Question

Derive the one-dimensional wave
equation.

Answer 1

Consider the relation between Newton’s law that is applied to the volume $\Delta V$ in the direction $x$ :

$\Delta F=\Delta m\frac{dv_{x}}{dt}$ (Newton' law)

Where, $F$ : force acting on the element with volume $\Delta V$

$\Delta F_{x}=-\Delta p_{x}\Delta S_{x}$

$=(\frac{\partial p}{\partial x}\Delta x+\frac{\partial p}{\partial x}dt)\Delta S_{x}$

$\simeq -\frac{\partial p}{\partial x}\Delta V-\Delta V\frac{\partial p}{\partial x}=\Delta m\frac{dv_{x}}{dt}$

as dt is small, it is not considered and $\Delta Sx$ is in $x$ direction so ΔyΔz and from Newton’s law

$=\rho \Delta V\frac{dv_{x}}{dt}$

From, $\frac{dv_{x}}{dt}\ \text{as}\ \frac{\partial v_{x}}{\partial t}$

$\frac{dv_{x}}{dt}=\frac{\partial v_{x}}{\partial t}+v_{x}\frac{\partial v_{x}}{\partial x}\approx \frac{\partial v_{x}}{\partial x}-\frac{\partial p}{\partial x}=\rho \frac{\partial v_{x}}{\partial t}$

Above equation is known as equation of motion.

$-\frac{\partial }{\partial x}(\frac{\partial p}{\partial x})=\frac{\partial }{\partial x}(\rho \frac{\partial v_{x}}{\partial t})$

$-\frac{\partial^2 p}{\partial x^2}=\rho \frac{\partial }{\partial t}(-\frac{1}{K}\frac{\partial p}{\partial t})$ (From conservation of mass)

$\frac{\partial p^{2}}{\partial x^{2}}-\frac{\rho }{K}\frac{\partial^2 p}{\partial t^2}=0$

Where, K: bulk modulus

Rewriting the above equation:

$\frac{\partial p^{2}}{\partial x^{2}}-\frac{1}{c}^{2}\frac{\partial^2 p}{\partial t^2}=0$

Where, c: velocity of sound given as $c=\sqrt{\frac{K}{\rho }}$

Thus, above is the one-dimensional wave equation derivation.