A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in fourth harmonic, and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 m/s, the mass of string is

Step-by-step solution

The fundamental frequency of a closed organ pipe of length L_p is given by f_p = v 4 L_p . Substituting the given values, v = 320 m/s and L_p = 0.8 m: f_p = 320 4 0.8 = 320 3.2 = 100 Hz The frequency of the fourth harmonic of a string of length L_s fixed at both ends is given by f_s = 4 2 L_s T , where T is the tension and is the linear mass density. Given that the string resonates with the fundamental frequency of the pipe, f_s = f_p = 100 Hz. Substituting L_s = 0.5 m and T = 50 N: 100 = 4 2 0.5 50 100 = 4 50 25 =