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

Step-by-step solution

For a closed organ pipe, the fundamental frequency is given f= v 4 L , where v=320 ~m / s is the velocity of sound in the medium of organ pipe and L=0.8 ~m being the length of pipe. Now, we are given second harmonic frequency of wire is equal and in resonance with fundamental frequency of pipe, thus; f= v 4 L = 1 l T Putting, T=50 ~N and l=0.8 ~m aligned & 320 4(0.8) = 1 0.5 50 & 50 = 1 50 aligned The length of string =0.5 ~m Thus, mass of string =0.02 0.5=0.01 ~kg =10 ~g