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

Step-by-step solution

The fundamental frequency of a stretched wire is given by f_ wire = 1 2L T / , and that of a closed pipe is f_ pipe,0 = v / (4L) . Initially, resonance occurs between the wire's fundamental and the pipe's fundamental: 1 2L T₁ / = v / (4L) . Simplifying yields T₁ / = v / 2 , so T₁ = v^2 / 4 . After increasing tension by 8 N, T₂ = T₁ + 8 , resonance occurs with the pipe's first overtone at 3v / (4L) : 1 2L T₂ / = 3v / (4L) . This simplifies to T₂ / = 3v / 2 , giving T₂ = 9 v^2 / 4 . Dividing the expressions for T₂ an