Speed of Sound
Knowing the speed of sound through any particular medium is important for understanding the speed at which mechanical (vs. electromagnetic) disturbances propagate. As a sound wave travels, it perturbs the properties of the medium slightly. Such properties include the speed of sound itself ($latex a$), pressure ($latex p$), density ($latex \rho$), and temperature ($latex T$).
Terms such as $latex dp$ represent infinitesimal (very small) changes in the properties. The speed of sound as a function of the fluid properties is simply:
$latex a = \sqrt{\gamma \frac{p}{\rho}}$
How do we derive this equation?
Conservation of Mass: $latex \rho a = \text{constant}$
Conservation of Momentum: $latex p + \rho a^2 = \text{constant}$
The idea here is to start with conservation of mass to describe the properties on either side of the sound wave and solve for the speed of sound.
$latex \rho a = (\rho + d\rho)(a + da)$
$latex \rho a = \rho a + \rho da + d\rho da$
During the derivation process, we can neglect terms such as $latex d\rho da$ because the produce of two infinitesimal terms is negligible in terms of order of magnitude. Now rearranging:
$latex \rho da + a d\rho = 0$
$latex a = -\rho \frac{da}{d\rho}$ (*)
Equation (*) is the final result from conservation of mass. However, we are not yet done. Now we go through a similar procedure starting with conservation of momentum, neglecting products of infinitesimals along the way:
$latex p + \rho a^2 = (p + dp) + (\rho d\rho)(a + da)^2$
$latex p + \rho a^2 = p + dp + (\rho + d\rho)(a^2 + 2a da + da^2)$
$latex dp + 2a\rho da + a^2 d\rho = 0$</p>tex p + \rho a^2 = (p + dp) + (\rho d\rho)(a + da)^2$
$latex da = -\frac{dp + a^2 d\rho}{2 a \rho}$ (**)
Now, substituting eq. (*) into ():
$latex a = \rho \left(\frac{dp + a^2 d\rho}{2 a \rho}\right)\frac{1}{d\rho}$
$latex a = \frac{1}{2a} \left(\frac{dp}{d\rho} + a^2\right)$
$latex a^2 = \frac{dp}{d\rho}$ (***)
Equation (***) describes the speed of sound as a function of an isentropic change in pressure and density. However, it does not quite resemble the equation first given, which describes the speed of sound as a function of the gas properties. To do so, we use the equation, assuming a calorically perfect gas:
$latex p \left(\frac{1}{\rho}\right)^\gamma = \text{constant} = c$
$latex p = \rho^\gamma c$
This equation can be differentiated with respect to density:
$latex \frac{dp}{d \rho} = \gamma c \rho^{\gamma - 1}$
$latex dp = \gamma c \frac{\rho^\gamma}{\rho} d\rho $
$latex dp = \gamma p \left(\frac{1}{\rho}\right)^\gamma \rho^\gamma \frac{1}{\rho} d\rho$
$latex \frac{dp}{d \rho} = \frac{\gamma p}{\rho}$ (**)
Finally, substituting (**) into (*) and solving for $latex a$, we arrive at the equation
$latex a = \sqrt{ \gamma \frac{p}{\rho} }$
This equation of the speed of sound is valid for sound traveling through air due to the fact that we used an equation assuming a calorically perfect gas during the derivation process.