A hollow pipe of length `0.8m` 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

Options

A. 5 grams
B. 10 grams
C. 20 grams
D. 40 grams

Answer

B

Explanation

Fundamental frequency of the pipe (closed organ pipe)
`f_(0) = (upsilon_("sound"))/(4l_("pipe")) = (320)/(4 xx 0.8) = 100` Hz
If `upsilon` is speed of wave on string and l is length of the string then frequency of the second harmonics can be written as follows :
`f = 2 (upsilon)/(2l) = (upsilon)/(0.5) = 2upsilon`
Fundamental frequency of pipe and second harmonic frequency of string are same.
`rArr 2upsilon = 100 rArr upsilon = 50` m/s
Speed of wave on string can be written as follows :
`upsilon = sqrt((F)/(mu)) rArr mu = (F)/(upsilon^(2)) = (50)/((50)^(2)) = 0.02` kg/m
Let m is the mass of string and length is given to be 0.5 m
`rArr" "mu = (m)/(0.5) = 0.02 rArr m = 0.01 kg = 10 g`
Hence option (b) is correct.