
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
175.2k+ 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
JEE Main Physics Mock Test 2025

JEE Main Maths Mock Test 2025: FREE Online Mock Test Series

JEE Main Chemistry Mock Test 2025

JEE Main Hydrocarbons Mock Test 2025-26: Free Practice Online

JEE Main 2025-26 Mock Test: Organic Compounds Containing Nitrogen

JEE Main 2025-26 Mock Test: Organic Compounds Containing Halogens

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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Uniform Acceleration

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26

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

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

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

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2025-26

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26
