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

Given:
The absolute temperature of air in a region increases linearly from T₁ to T₂  in a space of width d.
The speed of sound  at 273 K is v.
v_T is the velocity of the sound at temperature T.
Let us find the temperature variation at a distance x in the region.
Temperature variation is given by:

\[T =  T_1  + \frac{\left( T_2 - T_1 \right)}{d}x        \] 

\[  v \propto \sqrt{T}\] 

\[ \Rightarrow \frac{v_T}{v} = \sqrt{\left( \frac{T}{273} \right)}\] 

\[ \Rightarrow    v_T  = v\sqrt{\left( \frac{T}{273} \right)}\] 

\[ \Rightarrow dt = \frac{dx}{v_T} = \frac{du}{v} \times \sqrt{\left( \frac{273}{T} \right)}\] 

\[ \Rightarrow   t = \frac{\sqrt{273}}{v} \int\limits_0^d \frac{dx}{\left[ T_1 + \frac{\left( T_2 - T_1 \right)}{d}x \right]^\frac{1}{2}}\] 

\[ \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2d}{T_2 - T_1} \left[ T_1 + \frac{\left( T_2 - T_1 \right)}{d} \right]_0^d \] 

\[ \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2d}{T_2 - T_1}\left( \sqrt{T_2} - \sqrt{T_1} \right)\] 

\[ \Rightarrow t = \left( \frac{2d}{v} \right)\left( \frac{\sqrt{273}}{T_2 - T_1} \right) \times \sqrt{T_2} - \sqrt{T_1}      \left( \because A^2 - B^2 = \left( A - B \right)\left( A + B \right) \right)\] 

\[ \Rightarrow T = \frac{2d}{v}\frac{\sqrt{273}}{\sqrt{T_2} + \sqrt{T_1}}         .  .  . (i)\]

Evaluating this time:
Initial temperature T₁ = 280 K
Final temperature T₂ = 310 K
Space width d = 33 m
v = 330 m s⁻¹

On substituting the respective values in the above equation, we get:

\[T = \frac{2 \times 33}{330}\frac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96  \text { ms }\]