For a wave described by $y=A\sin(\omega t+kx)$, the direction of propa…

Question

For a wave described by $y=A\sin(\omega t+kx)$, the direction of propagation is

Accepted Answer

B. along the $-x$-direction

Keeping phase constant, $\omega t+kx=	ext{constant}$ gives $x=	ext{constant}-\frac{\omega}{k}t$.
As time increases, $x$ decreases, so the wave moves in the negative $x$-direction.
**Hence B is correct.**

Answer

along the $+x$-direction

Answer

perpendicular to the $x$-axis

Answer

indeterminate from the equation

For a wave described by $y=A\sin(\omega t+kx)$, the direction of propagation is

Keep practising Unit 2B: Friction, Gravitational and Normal Forces