An electronically driven loudspeaker is positioned near the open end of a resonance column apparatus. The air column in the tube has a length of 80 cm . The loudspeaker's frequency can be adjusted between 20 Hz and 2 kHz . Determine the frequencies at which the column will resonate. The speed of sound in air is 320 m s ⁻¹ .

Step-by-step solution

A resonance column apparatus acts as a pipe closed at one end (by the water surface) and open at the other. The resonant frequencies for a closed pipe are given by: f = (2n + 1)v 4L where v is the speed of sound, L is the length of the air column, and n = 0, 1, 2, Given v = 320 m s ⁻¹ and L = 80 cm = 0.8 m , we get: f = (2n + 1) 320 4 0.8 = 100(2n + 1) Hz The loudspeaker's frequency range is 20 Hz to 2000 Hz . Therefore, 100(2n + 1) 2000 2n + 1 20 n 9.5 Since n must be an integer, the possible values are n = 0, 1,