Question

A glass tube (open at both ends) of length L is positioned near an audio speaker of frequency f = 600 Hz. For what values of L will the tube resonate with the speaker? (Assume that the speed of sound in air is 343 m/s.)
m (lowest possible value)
m (second lowest possible value)
m (third lowest possible value)

Answer 1

When glass tube open at both ends, then the value of L will be given as :

using an equation,     f_n = n v / 2 L

For n = 1, we have

f₁ = (1) v / 2 L₁

v = speed of sound in air = 343 m/s

f = frequency of audio-speaker = 600 Hz

then, we get  

f₁ = (1) (343 m/s) / 2 L₁

L₁ = (343 m/s) / 2 (600 s^-1)

L₁ = 0.285 m                                     (lowest possible value)

For n = 2, we have

f₂ = (2) (343 m/s) / 2 L₂

L₂ = (686 m/s) / 2 (600 s^-1)

L₂ = 0.571 m                                        (second lowest possible value)

For n = 3, we have

f₃ = (3) (343 m/s) / 2 L₃

L₃ = (1029 m/s) / 2 (600 s^-1)

L₃ = 0.857 m                                               (third lowest possible value)