wave_mechanics organ_pipe_vibration_of_air_column Question 35
Question: 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
[Roorkee 1994]
Options:
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) $
Show Answer
Answer:
Correct Answer: B
Solution:
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.
BETA
