Question

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

Answer 1

Step1: Given data

The initial temperature of the air $\left(T_1\right)=280K$

Final temperature $\left(T_2\right)=310K$

Width of space$\left(d\right)=33m$

Speed of wave $\left(V\right)=330m{s}^{-1}$

Let us suppose the time taken by a sound wave to go through the region is $t$.

Step2: Formula Used

Using interpolation,

The temperature variation at a given position $x$ is,

$T=T_1+\frac{T_2-T_1}{d}\times x.............\left(1\right)$

Speed of sound at any temperature is given by-

$V=\sqrt{\frac{\gamma RT}{M}}$

Where $\gamma$ is the ratio of specific heat at constant pressure to the specific heat at constant volume,

$M$ is the molar mass of gas,

$R$ is the universal gas constant.

Step 3:Calculating the relation between $V\mathrm{and}T$

It is understood that,

$V\propto \sqrt{T}$. So,

$V$ at $273K$

$V=\sqrt{\frac{\gamma R\times 273}{M}}$

$V$ at temperature $TK$

$V_T=\sqrt{\frac{\gamma R\times T}{M}}$

Thus the ratio will be-

$\begin{array}{l}\frac{V_T}{V}=\sqrt{\frac{T}{273}}\\ V_T=V\sqrt{\frac{T}{273}}..................\left(2\right)\end{array}$

Where $V_T$ is the speed of the sound wave at any temperature $T$.

Step 4: Calculating the required time

We know that, $Speed=\frac{dis\tan ce}{time}$

$V_T=\frac{dx}{dt}$.

So,

$dt=\frac{dx}{V_T}............\left(3\right)$

Substitute value of $V_T$ from equation (2) to equation (3)

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

On integrating,

${\int }_0^{t}dt=\frac{\sqrt{273}}{V}{\int }_0^d\frac{dx}{\sqrt{T}}$

Now putting the value of $T$ from equation (1)-

$$\begin{gathered} {\int }_0^{t}dt=\frac{\sqrt{273}}{V}{\int }_0^d\frac{dx}{\sqrt{T_1+\frac{T_2-T_1}{d}\times x}} \\ t=\frac{\sqrt{273}}{V}{\int }_0^d\frac{dx}{\sqrt{T_1+\frac{T_2-T_1}{d}\times x}} \end{gathered}$$

Using the property rule, $\int X^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx=2X^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ here, $X=\sqrt{T_1+\left(\frac{T_2-T_1}{d}\right)x}$ applying the limits also put the given values we get-

$t=\frac{2\sqrt{273}}{V}\frac{{\left[\sqrt{T_1+\left({\displaystyle \frac{T_2-T_1}{d}}\right)x}\right]}_0^d}{\left({\displaystyle \frac{T_2-T_1}{d}}\right)}$

$$\begin{gathered} t=\frac{\sqrt{273}}{V}\times \frac{2d}{\left(T_2-T_1\right)}\times \left(\sqrt{T_2}-\sqrt{T_1}\right) \\ t=\frac{\sqrt{273}}{330}\times \frac{2\times 33}{\left(310-280\right)}\times \left(\sqrt{310}-\sqrt{280}\right) \\ t=0.05\times 2.2\times 0.8736 \\ t=0.09609sec \end{gathered}$$

Hence, the evaluated time will be $0.09609sec$.