Question:
A progressive wave travelling along the positive x-direction is represented by \(y(x,t)=A\sin(kx-\omega t+\phi)\). Its snapshot at t = 0 is given in the figure.
For this wave, the phase \(\phi\) is :
A progressive wave travelling along the positive x-direction is represented by \(y(x,t)=A\sin(kx-\omega t+\phi)\). Its snapshot at t = 0 is given in the figure.
For this wave, the phase \(\phi\) is :
For this wave, the phase \(\phi\) is :
Show Hint
At \(t=0\), \(y=A\sin(kx+\phi)\); read from the graph that \(y(0)=0\) with a negative slope at x = 0.
Updated On: Nov 5, 2026
- \(\frac{\pi}{2}\)
- \(\pi\)
- 0
- \(-\frac{\pi}{2}\)
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The Correct Option is B
Solution and Explanation
At \(t=0\): \(y(x)=A\sin(kx+\phi)\).
From the figure, the curve passes through the origin (\(y(0)=0\)) while going downward (a crest lies just to the left, a trough just to the right), so \(\sin\phi=0\) and the slope is negative.
Slope at x = 0: \(\frac{dy}{dx}=Ak\cos\phi\lt0\Rightarrow\cos\phi=-1\).
Hence \(\phi=\pi\) (then \(y=-A\sin kx\), a trough just right of the origin, matching the figure).
Answer: (2) \(\pi\)
From the figure, the curve passes through the origin (\(y(0)=0\)) while going downward (a crest lies just to the left, a trough just to the right), so \(\sin\phi=0\) and the slope is negative.
Slope at x = 0: \(\frac{dy}{dx}=Ak\cos\phi\lt0\Rightarrow\cos\phi=-1\).
Hence \(\phi=\pi\) (then \(y=-A\sin kx\), a trough just right of the origin, matching the figure).
Answer: (2) \(\pi\)
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