A Modest Temperature Gradient

Relevant wiki: Applying Boundary Conditions to Standing Waves

Let's start with the equation for the velocity of sound/pressure waves through a certain medium, $v = \sqrt{\frac{K}{\rho}}$ , where $\rho$ is the density and $K$ is the elastic bulk modulus given by $K = \gamma P$ . Plugging these values into the wave velocity equation, our equation becomes $v=\sqrt { \frac { \gamma P }{ \rho } }$ .

Since $P$ stands for pressure, we have finally found a way to relate the temperature gradient to the speed of the wave! Well, after a few steps. Using the ideal gas law to solve for pressure, $PV=nRT$ becomes $P=\frac { nRT }{ V }$ . However, before we substitute this into our velocity equation, let's find some way to substitute the density into this rearranged ideal gas law equation as well.

Firstly, since $n$ stands for the number of moles of gas, by definition, $nM=m$ , where $M$ is the molar mass of the gas (in $\frac { kg }{ mol }$ ) and $m$ is the mass (in $kg$ ). Rewriting this equation to solve for moles of gas, $n=\frac { m }{ M }$ , and this can be plugged back into the ideal gas law. Thus, our ideal gas law becomes this: $P=\frac { \left( \frac { m }{ M } \right) RT }{ V } =\frac { mRT }{ MV } =\left( \frac { m }{ V } \right) \frac { RT }{ M }$ Density, $\rho$ , is equal to $\frac { m }{ V }$ , so $\frac { m }{ V }$ can be replaced as such: $\rho \left( \frac { RT }{ M } \right)$ . Going back to the wave velocity equation, pressure can now be substituted with the above expression and, thus, the density can be canceled out: $v=\sqrt { \frac { \rho \left( \frac { RT }{ M } \right) \gamma }{ \rho } } =\sqrt { \frac { \gamma RT }{ M } }$

Now, here comes the calculus. Since the observer stands at a distance $15km$ away from the shock wave and we are trying to find the time it takes the wave to travel this distance, we must set up a differential equation to solve for the distance from the velocity. Note that, in this equation, I have changed the variable for distance from $d$ to $x$ : $v=\frac { dx }{ dt } =\sqrt { \frac { \gamma RT\left( { x } \right) }{ M } }$ In preparation for integration, we must separate the variables, noting that only the temperature changes with distance (due to the gradient). All other variables can factored out as constants: $\frac { dx }{ \sqrt { \frac { \gamma RT\left( { x } \right) }{ M } } } =dt\\ \sqrt { \frac { M }{ \gamma RT\left( { x } \right) } } dx=dt\\ dt=\sqrt { \frac { M }{ \gamma R } } \left( \frac { dx }{ \sqrt { T\left( x \right) } } \right)$

To find $T\left( x \right)$ , we must refer to the temperature gradient. The temperature gradient measures how the temperature changes with the distance from the shock wave's source. Thus, this expression can be written as a differential equation which gives the rate at which the temperature (in Kelvins, $K$ ) changes with distance (in kilometers, $km$ ): $\frac { dT }{ dx } =-\frac { 2 }{ 75 } x$ . Through separation of variables, we can integrate and solve for a function, $T\left( x \right)$ , which directly relates temperature with distance: $\\ dT=-\frac { 2 }{ 75 } xdx\\ \int _{ { T }_{ i } }^{ { T }_{ f } }{ dT } =-\frac { 1 }{ 75 } \int _{ 0 }^{ { x }_{ f } }{ 2xdx } \\ T\left( { x } \right) -T\left( 0 \right) =-\frac { { x }^{ 2 } }{ 75 }$ Since the problem states that the temperature at the wave's origin is $300K$ , $T\left( 0 \right)=300K$ . Thus, $T\left( { x } \right) =300-\frac { { x }^{ 2 } }{ 75 }$ .

Now that we have found $T\left( x \right)$ , let's solve the differential equation for the wave's distance traveled versus distance. However, in the wave equation, the velocity is in $\frac { m }{ s }$ instead of $\frac { km }{ s }$ . Therefore, $T\left( { x } \right)$ becomes $300-\frac { { \left( \frac { x }{ 1000 } \right) }^{ 2 } }{ 75 } =300-\frac { { x }^{ 2 } }{ 75000000 }$ , with $1000$ used as a conversion factor from $x$ being in $km$ to $m$ .

From here, the problem is purely computational. Solving the integral: $\int _{ 0 }^{ { t } }{ dt } =\sqrt { \frac { M }{ \gamma R } } \int _{ 0 }^{ { x } }{ \frac { dx }{ \sqrt { 300-\frac { { x }^{ 2 } }{ 75000000 } } } } \\ t=\sqrt { \frac { M }{ \gamma R } } \cdot \frac { 1 }{ \sqrt { 300 } } \int _{ 0 }^{ { x } }{ \frac { dx }{ \sqrt { 1-\frac { { x }^{ 2 } }{ 22500000000 } } } \\ } \\ t=\sqrt { \frac { M }{ \gamma R } } \cdot \frac { 1 }{ \sqrt { 300 } } \int _{ 0 }^{ { x } }{ \frac { dx }{ \sqrt { 1-{ \left( \frac { x }{ 150000 } \right) }^{ 2 } } } \\ }$ If we let $u=\frac { x }{ 150000 }$ , then $dx=150000du$ . $t=\sqrt { \frac { M }{ \gamma R } } \cdot \frac { 150000 }{ \sqrt { 300 } } \int _{ 0 }^{ \frac { x }{ 150000 } }{ \frac { du }{ \sqrt { 1-u^{ 2 } } } } \\ t=\sqrt { \frac { M }{ \gamma R } } \cdot 5000\sqrt { 3 } { \left[ \arcsin { u } \right] }_{ u=0 }^{ u=\frac { x }{ 150000 } }\\ t=\sqrt { \frac { M }{ \gamma R } } \cdot 5000\sqrt { 3 } \arcsin { \frac { x }{ 150000 } }$ The observer is $15km=15000m$ away. Knowing this fact and using the allowed approximations: $t=\sqrt { \frac { M }{ \gamma R } } \cdot 5000\sqrt { 3 } \arcsin { \frac { 15000 }{ 150000 } } \\ t\approx \frac { 1 }{ 20 } \cdot 5000\sqrt { 3 } \arcsin { \frac { 1 }{ 10 } } \\ t\approx 250\sqrt { 3 } \left( \frac { 1 }{ 10 } \right) \\ t\approx 25\sqrt { 3 } seconds\Rightarrow a\sqrt { b } s\\ a+b=25+3=\boxed { 28 }$