Displacement Wave And Pressure Wave

The relation between displacement waves and pressure waves is a fundamental concept in the study of sound and wave mechanics. Displacement waves describe the movement of particles in a medium as a wave passes through, while pressure waves represent the variations in pressure within the medium caused by the wave. In real life, this relationship is evident in how we perceive sound. For instance, when a guitar string vibrates, it creates displacement waves in the air, which in turn generate pressure waves that travel to our ears, allowing us to hear music. The interplay between these two types of waves is crucial in various technologies, such as in ultrasound imaging, where precise control of these waves enables us to create detailed images of the inside of the human body. In this article, we will discuss the concept of the relationship between displacement Waves and pressure Waves, with some important examples.

Relation Between Displacement Wave and Pressure Wave

A displacement wave refers to the oscillation of particles in a medium as a wave passes through it, while a pressure wave represents the resulting variations in pressure within that medium. These two types of waves are intimately connected: as particles move (displacement), they cause compressions and rarefactions in the surrounding medium, leading to pressure changes. This dynamic is observed in everyday life, such as when a tuning fork is struck, causing the air around it to oscillate. These oscillations (displacement waves) create regions of higher and lower pressure (pressure waves), which travel through the air to our ears, allowing us to hear sound.

As we have studied, when a longitudinal wave propagates in a gaseous medium, it produces rarefaction and compression in the medium, periodically. In the region where compression occurs, the pressure is higher than the normal pressure of the medium. In the region where rarefaction occurs, the pressure is lesser than the normal pressure of the medium. Thus we can also describe any longitudinal waves in a gaseous medium as pressure waves and these are also termed compressional waves.

Let us consider a longitudinal wave propagating in a positive 'x' direction as shown in the given figure. This figure shows a segment AB
of the medium of width 'dx'. Let a longitudinal wave propagate in this medium whose equation is given as -

$y=A \sin (k x-\omega t)$

In this equation, 'y' is the displacement of a medium particle situated at a distance 'x' from the origin along the direction of propagation
of the wave. From the figure, AB is the medium segment such that A is located at position x = x and B is at x=x+d x at an instant. If after some time t medium particle at A reaches a point A' which is displaced by y and the medium particle at B reaches point B' which is at a displacement y+dy from B. Here dy is given by equation as

$\begin{aligned}
& d y=A k \cos (k x-\omega t) d x \\
& d V=S d y=-S A k \cos (k x-\omega t) d x
\end{aligned}$

Where, S = Area of cross-section and V = Volume of section AB

$ \begin{gathered}
\frac{d V}{V}=\frac{d y}{d x}=\frac{S A k \cos (k x-\omega t) d x}{S d x} \\
\frac{d V}{V}=A k \cos (k x-\omega t)
\end{gathered}$

If B is the bulk modulus of the medium, then the excess pressure in section AB can be given as -

$ \begin{aligned}
& \Delta P=-B\left(\frac{d V}{V}\right)=-B\left(\frac{d y}{d x}\right) \\
& \Delta P=-B A k \cos (k x-\omega t) \\
& \Delta P=-\Delta P_{\max } \cos (k x-\omega t)
\end{aligned}$

\text { Here } \Delta P_{\max } \text { is the pressure amplitude at a medium particle at position } x \text { from origin and } \Delta P \text { is the excess pressure at that point }

So,

$ \Delta P_{\max }=B A k=\frac{2 \pi}{\lambda} A B$

In the compression zone, more particles stay in a unit volume of the medium. So, the density and pressure of the region will be higher. In the refracted zone, lesser particles stay in any unit volume. Let a sound wave propagate in a medium of Bulk modulus B and density $\rho$.

So,

$
B=\left(-\frac{d p}{d V / V}\right)
$

Also,
$
\frac{d V}{V}=-\frac{d p}{p}
$

From both Equation, we get, $d \rho=\frac{\rho}{B} d p$
The speed of sound is given by, $v=\sqrt{\frac{B}{\rho}} \Rightarrow \frac{\rho}{B}=\frac{1}{v^2}$ Hence, $d \rho=\frac{\rho}{B} \Delta p=\frac{1}{v^2} \Delta p$

So, this relation gives the relation between pressure with density. So the variation in density is like the variation in pressure

$
\Delta \rho=(\Delta \rho)_m \sin (k x-\omega t)
$
where, $(\Delta \rho)_m=\frac{\rho}{B}(\Delta p)_m=\frac{(\Delta p)_m}{v^2}$

Note - The density equation is in phase with the pressure equation and this is $\frac{\pi}{2}$ out of phase with the displacement equation.

