
The frequency of oscillation for a mass \[m\] will be proportional to ___
Answer
123.6k+ views
Hint: We will briefly look at the force balance equation for a harmonic oscillator to determine the different forces that act on the mass. Then we will assign the value of frequency of oscillation and look at its dependence on mass.
Complete step by step answer:
When a mass is tied to spring, it will execute a simple harmonic motion. The frequency of oscillation of this mass depends on the mass as well as the spring constant of the spring. Let the mass of the block be $m$ and the spring constant be $k$.
Then the only force acting on the block will be the restoring force of the spring which can be calculated from Hooke’s law as
$F = - kx$
From Newton’s second law, we know that this force will cause a net acceleration on the block which can be written mathematically as
$F = ma$
$ \Rightarrow - kx = ma$
Since we know that $a = \dfrac{{{d^2}x}}{{d{t^2}}}$, we can write
$m\dfrac{{{d^2}x}}{{d{t^2}}} + kx = 0$
Dividing both sides by $m$, we get
\[\dfrac{{{d^2}x}}{{d{t^2}}} + \dfrac{k}{m}x = 0\]
The term $k/m$ in the above equation is replaced by ${\omega ^2} = \dfrac{k}{m}$ where $\omega $ is the angular frequency of the oscillation of the mass.
Then the value of $\omega $ will be
$\omega = \sqrt {\dfrac{k}{m}} $
Since the angular frequency is related to the frequency as $\omega = 2\pi f$, we can write
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Hence we can see that the frequency of oscillation is proportional to
$f \propto \sqrt {\dfrac{k}{m}} $
Note: We should be aware of the force balance equation for a harmonic oscillator and its general form. While the angular frequency and the frequency of oscillation have the same dependence on the spring constant and mass, they’re different quantities and must not be confused.
Complete step by step answer:
When a mass is tied to spring, it will execute a simple harmonic motion. The frequency of oscillation of this mass depends on the mass as well as the spring constant of the spring. Let the mass of the block be $m$ and the spring constant be $k$.
Then the only force acting on the block will be the restoring force of the spring which can be calculated from Hooke’s law as
$F = - kx$
From Newton’s second law, we know that this force will cause a net acceleration on the block which can be written mathematically as
$F = ma$
$ \Rightarrow - kx = ma$
Since we know that $a = \dfrac{{{d^2}x}}{{d{t^2}}}$, we can write
$m\dfrac{{{d^2}x}}{{d{t^2}}} + kx = 0$
Dividing both sides by $m$, we get
\[\dfrac{{{d^2}x}}{{d{t^2}}} + \dfrac{k}{m}x = 0\]
The term $k/m$ in the above equation is replaced by ${\omega ^2} = \dfrac{k}{m}$ where $\omega $ is the angular frequency of the oscillation of the mass.
Then the value of $\omega $ will be
$\omega = \sqrt {\dfrac{k}{m}} $
Since the angular frequency is related to the frequency as $\omega = 2\pi f$, we can write
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Hence we can see that the frequency of oscillation is proportional to
$f \propto \sqrt {\dfrac{k}{m}} $
Note: We should be aware of the force balance equation for a harmonic oscillator and its general form. While the angular frequency and the frequency of oscillation have the same dependence on the spring constant and mass, they’re different quantities and must not be confused.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Mark the correct statements about the friction between class 11 physics JEE_Main

Find the acceleration of the wedge towards the right class 11 physics JEE_Main

A standing wave is formed by the superposition of two class 11 physics JEE_Main

Derive an expression for work done by the gas in an class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

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