
A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system if a mass 4 m is suspended from the same spring?
A. \[\dfrac{n}{4}\]
B. \[4n\]
C. \[\dfrac{n}{2}\]
D. \[2n\]
Answer
161.4k+ views
Hint: The frequency of the spring-mass oscillator is inversely proportional to the square root of the mass of the block attached to the spring. When spring constant remains constant then on increasing the mass of the block will result the decrease in the frequency of oscillation.
Formula used:
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \], here f is the frequency, k is the spring constant and m is the mass of the block.
Complete step by step solution:
The initial mass of the block is given as m. Let the spring constant of the given spring through which the mass is suspended to form the vertical spring mass oscillator is k.
Then using the formula of frequency of spring-mass oscillator is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
The initial frequency is given as n,
So, \[n = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
Now, the initial mass which is suspended to the spring is replaced with another block of mass 4m and the spring is same as used in the initial case.
So, using the formula of frequency of spring-mass oscillator, the final frequency will be;
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{k_2}}}{{{m_2}}}} \]
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{4m}}} \]
\[{f_2} = \dfrac{1}{2}\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)\]
On dividing the first frequency with the second frequency, we get
\[\dfrac{n}{{{f_2}}} = \dfrac{{\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)}}{{\dfrac{1}{2}\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)}}\]
\[\dfrac{n}{{{f_2}}} = 2\]
\[{f_2} = \dfrac{n}{2}\]
So, the frequency of the oscillator becomes \[\dfrac{n}{2}\]when the mass is increased four times keeping the spring same.
Therefore, the correct option is (C).
Note: We should be careful while using the formula of frequency of suspended mass-spring system. If the spring given in the question was not massless than we can’t use that formula as mass of the spring will also contribute to the characteristic of the motion of the system.
Formula used:
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \], here f is the frequency, k is the spring constant and m is the mass of the block.
Complete step by step solution:
The initial mass of the block is given as m. Let the spring constant of the given spring through which the mass is suspended to form the vertical spring mass oscillator is k.
Then using the formula of frequency of spring-mass oscillator is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
The initial frequency is given as n,
So, \[n = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
Now, the initial mass which is suspended to the spring is replaced with another block of mass 4m and the spring is same as used in the initial case.
So, using the formula of frequency of spring-mass oscillator, the final frequency will be;
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{k_2}}}{{{m_2}}}} \]
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{4m}}} \]
\[{f_2} = \dfrac{1}{2}\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)\]
On dividing the first frequency with the second frequency, we get
\[\dfrac{n}{{{f_2}}} = \dfrac{{\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)}}{{\dfrac{1}{2}\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)}}\]
\[\dfrac{n}{{{f_2}}} = 2\]
\[{f_2} = \dfrac{n}{2}\]
So, the frequency of the oscillator becomes \[\dfrac{n}{2}\]when the mass is increased four times keeping the spring same.
Therefore, the correct option is (C).
Note: We should be careful while using the formula of frequency of suspended mass-spring system. If the spring given in the question was not massless than we can’t use that formula as mass of the spring will also contribute to the characteristic of the motion of the system.
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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Class 11 JEE Main Physics Mock Test 2025

Differentiate between audible and inaudible sounds class 11 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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 1 Units and Measurements

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