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