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

Open in App

Solution

At any position $x$ , the temperature variation can be given by
$T = T_{1} + \frac{T_{2} - T_{1}}{d} \times x$

also, $v \propto \sqrt{T}$

So, $\frac{v_{T}}{v} = \sqrt{\frac{T}{273}}$

or $v_{T} = v \sqrt{\frac{T}{273}}$

as we know that, $v_{T} = \frac{d x}{d t}$

$d t = \frac{d x}{v t} = \frac{d x}{v} \sqrt{\frac{273}{T}}$

$t = \frac{273}{v} \int_{0}^{d} \frac{d x}{{[T_{1} + \frac{[T_{2} - T_{1}] x}{d}]}^{\frac{1}{2}}} [T = T_{1} + \frac{[T_{2} - T_{1}] x}{d}] t = \frac{2 d}{v} [\frac{\sqrt{273}}{\sqrt{T_{2}} + \sqrt{T_{1}}}] t = \frac{2 \times 33}{330} \times \frac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96 \times 10^{- 3} s e c$

Suggest Corrections

Similar questions

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 $v = 330 m s^{- 1}$ .

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 v of sound at 273 K.