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

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(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:

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