Question

A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ-7.20 g/m and is placed under a tension of 160.00 N. The string is placed next to a tube, open at both ends, of length L. The string is plucked and the tube resonates at the n-1 mode. The speed of sound is 343 m/s. What is the length of the tube? Select the correct answer O 2.3 m O 4.6 m O 7.2 m

Answer 1

First we will determine speed of wave in stretched stones U = squareroot T/mu Given: T rightarrow integral string = 160.00 n Linear mass density mu = 7.20 g/m = 7.2 times 10^-3 kg/m The speed U = squareroot 160.00 N/7.2 times 10^-3 kg/m = 149.07 m/km Now Frequency of wave produce d, n string f_string = U/2 L Given: length of string: L = alpha m f_string = 149.07/2 times 2m = 37.26 H_z In resonance condition, this frequency is to frequency produced in the to be open at both end \r\n f_tube = n^U sound/2 L_tube Given: - n = 1 U_sound = 343 m/s f_tube = f_string = 37.26 H_z and L_tube = 2} The using these Values, we get 37.26 H_z = 343 m/s/2 times L_tube L_tube = 4.60 m