Two tuning forks have frequencies 450Hz and 454Hz respectively. On sounding these together, the time interval between successive maximum intensities will be:
A) $\dfrac{1}{4}second$
B) $\dfrac{1}{2}second$
C) $1 second$
D) $2 second$
Answer
266.1k+ views
Hint: Here indirectly beat frequency is asked.
The beat is a word used for sound waves. The beat frequency is the difference in frequency of two waves. It is because of constructive and destructive interference.
Beat frequency gives us no maxima in beat and its time period gives us a time interval between two successive maxima.
Complete step by step solution:
When two waves of nearly equal frequencies traveling in a medium along the same direction meet at a point, beats are produced. When two sound waves of dissimilar frequency approach your ear, the alternating productive and damaging interference causes the sound to be alternatively loud and soft this is a phenomenon, which is termed as beating.
The beat frequency is equal to the complete value of the alteration in the frequency of the two waves. The count of beats per second is equivalent to the difference in frequencies of two waves is called beat frequency.
Beat Frequency Formula:
\[{f_b} = \left| {{f_2} - {f_1}} \right|\]
Where,
${f_b}$ is the beat frequency
${f_1}$ frequency of the first tuning fork
${f_2}$frequency of the second fork
So, the beat frequency will be,
Given data is
$
{f_1} = 450Hz \\
{f_2} = 454Hz$
So, the beat frequency will be
\[
\because {f_b} = \left| {{f_2} - {f_1}} \right|\, \\
\therefore {f_b} = \left| {454 - 450} \right| \\
\Rightarrow {f_b} = 4Hz
\]
Now as we know that the time period is the reciprocal of the frequency
So, the beat time period will be \[T = \dfrac{1}{{{f_b}}}\]
$T = \dfrac{1}{4} seconds$
Final answer is (A), the time interval between two maxima will be $\dfrac{1}{4}\sec.$
Note: The beat frequency gives the no of maxima in the beat.
Whenever two waves of different frequencies are imposed over each other they produce beats.
There is a lower difference between the frequencies of two waves; the waves strike our eardrum at a small interval of time; this phenomena of two maxima striking our eardrums in a nearly low time interval leads to the production of beats.
The beat is a word used for sound waves. The beat frequency is the difference in frequency of two waves. It is because of constructive and destructive interference.
Beat frequency gives us no maxima in beat and its time period gives us a time interval between two successive maxima.
Complete step by step solution:
When two waves of nearly equal frequencies traveling in a medium along the same direction meet at a point, beats are produced. When two sound waves of dissimilar frequency approach your ear, the alternating productive and damaging interference causes the sound to be alternatively loud and soft this is a phenomenon, which is termed as beating.
The beat frequency is equal to the complete value of the alteration in the frequency of the two waves. The count of beats per second is equivalent to the difference in frequencies of two waves is called beat frequency.
Beat Frequency Formula:
\[{f_b} = \left| {{f_2} - {f_1}} \right|\]
Where,
${f_b}$ is the beat frequency
${f_1}$ frequency of the first tuning fork
${f_2}$frequency of the second fork
So, the beat frequency will be,
Given data is
$
{f_1} = 450Hz \\
{f_2} = 454Hz$
So, the beat frequency will be
\[
\because {f_b} = \left| {{f_2} - {f_1}} \right|\, \\
\therefore {f_b} = \left| {454 - 450} \right| \\
\Rightarrow {f_b} = 4Hz
\]
Now as we know that the time period is the reciprocal of the frequency
So, the beat time period will be \[T = \dfrac{1}{{{f_b}}}\]
$T = \dfrac{1}{4} seconds$
Final answer is (A), the time interval between two maxima will be $\dfrac{1}{4}\sec.$
Note: The beat frequency gives the no of maxima in the beat.
Whenever two waves of different frequencies are imposed over each other they produce beats.
There is a lower difference between the frequencies of two waves; the waves strike our eardrum at a small interval of time; this phenomena of two maxima striking our eardrums in a nearly low time interval leads to the production of beats.
Recently Updated Pages
JEE Main Mock Test 2025-26: Principles Related To Practical

JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26: Magnetic Effects of Current & Magnetism Mock Test

JEE Main Statistics and Probability Mock Test 2025-26

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

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

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

JEE Advanced 2026 Marks vs Rank: Estimate IIT Rank from Your Score

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

