In a pipe open at two ends a diatomic gas is oscillating in 1st harmonic. The length of the tube is π  m and the maximum pressure variation is  1.4Pa. The  maximum displacement of gas particles which are a distance of π3m  from one end is  x×10−6m. The value of x  is (Take 1atm= 105 Pa)

Open end pipe has pressure Node and displacement  Antinode, so, let open end be  x=0, then particle  displacement amplitude will bes=s0coskx=ΔPmBkcoskx here , k=2πλ For sound  wave in a gas,  B=γP (adiabatic process)So,  s=1.4Pa(γ×105Pa)×2πλcos(2πλx)Diatomic gas,  γ=751st harmonic,  λ2=L=π or  λ=2πPut x=π3m We get,  s=5×10−6m