Spread the love

Relation between Displacement Wave and Pressure Wave

Relation between Displacement Wave and Pressure Wave :- A sound wave can be expressed either by displacement of its particles from mean position or by excess pressure produced by compression and rarefaction. Consider a harmonic displacement wave moving through air contained in a long tube of cross sectional area S as shown in figure below :

Here we consider the motion of an element of medium having initial volume :

$\displaystyle {{V}_{i}}=S({{x}_{2}}-{{x}_{1}})=S\Delta x$ …..(1)

After some time, this volume of the medium changes its position and its new volume becomes :

$\displaystyle {{V}_{i}}+\Delta V=S[({{x}_{2}}+{{y}_{2}})-({{x}_{1}}+{{y}_{1}})]$

$\displaystyle \Rightarrow {{V}_{i}}+\Delta V=S({{x}_{2}}-{{x}_{1}})+S({{y}_{2}}-{{y}_{1}})$

So change in volume of the medium is given by

$\displaystyle \Delta V=S({{y}_{2}}-{{y}_{1}})$

$\displaystyle \Delta V=S\Delta y$ …..(2)

From the definition of Bulk Modulus, the pressure variation in the air is given by

$\displaystyle \Delta P=-B\left( \frac{\Delta V}{{{V}_{i}}} \right)$

$\displaystyle \Rightarrow \Delta P=-B\left( \frac{S\Delta y}{S\Delta x} \right)$

$\displaystyle \Rightarrow \Delta P=-B\left( \frac{\Delta y}{\Delta x} \right)$

Now when Δx approaches to zero, the ratio $\displaystyle \left( \frac{\Delta y}{\Delta x} \right)$ becomes $\displaystyle \left( \frac{\partial y}{\partial x} \right)$ (The partial derivative indicates that we are interested in the variation of y with position x at a fixed time).

$\displaystyle \therefore \Delta P=-B\frac{\partial y}{\partial x}$ …..(3)

The above equation relates the displacement equation with the pressure equation. The displacement equation in a progressive wave is given by :

$\displaystyle y=A\sin (\omega t-kx)$ …..(4)

$\displaystyle \Rightarrow \frac{\partial y}{\partial x}=-kA\cos (\omega t-kx)$ …..(5)

From equation (3) and (5),

$\displaystyle \Delta P=BAk\cos (\omega t-kx)$

$\displaystyle \Delta P={{(\Delta P)}_{m}}\cos (\omega t-kx)$ …..(6)

Where $\displaystyle {{(\Delta P)}_{m}}=BAk$ is the amplitude of pressure variations.

From equations (4) and (6), we observe that the pressure equation is 90° out of phase with displacement equation. When the displacement is zero, the pressure variation is either maximum or minimum and vice-versa.

Below figure shows displacement of air molecules from their equilibrium position in a harmonic wave and the corresponding pressure variations.

Points N₁, N₂, N₃, N₄ and N₅ are the positions of zero displacements (no displacement) and points M₁, M₂, M₃, M₄, M₅are the positions of maximum/minimum pressure variations. Similarly points A₁, A₂, A₃, A₄, A₅ are points of maximum displacements of air molecules from their equilibrium positions and also the points of zero pressure variation.

Now for example, have a look at points N₂ and N₃ in above figure. Just to the left of N₃, the displacement is negative and to the right of N₃ displacement is positive. This indicates that the molecules are displaced away from point N₃. So at point N₃ the pressure of air is minimum. If P_o is the atmospheric pressure (normal pressure), then pressure at point N₃ will be,

$\displaystyle P({{N}_{3}})={{P}_{0}}-{{(\Delta P)}_{m}}={{P}_{0}}-BAk$

At point N₂, the molecules of the air are displaced towards it, hence pressure (and hence density) is maximum at N₂ . So we can write,

$\displaystyle P({{N}_{2}})={{P}_{0}}+{{(\Delta P)}_{m}}={{P}_{0}}+BAk$

Finally, at points A₁, A₂, A₃, A₄, A₅ the pressure (and hence the density) does not change because the gas molecules on both sides of points A₁, A₂, A₃, A₄, A₅ have equal displacements in the same direction and hence,

$\displaystyle P({{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}},{{A}_{5}})={{P}_{0}}$

Relation between Displacement Wave and Pressure Wave

To understand the above figure and relation between displacement wave and pressure wave, air molecules oscillating in a long tube is shown in video below. In the tube as the piston(red colour bar) moves back and forth, a standing wave is produced in it.

Example :

Find –

(a) Displacement amplitude of a sound wave of frequency 100 Hz and pressure amplitude 10 Pa,

(b) Pressure amplitude of a sound wave of frequency 300 Hz and displacement amplitude 10^-7 m

(Given speed of sound in air = 340m/s and density of air = 1.29kg/m³)

Solution :

(a) Since $\displaystyle {{(\Delta P)}_{m}}=BAk$

$\displaystyle \Rightarrow A=\frac{{{(\Delta P)}_{m}}}{Bk}=\frac{{{(\Delta P)}_{m}}}{\left( \rho {{v}^{2}} \right)\left( \frac{\omega }{v} \right)}=\frac{{{(\Delta P)}_{m}}}{\rho v\omega }=\frac{{{(\Delta P)}_{m}}}{\rho v(2\pi f)}$