Question

The consecutive overtones of an air column closed at one end are 405 Hz and 675 Hz respectively. Find the fundamental frequency of a similar air column but open at both ends.

Answer 1

For the air column closed at one end, let 

L = the length of the air column, 

n_c = the fundamental frequency,

n_q, n_{q + 1} = the frequencies of the qth and (q + 1)th overtones, where q = 1, 2, 3, … Since only odd harmonics are present as overtones, n_q = (2q + 1)n_c and n_{q+ 1} =  [2(q + 1) + n]n_c

= (2q + 3)n_c

Data : n_q = 405 Hz, n_{q + 1} = 675 Hz

Solving for q, q = 1

Therefore, the two given frequencies correspond to the first and second overtones, i.e., the third and fifth harmonics.

∴ 3n_c = 405 Hz 

∴ n_c = 135 Hz

This gives the fundamental frequency of the air column closed at one end.

The fundamental frequency (n_o) of an air column of same length but open at both ends is double that of the air column closed at one end (ignoring the end correction).

∴ n_o = 2n_c = 2 × 135 = 270 Hz

This gives the fundamental frequency of a similar air column but open at both ends.

Data : n = 480 Hz, l₁ = 16.8 cm, l₂ = 51.8 cm 

(1) The speed of sound in air is v = 2n (l₂ – l₁) = 2 × 480 × (51.8 – 16.8) = 33600 cm/s = 336 m/s

(2) Let λ be the wavelength of sound waves and e be the end correction. For the first resonance, l₁ + e = \(\frac λ4\) … (1)

For the second resonance, l₂ + e = \(\frac {3λ}4\)… (2)

From Eq. (1), λ = 4(l₁ + e).

Substituting this value in Eq. (2), we get,