In an open organ pipe $
u_3$ and $
u_6$ are $3^{ ext {rd }}$ and $6^{ ext {th }}$ harmonic frequ

Question

In an open organ pipe $
u_3$ and $
u_6$ are $3^{	ext {rd }}$ and $6^{	ext {th }}$ harmonic frequencies, respectively. If $
u_6-
u_3=2200 \mathrm{~Hz}$ then length of the pipe is $\_\_\_\_$ mm . (Take velocity of sound in air is $330 \mathrm{~m} / \mathrm{s}$.)

Professor PC Thomas Classes · Answer

For an open organ pipe, the frequency of the $n^{	ext{th}}$ harmonic is given by $
u_n = \frac{nv}{2L}$. Thus, the 3rd and 6th harmonics are $
u_3 = \frac{3v}{2L}$ and $
u_6 = \frac{6v}{2L}$. The difference is $
u_6 - 
u_3 = \frac{6v}{2L} - \frac{3v}{2L} = \frac{3v}{2L}$. We are given that this difference is $2200 	ext{ Hz}$. So, $\frac{3 	imes 330}{2L} = 2200 \Rightarrow L = \frac{990}{4400} 	ext{ m} = 0.225 	ext{ m} = 225 	ext{ mm}$.