The absolute temperature of air in 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 $273K$. Evaluate this time for ${T_1} = 280K$, ${T_2} = 310K$, $d = 33m$ and $u = 330m{s^{ - 1}}$.

Hint: In this question, we are given the values ${T_1} = 280K$, ${T_2} = 310K$, $d = 33m$ and $u = 330m{s^{ - 1}}$. We have to find the time taken by the sound wave to go through the region (where absolute temperature of air in region linearly increases from ${T_1}$ to ${T_2}$) in terms of ${T_1}$, ${T_2}$, d and the speed u at $273K$. First step is to write the temperature variation at any position between the region. Then, apply $v \propto \sqrt T $. As the temperature is given it will be converted to $\dfrac{{{v_T}}}{v} = \sqrt {\dfrac{T}{{273}}} $. In last apply ${v_T} = \dfrac{{dx}}{{dt}}$ this formula. Integrate the terms by putting all the formulas and solve further.

Formula used:
Velocity of the temperature ${v_T} = \dfrac{{dx}}{{dt}}$
Formula for temperature variation $T = {T_1} + \dfrac{{{T_2} - {T_1}}}{d} \times x$

Complete answer:
Given that,
${T_1} = 280K$
${T_2} = 310K$
Distance between ${T_1}$ and ${T_2}$, $d = 33m$
Speed of the sound at $273K$, $u = 330m{s^{ - 1}}$

The temperature variation at any position $x$ is :
$T = {T_1} + \dfrac{{{T_2} - {T_1}}}{d} \times x - - - - - - \left( 1 \right)$
Also, velocity $v \propto \sqrt T $
Therefore, $\dfrac{{{v_T}}}{v} = \sqrt {\dfrac{T}{{273}}} $

As we know that, velocity of the temperature ${v_T} = \dfrac{{dx}}{{dt}}$
It implies that:
$dt = \dfrac{{dx}}{{{v_T}}} = \dfrac{{dx}}{v}\sqrt {\dfrac{{273}}{T}} $

Integrating both the sides,
$t = \dfrac{{\sqrt {273} }}{v}\int\limits_0^d {\dfrac{1}{{\sqrt T }}dx} $

From equation (1)
$t = \dfrac{{\sqrt {273} }}{v}\int\limits_0^d {\dfrac{1}{{\sqrt {{T_1} + \dfrac{{{T_2} - {T_1}}}{d} \times x} }}dx} $
$t = \dfrac{{\sqrt {273} }}{v} \times \dfrac{{2d}}{{{T_2} - {T_1}}}\left[ {{T_1} + \dfrac{{{T_2} - {T_1}}}{d} \times x} \right]_0^d$
$t = \dfrac{{\sqrt {273} }}{v} \times \dfrac{{2d}}{{{T_2} - {T_1}}}\left( {\sqrt {{T_2}} - \sqrt {{T_2}} } \right)$
$t = \dfrac{{2d\sqrt {273} }}{{v\left( {\sqrt {{T_2}} + \sqrt {{T_1}} } \right)}}$

Putting all the given values,
$t = \dfrac{{2\left( {33} \right)\sqrt {273} }}{{330\left( {\sqrt {310} + \sqrt {280} } \right)}}$
$t = 96 \times {10^{ - 3}}\sec $

Hence, the time taken by a sound wave to go through the region is $96 \times {10^{ - 3}}\sec $.

Note: A sound wave is a pattern of disturbance which is caused by the movement of energy as it propagates away from the source of the sound through a medium (such as air, water, or any other liquid or solid matter). Sound waves are produced by object vibrations and pressure waves, such as a ringing cell phone.