The two nearest harmonics of a tube closed at one end and open at the other end are $220 ext{ Hz}$ and $260

Question

The two nearest harmonics of a tube closed at one end and open at the other end are $220 	ext{ Hz}$ and $260 	ext{ Hz}$. What is the fundamental frequency of the system?

Accepted Answer

1. **Identify the Properties of the Tube:** For a tube closed at one end and open at the other (closed organ pipe), only odd harmonics are present . The frequencies of these harmonics are given by $f_n = n f_1$, where $n = 1, 3, 5, \dots$ and $f_1$ is the fundamental frequency.
2. **Set up the Equation:** Let the two nearest (consecutive) harmonics be $n f_1$ and $(n+2) f_1$.
   Given:
   $$n f_1 = 220 	ext{ Hz}$$
   $$(n+2) f_1 = 260 	ext{ Hz}$$
3. **Calculate the Fundamental Frequency:** Subtracting the first equation from the second gives the difference between consecutive harmonics:
   $$(n+2) f_1 - n f_1 = 260 - 220$$
   $$2 f_1 = 40 \implies f_1 = 20 	ext{ Hz}$$
Therefore, the fundamental frequency of the system is $20 	ext{ Hz}$.

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