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

The figure shows a sinusoidal wave at a given instant. Which points are in the same phase?

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$. This represents a

Two waves are given by $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \cos(\omega t - kx)$. The phase difference between the two waves is

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$. This represents:

The equation of a wave is given by (all quantities expressed in $M.K.S.$ units) $Y = 5 \sin(10\pi t - 0.1\pi x)$ along the $x$-axis. The phase difference between two points separated by a distance of $10 \ m$ along the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module