The stationary wave $y = 2a \sin kx \cos \omega t$ in a clo…

Physics Wave and Sound Organ Pipe (Vibration of Air Column) Single Correct MCQ

Published on: September 12, 2026

The stationary wave $y = 2a \sin kx \cos \omega t$ in a closed organ pipe is the result of the superposition of $y = a \sin(\omega t - kx)$ and

$y = -a \cos(\omega t + kx)$

$y = -a \sin(\omega t + kx)$

$y = a \sin(\omega t + kx)$

$y = a \cos(\omega t + kx)$

Text Solution

Verified by Experts

The correct answer is:

In closed organ pipe. If $y_{\text{incident}} = a \sin(\omega t - kx)$

then $y_{\text{reflected}} = a \sin(\omega t + kx + \pi) = -a \sin(\omega t + kx)$

Superimposition of these two waves give the required stationary wave.

Previous Question Next Question

Get practice questions, solutions, and test series in one place.

Share your experience with this question and solution.

Send your comment, doubt, correction, or feedback to admin.

What students say about this solution

No reviews yet. Be the first to write a review.