In an open organ pipe, v_3 and v_6 are the 3^ | vedclass

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(B) For an open organ pipe, the frequency of the $n^{	ext{th}}$ harmonic is given by $f_n = n \left( \frac{v}{2L} ight)$, where $v$ is the speed of

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$225$

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(B) For an open organ pipe, the frequency of the $n^{	ext{th}}$ harmonic is given by $f_n = n \left( \frac{v}{2L} ight)$, where $v$ is the speed of sound and $L$ is the length of the pipe.Given $v_3 = 3 \left( \frac{v}{2L} ight)$ and $v_6 = 6 \left( \frac{v}{2L} ight)$.According to the problem, $v_6 - v_3 = 2200 	ext{ Hz}$.Substituting the expressions: $6 \left( \frac{v}{2L} ight) - 3 \left( \frac{v}{2L} ight) = 2200$.$3 \left( \frac{v}{2L} ight) = 2200$.Given $v = 330 	ext{ m/s}$, we have $3 \left( \frac{330}{2L} ight) = 2200$.$990 / (2L) = 2200$.$2L = 990 / 2200 = 0.45 	ext{ m}$.$L = 0.225 	ext{ m} = 225 	ext{ mm}$.

In an open organ pipe, $v_3$ and $v_6$ are the $3^{\text{rd}}$ and $6^{\text{th}}$ harmonic frequencies, respectively. If $v_6 - v_3 = 2200 \text{ Hz}$, then the length of the pipe is . . . . . . mm. (Take the velocity of sound in air as $330 \text{ m/s}$.)

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