To solve the problem of finding the number of possible natural oscillations of an air column in a pipe closed at one end, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Length of the pipe, \( L = 85 \, \text{cm} = 0.85 \, \text{m} \)
- Velocity of sound, \( v = 340 \, \text{m/s} \)
- Maximum frequency, \( f_{\text{max}} = 1250 \, \text{Hz} \)
2. **Determine the Fundamental Frequency:**
- For a pipe closed at one end, the fundamental frequency \( f_1 \) is given by the formula:
\[
f_1 = \frac{v}{4L}
\]
- Substituting the values:
\[
f_1 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \, \text{Hz}
\]
3. **Determine the Frequencies of Higher Harmonics:**
- The frequencies of the harmonics for a pipe closed at one end are given by:
\[
f_n = (2n - 1) \frac{v}{4L}
\]
- Where \( n \) is the harmonic number (1, 2, 3,...).
4. **Calculate the Frequencies:**
- For \( n = 1 \):
\[
f_1 = 100 \, \text{Hz}
\]
- For \( n = 2 \):
\[
f_2 = 3 \times 100 = 300 \, \text{Hz}
\]
- For \( n = 3 \):
\[
f_3 = 5 \times 100 = 500 \, \text{Hz}
\]
- For \( n = 4 \):
\[
f_4 = 7 \times 100 = 700 \, \text{Hz}
\]
- For \( n = 5 \):
\[
f_5 = 9 \times 100 = 900 \, \text{Hz}
\]
- For \( n = 6 \):
\[
f_6 = 11 \times 100 = 1100 \, \text{Hz}
\]
- For \( n = 7 \):
\[
f_7 = 13 \times 100 = 1300 \, \text{Hz}
\]
5. **Count the Frequencies Below 1250 Hz:**
- The frequencies calculated are:
- \( f_1 = 100 \, \text{Hz} \)
- \( f_2 = 300 \, \text{Hz} \)
- \( f_3 = 500 \, \text{Hz} \)
- \( f_4 = 700 \, \text{Hz} \)
- \( f_5 = 900 \, \text{Hz} \)
- \( f_6 = 1100 \, \text{Hz} \)
- \( f_7 = 1300 \, \text{Hz} \) (not counted as it is above 1250 Hz)
- The valid frequencies below 1250 Hz are \( f_1, f_2, f_3, f_4, f_5, \) and \( f_6 \).
6. **Conclusion:**
- The total number of possible natural oscillations (harmonics) below 1250 Hz is 6.
### Final Answer:
The number of possible natural oscillations of the air column in the pipe is **6**.
