MCQ
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
    $y = - a\cos (\omega \,t + kx)$
  • $y = - a\sin (\omega \,t + kx)$
  • C
    $y = a\sin (\omega \,t + kx)$
  • D
    $y = a\cos (\omega \,t + kx)$

Answer

Correct option: B.
$y = - a\sin (\omega \,t + kx)$
b
(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.

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