The wave described by y = 0.25 sin(10pi x - 2 | vedclass

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

Question

(A) The given wave equation is $y = 0.25 \sin(10\pi x - 2\pi t)$.Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$:$1$. The a

Vedclass · Accepted Answer

$+ve$ $x$-direction with frequency $1 \, Hz$ and wavelength $\lambda = 0.2 \, m$.

Vedclass · Answer

(A) The given wave equation is $y = 0.25 \sin(10\pi x - 2\pi t)$.Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$:$1$. The amplitude $A = 0.25 \, m$.$2$. The wave number $k = 10\pi$. Since $k = \frac{2\pi}{\lambda}$,we have $\lambda = \frac{2\pi}{10\pi} = 0.2 \, m$.$3$. The angular frequency $\omega = 2\pi$. Since $\omega = 2\pi f$,we have $f = \frac{2\pi}{2\pi} = 1 \, Hz$.$4$. Since the sign between $kx$ and $\omega t$ is negative,the wave travels in the positive $x$-direction.Therefore,the wave travels in the $+ve$ $x$-direction with frequency $1 \, Hz$ and wavelength $\lambda = 0.2 \, m$.

The wave described by $y = 0.25 \sin(10\pi x - 2\pi t)$,where $x$ and $y$ are in $meters$ and $t$ in $seconds$,is a wave travelling along:

Similar Questions

The equation for a spherical progressive wave is (where $r$ is the distance from the source):

In the wave equation $y = 0.5 \sin \frac{2 \pi}{\lambda}(400 t - x ) \, m$,the velocity of the wave will be ......... $m/s$.

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

$A$ sine wave has an amplitude $A$ and wavelength $\lambda$. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module