A wire of length 'L' and linear density 'm' is stretched between two rigid supports with tension 'T'. It is observed that wire resonates in the P^ th harmonic at a frequency of 320 Hz and resonates again at next higher frequency of 400 Hz in two successive modes. The value of 'P' is

Step-by-step solution

The frequency of the n^ th harmonic of a stretched wire fixed at both ends is given by f_n = n f₁ , where f₁ is the fundamental frequency. For the P^ th harmonic, the frequency is given as: f_P = P f₁ = 320 The next higher resonant frequency corresponds to the (P+1)^ th harmonic: f_ P+1 = (P+1) f₁ = 400 Dividing the two equations, we get: P+1 P = 400 320 P+1 P = 5 4 5P = 4P + 4 P = 4 Answer: 4