Q. 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
A) $y = - a\cos{(\omega t + kx)}$
B) $y = - a\sin{(\omega t + kx)}$
C) $y = a\sin{(\omega t + kx)}$
D) $y = a\cos{(\omega t + kx)}$
Explanation
(b) In closed organ pipe. If $y_{incident} = a\sin{(\omega t - kx)}$ then $y_{reflected} = a\sin{(\omega t + kx + \pi)} = - a\sin{(\omega t + kx)}$
Superimposition of these two waves give the required stationary wave
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.