A wave travelling in the +ve x-direction havi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The standard equation of a wave travelling along the $+ve$ $x$-direction is given by:$y = A \sin(kx - \omega t)$where:$A$ is the amplitude of the

Vedclass · Accepted Answer

$y = \sin(x - 2t)$

Vedclass · Answer

(A) The standard equation of a wave travelling along the $+ve$ $x$-direction is given by:$y = A \sin(kx - \omega t)$where:$A$ is the amplitude of the wave.$k$ is the angular wave number.$\omega$ is the angular frequency of the wave.Given: $A = 1\, m$,$\lambda = 2\pi\, m$,and $f = \frac{1}{\pi}\ Hz$.Calculating the wave number $k$:$k = \frac{2\pi}{\lambda} = \frac{2\pi}{2\pi} = 1\, m^{-1}$.Calculating the angular frequency $\omega$:$\omega = 2\pi f = 2\pi 	imes \frac{1}{\pi} = 2\, rad/s$.Substituting these values into the standard equation:$y = 1 \sin(1x - 2t) = \sin(x - 2t)$.

$A$ wave travelling in the $+ve$ $x$-direction having displacement along $y$-direction as $1\, m$,wavelength $2\pi\, m$,and frequency of $\frac{1}{\pi}\ Hz$ is represented by

Similar Questions

When two sound waves travel in the same direction in a medium,the displacements of a particle located at $x$ at time $t$ are given by:
$y_1 = 0.05 \cos(0.50 \pi x - 100 \pi t)$
$y_2 = 0.05 \cos(0.46 \pi x - 92 \pi t)$
Then the velocity of the waves is..... $m/s$.

The path difference between the two waves $y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$ and $y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$ is

Write the equation of displacement of an element of a medium in a harmonic wave.

Select the correct statement about the reflection and refraction of a wave at the interface between medium $1$ and medium $2$.

What is the phase difference between two successive crests in a wave?

Vedclass Test Series

Exam Paper Generator

Online Exam Module