Sound WavesHard
Question
In an open organ pipe $v_{3}$ and $v_{6}$ are $3^{\text{rd~}}$ and $6^{\text{th~}}$ harmonic frequencies, respectively.
If $v_{6} - v_{3} = 2200\text{ }Hz$ then length of the pipe is
$\_\_\_\_$ mm.
(Take velocity of sound in air is $330\text{ }m/s$.)
Options
A.275
B.225
C.200
D.250
Solution
$f = n\left( \frac{V_{0}}{2L} \right)$
$${\frac{6{\text{ }V}_{0}}{2\text{ }L} - \frac{3{\text{ }V}_{0}}{2\text{ }L} = 2200 }{\frac{3{\text{ }V}_{0}}{2\text{ }L} = 2200 }{\frac{3 \times 330}{2 \times L} = 2200 }{L = \frac{3 \times 330}{2 \times 2200} }{L = 0.225\text{ }m }{L = 225\text{ }mm }$$
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