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Factors That do Not Affect Speed of Sound

The absolute ...

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 T₂in 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:
$T = T_{1} + \frac{(T_{2} - T_{1})}{d} x v \propto \sqrt{T} \Rightarrow \frac{v_{T}}{v} = \sqrt{(\frac{T}{273})} \Rightarrow v_{T} = v \sqrt{(\frac{T}{273})} \Rightarrow d t = \frac{d x}{v_{T}} = \frac{d u}{v} \times \sqrt{(\frac{273}{T})} \Rightarrow t = \frac{\sqrt{273}}{v} \int_{0}^{d} \frac{d x}{{[T_{1} + \frac{(T_{2} - T_{1})}{d} x]}^{\frac{1}{2}}} \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2 d}{T_{2} - T_{1}} {[T_{1} + \frac{(T_{2} - T_{1})}{d}]}_{0}^{d} \Rightarrow t = \frac{\sqrt{273}}{v} \times \frac{2 d}{T_{2} - T_{1}} (\sqrt{T_{2}} - \sqrt{T_{1}}) \Rightarrow t = (\frac{2 d}{v}) (\frac{\sqrt{273}}{T_{2} - T_{1}}) \times \sqrt{T_{2}} - \sqrt{T_{1}} (∵ A^{2} - B^{2} = (A - B) (A + B)) \Rightarrow T = \frac{2 d}{v} \frac{\sqrt{273}}{\sqrt{T_{2}} + \sqrt{T_{1}}} . . . (i)$

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:
$T = \frac{2 \times 33}{330} \frac{\sqrt{273}}{\sqrt{280} + \sqrt{310}} = 96 ms$

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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 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.