The equation $\overrightarrow {\phi \,} (x,\,t) = \overrightarrow {j\,} \sin \,\left( {\frac{{2\pi }}{\lambda }v\,t} \right)\cos \,\left( {\frac{{2\pi }}{\lambda }x} \right)$ represents
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Let's first find if the given wave equation is stationary or progressive. If we can define certain points for which the amplitude of the wave will always be zero, i.e if the wave has nodal points, then the wave will be considered as stationary. If we put $x =\frac{ n \lambda}{4}$, then the cosine component of the wave becomes zero at, $n =$ $1,2,3,4 \ldots$. So, we can conclude that that their are points in the wave where the amplitude always remains zero as it doesn't depend on time.
Hence, the wave is stationary.
Looking at the equation we see that the wave has an $x$ component. Which means that the wave is propagating in $x$-direction. The amplitude of the wave is $\vec{j}$, which means that the vibration of the wave is in y-direction. We, conclude that the direction of vibration is perpendicular to the direction of propagation. This is the characteristic of a transverse wave, and hence the wave is transverse.
The correct answer is option $(D)$ i.e the wave is transverse and stationary.
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