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For a wave described by $y=A\sin(\omega t+kx)$, the direction of propagation is
Aalong the $+x$-direction
Balong the $-x$-direction
Cperpendicular to the $x$-axis
Dindeterminate from the equation
Answer & Solution
Correct answer: B. along the $-x$-direction
Keeping phase constant, $\omega t+kx=\text{constant}$ gives $x=\text{constant}-\frac{\omega}{k}t$. As time increases, $x$ decreases, so the wave moves in the negative $x$-direction. Hence B is correct.
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