Pressure wave speed $v=\sqrt{\frac{B}{\rho}}$ rearranged $\rho v^2=B$ is similar to kinetic energy: $E_k=\frac{1}{2}mv^2$. Is this a coincidence?

Question

The speed of a Pressure wave speed is given by: $v = \sqrt{ \frac{B}{\rho}}$, which seems rearranged very similar to kinetic energy: $\rho v^2 = B$. How can this be understood, is this a coincidence?

This seems almost like a kinetic energy density.

For a wave in a rope under tension it is similar as well:

$$ \mu v^2 = T$$

Kinetic energy is given by: $$ E_K = \frac{1}{2} mv^2 $$

Is this a coincidence? How can this intuitively be understood? Why is it missing the $ \frac{1}{2}$? Can you use this to derive wave speed with conservation of energy?

I mean it could be almost like the work done by a rope per meter is:

$$ \frac{W}{\Delta x} = \frac{T\Delta x}{\Delta x} = T = E_{k,f}-E_{k,i} = \frac{1}{2} \frac{m}{\Delta x} v^2 \propto \mu v^2 $$

Answer 1

I think I found an answer:

For a string of linear mass density $\rho$ under tension $F_T$ the wave speed relation is:

$$ \rho v^2 = F_T$$

The kinetic energy of the transverse displacement $y(x,t)$ is given by:

$$ \text{KE}= \frac{1}{2} m \dot y^2= \int \frac{1}{2} \rho \dot y^2$$

The potential energy is the work done by stretching the string:

$$ \text{PE} = F_T \Delta l = \int F_T (\sqrt{1+y'^2}-1)dx \approx \int \frac{1}{2} F_T y'^2 dx $$ Where $y'= dy/dx$

Then since $$ \dot y = vy', \dot y^2 = v^2y'^2=\frac{F_T}{\rho}y'^2$$

Thus $$ U = \text{KE}+\text{PE}= 2\text{PE}= \boxed{2\text{KE} = \rho v^2} $$

There can be made an analogous derivation for a gas probably.

There is also a similarity for an electromagnetic wave:

$$ u_E = \frac{1}{2}\epsilon_0 E^2 $$ $$ u = u_E + u_B = 2u_E = 2u_B= \epsilon_0 E^2$$

Source: https://www.physics.princeton.edu/~mcdonald/examples/destructive.pdf