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Shashank Borah
Asked: 2022-10-29T16:37:43+05:30 In:

Two narrow organ pipes, one open (length `l_(1)`) and the other cloed (length `l_(2)`) are sounded in their respective fundamental modes. The beat frequency heard is `5 Hz`. If now the pipes are sounded in their first overtones, then also the beat frequency heard is `5 Hz`. Then:
A. `(l_(1))/(l_(2)) = (1)/(2)`
B. `(l_(1))/(l_(2)) = (1)/(1)`
C. `(l_(1))/(l_(2)) = (3)/(2)`
D. `(l_(1))/(l_(2)) = (2)/(3)`

Two narrow organ pipes, one open (length `l_(1)`) and the other cloed (length `l_(2)`) are sounded in their respective fundamental modes. The beat frequency heard is `5 Hz`. If now the pipes are sounded in their first overtones, then also the beat frequency heard is `5 Hz`. Then:
A. `(l_(1))/(l_(2)) = (1)/(2)`
B. `(l_(1))/(l_(2)) = (1)/(1)`
C. `(l_(1))/(l_(2)) = (3)/(2)`
D. `(l_(1))/(l_(2)) = (2)/(3)`
Parabola
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1 Answer

  1. Correct Answer – B::C
    fundamental frequency of
    open pipe, `n_(0) = (V)/(2l_(1))` , closed pipe, `n_(C) = (V)/(4l_(2))`
    `(V)/(2l_(1)) – (V)/(4l_(2)) = 5`
    For first overtone , `n_(0) = (V)/(l_(1)) , n_(0) = (3V)/(4l_(2))`
    `(V)/(l_(1)) – (3V)/(4l_(2)) = 5` …….(2)
    on solving `(1)` and `(2) l_(1) = l_(2) rArr (l_(1))/(l_(2)) = (1)/(1)`

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