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Solved Examples Based on Displacement Wave and Pressure wave

Example 1: The maximum pressure variation that the human ear can tolerate is 30 N/m². The maximum displacement for a sound wave in the air having a frequency of 10³ kHz is? (Use density of air = $\rho=1.2 \frac{\mathrm{kg}}{\mathrm{m}^3}$ and speed of sound in air = $=v=343 \mathrm{~m} / \mathrm{s}$)

1) $\frac{2 \pi}{3} \times 10^{-2} \mathrm{~km}$
2) $\frac{2 \times 10^{-4}}{\pi} \mathrm{km}$
3) $\frac{\pi}{3} \times 10^{-2} \mathrm{~km}$
4) $\frac{10^{-4}}{3 \pi} \mathrm{km}$

Solution

Equation of sound wave
wherein

$\begin{aligned} & \Delta P=\Delta P_{\text {max }} \cdot \sin \left[\omega\left(t-\frac{x}{v}\right)\right] \\ & \text { wherein } \\ & \Delta P=\text { variation in pressure at a point } \\ & \Delta P_{\text {max }}=\text { maximum variation in pressure } \\ & \left(\Delta P_{\text {max }}\right)=B A K \Rightarrow A=\frac{\Delta P}{B k} \\ & v=\frac{\omega}{k}, \\ & B=\rho \times v^2 \\ & k=\omega \sqrt{\frac{\rho}{B}} \Rightarrow A=\frac{\Delta P_{\max }}{2 \pi f \rho v}=\frac{10^{-4}}{3 \pi} \mathrm{Km} \\ & \text { (Use } \rho=1.2 \frac{\mathrm{kg}}{\mathrm{m}^3} \text { and } v=343 \mathrm{~m} / \mathrm{s} \text { ) }\end{aligned}$

Hence, the answer is the option (4).

Example 2: The pressure wave, $P=0.01 \sin [1000 t-3 x] \mathrm{Nm}^{-2}$ corresponds to the sound produced by a vibrating blade on a day when the atmospheric temperature is $0^{\circ} \mathrm{C}$.On some other day when the temperature is T. the speed of sound produced by the same blade and at the same frequency is found to be an Approximate value of (in ):

1) 4

2) 11

3) 12

4) 15

Solution:

Equation of sound wave

$\begin{aligned} & \Delta P=\Delta P_{\max } \cdot \sin \left[\omega\left(t-\frac{x}{V}\right)\right] \\ & \text { wherein } \\ & \Delta P=\text { variation in pressure at a point } \\ & \Delta P_{\text {max }}=\text { maximum variation in pressure at } 0^{\circ} \mathrm{C} \\ & P=0.01 \sin (1000 t-3 x) \mathrm{Nm}^{-2} \\ & V_1=\frac{\omega}{k} \\ & V_1=\frac{1000}{3} \\ & \text { at temp } T \\ & V_2=336 \mathrm{~ms}^{-1} \\ & \frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}} \quad \text { (Where } \mathrm{T} \text { is in Kelvin) } \\ & \frac{1000}{3}=\sqrt{\frac{273}{T}} \\ & \Rightarrow 336=277.41 \mathrm{k} \\ & \Rightarrow=T=4.4^{\circ} \mathrm{C}\end{aligned}$

Hence, the answer is the option (4).

Example 3: Calculate the speed (in m/s) of the longitudinal wave in oxygen at $0^{\circ} \mathrm{C}$ and 1 atm $\left(10^5 \mathrm{~Pa}\right)$ having bulk modulus equal to $1.41 \times 10^5 \mathrm{~Pa}$ and density of 1.43 kg/m³.

1) 314

2) 612

3) 972

4) 0

Solution:

Speed of sound wave
$
v=\sqrt{\frac{B}{\rho}}
$
wherein
$B$ is the bulk modulus that represents the elastic property of the medium
$\rho=$ the density of the medium that represents the inertial property of the medium.
$
V_{o_2}=\sqrt{\frac{B}{\rho}}=\sqrt{\frac{1.41 \times 10^5}{1.43}}=314 \mathrm{~m} / \mathrm{s}
$

Hence, the answer is the option (1).

Example 4: Calculate the speed (in m/s) of the longitudinal wave in the helium gas of bulk modulus $1.7 \times 10^5 \mathrm{~Pa}$ and density is 0.18 kg/m³ at $0^{\circ} \mathrm{C}$ and 1 atm pressure.

1) 972

2) 413

3) 314

4) 600

Solution

Speed of sound wave

$
v=\sqrt{\frac{B}{\rho}}
$
wherein
$B$ is the bulk modulus that represents the elastic property of the medium
$\rho=$ the density of the medium that represents the inertial property of the medium.
$
V_{H e}=\sqrt{\frac{B}{\rho}}=\sqrt{\frac{1.7 \times 10^5}{0.18}}=972 \mathrm{~m} / \mathrm{s}
$

Hence, the answer is the option (1).

Example 5: A granite rod of 60 cm in length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×10³ kg/m3 and its Young’s modulus is 9.27×10¹⁰ Pa. What will be the fundamental frequency (in Hz) of the longitudinal vibrations?

1) 5

2) 7.5

3) 2.5

4) 10

Solution:

$\begin{aligned} \nu_o & =\frac{v}{2 l}=\frac{1}{2 l} \cdot \sqrt{\frac{\gamma}{\rho}}=\frac{1}{2 * 0.6} \sqrt{\frac{9.27 * 10^{10}}{2.7 * 10^3}} \\ & =4.9 * 10^3 \mathrm{HZ} \simeq 5 \mathrm{kHZ}\end{aligned}$

Hence, the answer is the option (1).

Summary

The relationship between displacement waves and pressure waves is key to understanding sound propagation. Displacement waves involve particle oscillation in a medium, while pressure waves correspond to the resulting pressure variations. These two waves are interconnected, with changes in displacement causing pressure fluctuations. The equations provided demonstrate how these waves behave in various scenarios, linking pressure, density, and sound speed.