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

Question

A steel wire with mass per unit length \( 7.0 \times 10^{-3} \, \text{kg/m} \) is under a tension of \( 70 \, \text{N} \). The speed of transverse waves in the wire will be:

Answer 1

To solve the problem of finding the fundamental frequency of a tube closed at one end and open at the other, we need to understand the harmonic frequencies of such a system.

For a tube closed at one end and open at the other, the frequencies of the harmonics are given by:

f_n = (2n - 1)\frac{v}{4L}

where n is the harmonic number, v is the speed of sound in the air, and L is the length of the tube. The harmonics occur at odd integer multiples of the fundamental frequency.

Given that the two nearest harmonics are 220 Hz and 260 Hz, these represent successive odd harmonics for a closed tube. Let's denote the odd harmonics as f_{n} and f_{n+1}. Hence, we have:

f_{n} = (2n - 1)f_1 = 220 \, \text{Hz}

f_{n+1} = (2n + 1)f_1 = 260 \, \text{Hz}

Subtracting these equations gives:

f_{n+1} - f_{n} = [2n + 1 - (2n - 1)]f_1 = 260 - 220 \, \text{Hz}

2f_1 = 40 \, \text{Hz}

Hence, solving for the fundamental frequency f_1:

f_1 = \frac{40}{2} = 20 \, \text{Hz}

Therefore, the fundamental frequency of the system is 20 Hz.

Given the options, the correct answer is 20 Hz.

Answer 2