
Therefore, the correct option is (D).
Note: We should assume that the spring’s mass is negligible in relative to the mass suspended through the spring. Otherwise the motion of the block will not be perfect simple harmonic.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Derivation of Equation of Trajectory Explained for Students

Understanding the Angle of Deviation in a Prism

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

