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 $1 	ext{ Hz}$ and wavelength $\lambda = 0.2 	ext{ m}$.

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 $A = 0.25 	ext{ m}$.$2$. The wave number $k = 10\pi$. Since $k = \frac{2\pi}{\lambda}$,we have $\frac{2\pi}{\lambda} = 10\pi$,which gives $\lambda = 0.2 	ext{ m}$.$3$. The angular frequency $\omega = 2\pi$. Since $\omega = 2\pi f$,we have $2\pi f = 2\pi$,which gives $f = 1 	ext{ Hz}$.$4$. Since the sign between the $kx$ and $\omega t$ terms is negative,the wave is travelling in the positive $x$-direction.Therefore,the wave travels in the $+ve$ $x$-direction with frequency $1 	ext{ Hz}$ and wavelength $\lambda = 0.2 	ext{ m}$.

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

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