The equation of a wave is represented by $y = {10^{ - 4}}\sin \left( {100t - \dfrac{x}{{10}}} \right)\,m$, then the velocity of wave will be,
(A) $100\,m{s^{ - 1}}$
(B) $4\,m{s^{ - 1}}$
(C) $1000\,m{s^{ - 1}}$
(D) $10\,m{s^{ - 1}}$
Answer
259.8k+ views
Hint: The given equation is compared with the general wave equation, then we get the angular frequency value and the wavenumber value, then by equating the angular frequency formula with the angular frequency value, the frequency can be determined and by using the wave number formula, then the wavelength can be determined by using these two values the velocity is determined.
Useful formula
The general wave equation is given by,
$y = A\sin \left( {\omega t - kx} \right)$
Where, $y$ and $A$is the amplitude of the wave, $\omega $ is the angular frequency of the wave, $t$ is the time taken by the wave, $kx$ is the wave number.
Complete step by step solution
Given that,
The equation of a wave is represented by $y = {10^{ - 4}}\sin \left( {100t - \dfrac{x}{{10}}} \right)\,m$,
Now,
The general wave equation is given by,
$y = A\sin \left( {\omega t - kx} \right)\,.................\left( 1 \right)$
By comparing the equation (1) with the given equation of the wave, then
$A\sin \left( {\omega t - kx} \right) = {10^{ - 4}}\sin \left( {100t - \dfrac{x}{{10}}} \right)$
Now the values of the terms are equated as,
$\omega = 100$ and $k = \dfrac{1}{{10}}$
Now, the formula of the angular frequency is given as,
$\omega = 2\pi \nu $
Where, $\nu $ is the frequency of the wave.
Now equating the value of the angular frequency with the angular frequency formula, then
$100 = 2\pi \nu $
By rearranging the terms in the above equation, then the above equation is written as,
$\nu = \dfrac{{100}}{{2\pi }}\,...............\left( 2 \right)$
Now, the formula of the wave number is given as,
$k = \dfrac{{2\pi }}{\lambda }$
Now equating the value of the wave number with the wavenumber formula, then
$\dfrac{1}{{10}} = \dfrac{{2\pi }}{\lambda }$
By rearranging the terms, then the above equation is written as,
$\lambda = 2\pi \times 10\,..................\left( 3 \right)$
Now, velocity is given by the product of the frequency and the wave length, then
By substituting the equation (2) and equation (3) in the above equation, then
$V = \dfrac{{100}}{{2\pi }} \times 2\pi \times 10$
By cancelling the terms in the above equation, then
$V = 100 \times 10$
By multiplying the terms in the above equation, then
$V = 1000\,m{s^{ - 1}}$
Hence, the option (C) is the correct answer.
Note: The velocity of the wave is directly proportional to the frequency of the wave and the wavelength of the wave. If the frequency of the wave or wavelength of the wave increases, the velocity of the wave also increases. If the frequency of the wave or wavelength of the wave increases, the velocity of the wave also decreases.
Useful formula
The general wave equation is given by,
$y = A\sin \left( {\omega t - kx} \right)$
Where, $y$ and $A$is the amplitude of the wave, $\omega $ is the angular frequency of the wave, $t$ is the time taken by the wave, $kx$ is the wave number.
Complete step by step solution
Given that,
The equation of a wave is represented by $y = {10^{ - 4}}\sin \left( {100t - \dfrac{x}{{10}}} \right)\,m$,
Now,
The general wave equation is given by,
$y = A\sin \left( {\omega t - kx} \right)\,.................\left( 1 \right)$
By comparing the equation (1) with the given equation of the wave, then
$A\sin \left( {\omega t - kx} \right) = {10^{ - 4}}\sin \left( {100t - \dfrac{x}{{10}}} \right)$
Now the values of the terms are equated as,
$\omega = 100$ and $k = \dfrac{1}{{10}}$
Now, the formula of the angular frequency is given as,
$\omega = 2\pi \nu $
Where, $\nu $ is the frequency of the wave.
Now equating the value of the angular frequency with the angular frequency formula, then
$100 = 2\pi \nu $
By rearranging the terms in the above equation, then the above equation is written as,
$\nu = \dfrac{{100}}{{2\pi }}\,...............\left( 2 \right)$
Now, the formula of the wave number is given as,
$k = \dfrac{{2\pi }}{\lambda }$
Now equating the value of the wave number with the wavenumber formula, then
$\dfrac{1}{{10}} = \dfrac{{2\pi }}{\lambda }$
By rearranging the terms, then the above equation is written as,
$\lambda = 2\pi \times 10\,..................\left( 3 \right)$
Now, velocity is given by the product of the frequency and the wave length, then
By substituting the equation (2) and equation (3) in the above equation, then
$V = \dfrac{{100}}{{2\pi }} \times 2\pi \times 10$
By cancelling the terms in the above equation, then
$V = 100 \times 10$
By multiplying the terms in the above equation, then
$V = 1000\,m{s^{ - 1}}$
Hence, the option (C) is the correct answer.
Note: The velocity of the wave is directly proportional to the frequency of the wave and the wavelength of the wave. If the frequency of the wave or wavelength of the wave increases, the velocity of the wave also increases. If the frequency of the wave or wavelength of the wave increases, the velocity of the wave also decreases.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

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 Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

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

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

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

