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Laplace Correction

The absolute ...

Question

The absolute temperature of air in a region linearly increases from $T_{1}$ , to $T_{2}$ , in a space of width d. Find the time taken by a sound wave to go through the region in terms of $T_{1} T_{2}$ , d and the speed u of sound at 273 K. Evaluate this time for $T_{1} \approx 280 K, T_{3} = 310 K$ , d = 33 m and $v = 330 m s^{- 1}$ .

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Solution

The variation of temperature is given by $T = T_{1} + \frac{(T_{2} - T_{1})}{d} x . . . . (i)$

We know that, $V \propto \sqrt{T}$

$\Rightarrow \frac{V T}{V} = \sqrt{(\frac{T}{273})}$

$\Rightarrow V T = V \sqrt{(\frac{T}{273})}$

$\Rightarrow d t = \frac{d x}{V T} = \frac{d u}{V} \times (\frac{273}{T})$

$\Rightarrow t = \frac{\sqrt{273}}{v} \int_{0}^{d} \frac{d x}{[T_{1} + \frac{(T_{2} - T_{1})}{d} x]}$

$= \frac{\sqrt{273}}{V} \times \frac{2 d}{T_{2} - T_{1}} (\sqrt{T_{2}} - \sqrt{T_{1}})$

$= (\frac{2 d}{V}) (\frac{\sqrt{273}}{T_{2} - T_{1}}) \sqrt{T_{2}} - \sqrt{T_{1}}$

$T = \frac{2 d}{V} \frac{\sqrt{273}}{\sqrt{T_{2}} + \sqrt{T_{1}}}$

[Since $T_{1} - T_{2} = (\sqrt{T_{2}} + \sqrt{T_{1}}) (\sqrt{T_{2}} - \sqrt{T_{1}})]$

Putting the given value we get $\frac{2 \times 33}{330} \frac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96 m s$

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The absolute temperature of air in a region linearly increases from T1, to T2, in a space of width d. Find the time taken by a sound wave to go through the region in terms of T1T2, d and the speed u of sound at 273 K. Evaluate this time for T1≈280K,T3=310K, d = 33 m and v=330ms−1.