Answer to Question #233201 in Differential Equations for Jyoti

Question #233201
Derive the one-dimensional wave
equation.
1
Expert's answer
2021-09-07T03:08:53-0400

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.


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