A progressive wave travelling along the positive x-direction is represented by y(x,t) = Asin(kx – ω t + φ). Its snapshot at t = 0 is given in the figure. For this wave, the phase φ is: (a) π/2 (b) π (c) 0 (d) -π/2

Answer: (b). At t = 0: y = Asin (kx + φ). From the graph, y = 0 at x = 0 and y becomes negative for small x > 0, so y = -Asin kx. -Asin kx = Asin (kx + π) ⇒ φ = π So the answer is (b) π.

A progressive wave travelling along the positive x-direction is… | JEE Main 2019 (12 Apr, Shift 1)

Q85JEE Main 2019 · 12 APR · SHIFT 1Difficult

Progressive Wave Equation

A progressive wave travelling along the positive x-direction is represented by y(x,t) = Asin(kx – $\omega$ t + $\phi$ ). Its snapshot at t = 0 is given in the figure.

For this wave, the phase $\phi$ is:

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Answer: (b)

At $t = 0$ : $y = A\sin(kx + \phi)$ .
From the graph, $y = 0$ at $x = 0$ and $y$ becomes negative for small $x > 0$ , so $y = -A\sin kx$ .
$-A\sin kx = A\sin(kx + \pi) \Rightarrow \phi = \pi$
So the answer is (b) $\pi$ .

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Progressive Wave Equation question from JEE Main 2019 (12 Apr, Shift 1)

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