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 be
$s = s_{0} \cos k x = \frac{Δ P_{m}}{B k} \cos k x h e r e, k = \frac{2 π}{λ}$
For sound wave in a gas, $B = γ P$ (adiabatic process)
So, $s = \frac{1.4 P a}{(γ \times 10^{5} P a) \times \frac{2 π}{λ}} \cos (\frac{2 π}{λ} x)$
Diatomic gas, $γ = \frac{7}{5}$
1st harmonic, $\frac{λ}{2} = L = π$ or $λ = 2 π$
$P u t x = \frac{π}{3} m$
We get, $s = 5 \times 10^{- 6} m$

Stationary waves

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Stationary waves

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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)

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