Question

A tube 1.0 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.3 m long and a mass of 0.01 kg . It is held fixed at both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration at its fundamental frequency by resonance. Find (a) the frequency of oscillation of the air column and (b) the tension in the wire. Speed of sound in air = 330 m//s .

Answer 1

(a) The frequency of oscillation of the air column is :

• Given : Air column length = 1 m, wire length = 0.3 m and mass = 0.01 kg

• Also given that air column is vibrating in its first overtone. So, frequency will be

F = 3v / 4l

Where v = velocity of sound in air and l = length of air column

F = 3×330 / 4×1 = 990 / 4

• F = 247.5 Hz

(b) The tension in the wire

For string, its fundamental frequency is given as

F = v / 2l = ( 1/2l )×(√T/µ)

247.5 Hz = ( 1/2×0.3 )×(√T/µ)

µ = 1/30

Substitute all values in above relation,

• We get T = 735 N