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 π$$3m$$ from one end is x×$$10$$−$$6m$$. The value of x is (Take$$ $$1atm=$$ $$105$$ $$Pa)

Question

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 π$$3m$$  from one end is  x×$$10$$−$$6m$$. The value of x  is (Take$$ $$1atm=$$ $$105$$ $$Pa)

Infinity Learn · Accepted Answer

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 , $$k=2$$πλ For sound  wave in a gas,  $$B=$$γP (adiabatic$$ $$process)So$$,  $$s=1.4Pa($$γ×$$105Pa)×$$2$$πλ$$cos$$$$(2$$πλ$$x)Diatomic$$ gas,  γ$$=751st$$ harmonic,  λ$$2=L=$$π or  λ$$=2$$$$π$$Put $$x=$$π$$3m$$ We get,  $$s=5$$×$$10$$−$$6m$$

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