A cylindrical metal tube is open at both ends and has a length of 50 cm . Determine the frequencies between 1000 Hz and 2000 Hz at which the air column inside the tube can resonate. The speed of sound in air is 340 m s ⁻¹ .

Step-by-step solution

For a tube open at both ends, the resonant frequencies are given by f_n = nv 2L , where n = 1, 2, 3, Given L = 50 cm = 0.5 m and v = 340 m s ⁻¹ , the fundamental frequency is: f₁ = 340 2 0.5 = 340 Hz The possible resonant frequencies are integral multiples of the fundamental frequency: f_n = n 340 Hz Calculating the frequencies for different values of n : For n = 3 , f₃ = 3 340 = 1020 Hz For n = 4 , f₄ = 4 340 = 1360 Hz For n = 5 , f₅ = 5 340 = 1700 Hz For n = 6 , f₆ = 6 340 = 2040 Hz The frequencies that lie stric