
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
162.3k+ 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
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

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
