Question

The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (velocity of sound = 340 ms$^{-1}$)

• A. 6
• B. 8
• C. 4
• D. 5

Hint:

In a pipe closed at one end, only odd harmonics are allowed. Use the formula for the fundamental frequency and multiply by odd integers to find the other possible natural frequencies.

Answer 1

The Correct Option is A

The natural frequencies of a pipe closed at one end are given by the formula: \[ f_n = \frac{n v}{4 L} \] Where: 
- \( f_n \) is the frequency of the \( n \)-th harmonic, 
- \( v = 340 \, \text{ms}^{-1} \) is the velocity of sound, 
- \( L = 85 \, \text{cm} = 0.85 \, \text{m} \) is the length of the pipe. The frequencies for a closed-end pipe are given by the odd multiples of the fundamental frequency, so: \[ f_1 = \frac{v}{4 L}, \quad f_3 = 3 \cdot \frac{v}{4 L}, \quad f_5 = 5 \cdot \frac{v}{4 L}, \quad \dots \] We need to find the maximum \( n \) such that \( f_n<1250 \, \text{Hz} \). Calculating \( f_1 \): \[ f_1 = \frac{340}{4 \times 0.85} = 100 \, \text{Hz} \] Now we calculate the higher harmonics: \[ f_3 = 3 \times 100 = 300 \, \text{Hz}, \quad f_5 = 5 \times 100 = 500 \, \text{Hz}, \quad f_7 = 7 \times 100 = 700 \, \text{Hz}, \quad f_9 = 9 \times 100 = 900 \, \text{Hz}, \quad f_{11} = 11 \times 100 = 1100 \, \text{Hz}, \quad f_{13} = 13 \times 100 = 1300 \, \text{Hz} \] Since the frequency \( f_{13} = 1300 \, \text{Hz} \) exceeds 1250 Hz, we consider harmonics up to \( f_{11} \), which gives 6 possible oscillations (i.e., \( f_1, f_3, f_5, f_7, f_9, f_{11} \)). 
Thus, the number of possible natural oscillations is \( 6 \).