Question

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 T1, T2, d and the speed v of sound at 273 K. Evaluate this time for T1 = 280 K, T2 = 310 K, d = 33 m and v = 330 m s−1.

Answer 1

The Time will be 96 ms

Explanation:

The variation of temperature is given by-

T= $T_1 + \frac {(T_2 - T_1)}{d}$"""(1)

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

$\frac{V_T}{V} = \sqrt \frac {T}{273}$

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

dt = $\frac {dx}{V_T}$ = $\frac {du}{V} \times\sqrt \frac {273}{T}$

t = $\frac {273}{V}\int_{0}^{d}\frac {dx}{[T_1 +(T_2 -T_1)x/d]^{1/2}}$

$\frac {\sqrt273}{V} \times \frac {2d}{T_2 - T_1}[T_1 + \frac {T_2 - T_1}{d}x]$

= $(\frac {2d}{V}(\frac {\sqrt 273}{\sqrt T_2 - \sqrt T_1})) \times (\sqrt T_2 - \sqrt T_1)$

T = $\frac {2d}{V} \frac {\sqrt 273}{\sqrt T_2 + \sqrt T_1}$

putting the given value we will get -

= $\frac {2 \times 33}{330}$

= $\frac {\sqrt 273}{\sqrt 280 + \sqrt 310}$

= 96 ms