
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
241.2k+ 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
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Clemmensen and Wolff Kishner Reductions Explained for JEE & NEET

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Inductive Effect and Its Role in Acidic Strength

Understanding Average and RMS Value in Electrical Circuits

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

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

CBSE Notes Class 11 Physics Chapter 10 - Thermal Properties of Matter - 2025-26

NCERT Solutions For Class 11 Physics Chapter 14 Waves - 2025-26

Understanding Electromagnetic Waves and Their Importance

NCERT Solutions For Class 11 Physics Chapter 12 Kinetic Theory - 2025-26

Free Radical Substitution and Its Stepwise Mechanism

