The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by 8 N , it is in resonance with the first overtone of the closed tube. The initial tension in the wire is

Options

A. 4 N
B. 8 N
C. 16 N
D. 1 N

Answer

D

Explanation

`n = 1/(2L) sqrt(T/m) `
and the fundamental frequency of a closed pipe isi
` n = v/(4L) `
then ` n = v/(4L) = 1/(2L) sqrt(T/m) ` …(1)
and in the second case, the first overtone of the closed pipe ` = 3n = (3v)/(4L) `
and for the string ` N = 1/(2L) sqrt((T + 8)/m) `
` :. (3v)/(4L) = 1/(2L) sqrt((T + 8)/m) ` ....(2)
Dividing (2) by (1) , we get
`3 = sqrt((T + 8)/T) :. 9 = (T + 8)/T`
` :. 9T = T + 8 :. 8T = 8:. T = 1 N `