Velocity Of Sound In Different Media

Introduction

Sound is a mechanical wave that requires a medium (like air, water, or solid materials) to propagate. The speed of sound depends on the properties of the medium through which it travels, such as density and elasticity. In gases, sound travels more slowly due to the larger distance between particles, while in liquids and solids, it moves faster because particles are closer together, allowing quicker transfer of energy. Temperature also plays a significant role in sound speed, especially in gases, where an increase in temperature causes faster molecular motion and, hence, a faster sound speed. The nature of the medium—whether it's a solid, liquid, or gas—affects how efficiently sound waves can propagate.

Speed of sound wave in a material medium -

For deriving the equation of speed let us consider a section AB of medium as shown in figure of cross-sectional area S. Let A and B be two cross-sections as shown. Let in this medium sound wave propagation be from left to right. If wave source is located at origin O and when it oscillates, the oscillations at that point propagate along the rod.

The stress at any cross section can be written as -
$
\delta_t=\frac{F}{S}
$
. . . . .. . . . (i)

Let us consider a section AB of the material as shown in the figure, of medium at a general instant of time t . The end A is at a distance 'x' from O and point B is at a distance 'x+d x' from O. Let in time duration 'dt' due to oscillations, medium particles at A be displaced along the length of medium by 'y' and those at B by 'y+d y' . The resulting positions of section are A' and B' as shown in figure. By this we can say that the section AB is elongated by a length 'dy' . Thus strain produced in it is -
$
E=\frac{d y}{d x} \ldots \ldots . .(i i)
$
If Young's modulus of the material of medium is $Y$, we have

$
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\delta_1}{E}
$

By using Hooke's law -

From Eqs. (i) and (ii), we have

$
\begin{aligned}
& Y=\frac{F / S}{d y / d x} \\
& F=Y S \frac{d y}{d x} \ldots . .(i i i)
\end{aligned}
$
Here, $F=$ Force
if $d m$ is the mass of section $A B$ and $a$ is its acceleration, which can be given as for a medium of density $\rho$ as

$
\begin{gathered}
d m=\rho S d x \\
a=\frac{d^2 y}{d t^2}
\end{gathered}
$
From Eq. (iv), we have

$
\begin{aligned}
& d F=(\rho S d x) \frac{d^2 y}{d t^2} \\
& \frac{d F}{d x}=\rho S \frac{d^2 y}{d t^2}
\end{aligned}
$
From Eq. (iii) on differentiating w.r.t. to $x$, we can write

$
\frac{d F}{d x}=Y S \frac{d^2 y}{d x^2}
$
From Eqs. (iv) and (v), we get

The above equation shows wave velocity

In the case of gas or liquid, which shows only volume elasticity, E = B, where B = Bulk modulus of elasticity.

For longitudinal wave for liquid or gas -
$
v=\sqrt{\frac{B}{\rho}}
$

where $\rho=$ Density of the medium

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Some Solved Examples

Example 1: The maximum pressure variation that the human ear can tolerate is $30 \mathrm{~N} / \mathrm{m}^2$. The maximum displacement for a sound wave in the air having a frequency of $10^3 \mathrm{kHz}$ 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 -

$
\Delta P=\Delta P_{\max } \cdot \sin \left[\omega\left(t-\frac{x}{v}\right)\right]
$

- wherein
$\Delta P=$ variation in pressure at a point
$\Delta P_{\text {max }}=$ maximum variation in pressure

$
\begin{aligned}
& \left(\Delta P_{\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} \\
& \left(\text { Use } \rho=1.2 \frac{\mathrm{kg}}{\mathrm{m}^3} \text { and } v=343 \mathrm{~m} / \mathrm{s}\right)
\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 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 $336 \mathrm{~ms}^{-1}$. Approximate value of $T$ (in ${ }^{\circ} \mathrm{C}$ ) is:
1) 4
2)11
3) 12
4)15

Solution
Equation of sound wave -

$
\Delta P=\Delta P_{\max } \cdot \sin \left[\omega\left(t-\frac{x}{V}\right)\right]
$

wherein
$\Delta P=$ variation in pressure at a point
$\Delta P_{\max }=$ maximum variation in pressure at $0^{\circ} \mathrm{C}$

$
\begin{aligned}
& P=0.01 \sin (1000 t-3 x) \mathrm{Nm}^{-2} \\
& V_1=\frac{\omega}{k}
\end{aligned}
$
$
\begin{aligned}
& 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 T is in Kelvin) } \\
& \frac{\frac{1000}{3}}{336}=\sqrt{\frac{273}{T}} \\
& \Rightarrow T=277.41 \mathrm{k} \\
& =T=4.4^{\circ} \mathrm{C}
\end{aligned}
$
Hence, the answer is (4).

Example 3: Calculate the speed (in $\mathrm{m} / \mathrm{s}$ ) of the longitudinal wave in the helium gas of bulk modulus $1.7 \times 10^5 \mathrm{~Pa}$ and density is $0.18 \mathrm{~kg} / \mathrm{m}^3$ at $0^{\circ} \mathrm{C}$ and 1 atm pressure.
1) 972
2) 413
3) 314
4) 600

Solution
As we learnt in
Speed of sound wave -

$
v=\sqrt{\frac{B}{\rho}}
$

- wherein
$B$ is bulk modulus that represents the elastic property of the medium
$\rho=$ the density of the medium that represents the inertial property of the medium.

$V_{o_{2}} = \sqrt{\frac{B}{\rho}} = \sqrt{\frac{1.41 \times 10^{5}}{1.43}} = 314 m/s$

Hence, the answer is 314.

Example 4: A granite rod of 60 cm 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

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

Summary

The speed of sound is influenced by the medium's density and elasticity. In solids, sound waves travel faster than in liquids and gases because the particles are more tightly packed. Factors like temperature and pressure can also affect sound speed, particularly in gases, where warmer conditions generally mean faster sound propagation. For example, the speed of sound in air at 20°C is about 343 m/s, but it increases with rising temperatures. The mathematical formula for sound speed involves the medium's bulk modulus and density, but conceptually, it reflects how fast energy can transfer between particles. Understanding sound speed is crucial in fields like acoustics, engineering, and meteorology.