# 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

Verified

54.3k+ 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.

Last updated date: 23rd May 2023

•

Total views: 54.3k

•

Views today: 0.13k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main