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 its second 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 ms⁻¹, the mass of the string is:

Frequency of the closed organ pipe is, f = v/4L = 320/(4*0.8) = 100 Hz
The frequency of the second harmonic of the string is given by:
f = 2(1/2l)√(T/μ) = √(T/μ)/l
where T is the tension, μ is the linear mass density, and l is the length of the string.
Since the string resonates with the fundamental frequency of the pipe, we have:
100 Hz = √(50/μ)/(0.5)
100 * 0.5 = √(50/μ)
50 = √(50/μ)
Squaring both sides:
2500 = 50/μ
μ = 50/2500 = 0.02 kg/m
The mass of the string is given by:
m = μ * l = 0.02 kg/m * 0.5 m = 0.01 kg = 10 g