The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe $L$ me

Question

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe $L$ metre long. The length of the open pipe will be

Accepted Answer

1. **Determine Frequency of Open Pipe:** For an open organ pipe, the resonant frequencies are all harmonics, given by $f = n\frac{v}{2L_o}$ where $n = 1, 2, 3, \dots$. The fundamental is $n=1$, the first overtone is $n=2$, and the second overtone is $n=3$. So, the frequency of the second overtone is $f_o = \frac{3v}{2L_o}$ .
2. **Determine Frequency of Closed Pipe:** For a closed organ pipe, only odd harmonics are present, given by $f = (2n+1)\frac{v}{4L_c}$ where $n=0, 1, 2, \dots$ is the overtone number. For the first overtone ($n=1$), the frequency is $f_c = \frac{3v}{4L_c}$ .
3. **Equate Frequencies:** The length of the closed pipe is given as $L_c = L$. Since the two overtones have the same frequency ($f_o = f_c$):
   $$\frac{3v}{2L_o} = \frac{3v}{4L}$$
4. **Calculate Length:** Solving for the length of the open pipe ($L_o$):
   $$\frac{1}{2L_o} = \frac{1}{4L} \implies 2L_o = 4L \implies L_o = 2L$$

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