Welcome to Lecture 110 of our comprehensive JEE Physics Master Series. In this lecture, we transition from transverse waves on strings to longitudinal mechanical waves in compressible fluid media: Sound Waves. We formulate the speed of sound from first principles, examine the historical breakdown of Newton’s isothermal formula and Laplace’s adiabatic correction, analyze the mathematical duality between displacement waves and pressure/density waves, investigate the physics of acoustic intensity and the logarithmic decibel ($\text{dB}$) scale, and conclude with 4 graded benchmark examples followed by a 10-question practice worksheet (DPP) with complete solutions.
1. Speed of Longitudinal Waves & The Laplace Correction
Sound travels through an elastic fluid medium as an alternating sequence of compressions and rarefactions. The general propagation speed of a longitudinal wave in any elastic medium of density $\rho$ governed by bulk modulus $B$ is given by:
$$v = \sqrt{\frac{B}{\rho}}$$
where the bulk modulus is defined as $B = -V \frac{dP}{dV}$.
1. Newton’s Formula (Isothermal Assumption):
Sir Isaac Newton assumed that compressions and rarefactions occur slowly enough for thermal conduction to keep the medium in constant thermal equilibrium with its surroundings ($T = \text{constant}$). For an ideal gas undergoing isothermal changes:
$$P V = \text{constant} \implies P dV + V dP = 0 \implies B_{\text{iso}} = -V \frac{dP}{dV} = P$$
This yields Newton’s formula for the speed of sound:
$$v_{\text{Newton}} = \sqrt{\frac{P}{\rho}}$$
At STP for air ($P = 1.013 \times 10^5\text{ N/m}^2$, $\rho = 1.293\text{ kg/m}^3$):
$$v_{\text{Newton}} = \sqrt{\frac{1.013 \times 10^5}{1.293}} \approx 280\text{ m/s}$$
This theoretical value falls approximately $16\%$ below the experimentally observed speed in air ($\approx 332\text{ m/s}$ at STP).
2. Laplace’s Correction (Adiabatic Process):
Pierre-Simon Laplace resolved the discrepancy by pointing out that acoustic compressions and rarefactions happen so rapidly that heat exchange between adjacent high-pressure (warm) and low-pressure (cool) layers is negligible. Furthermore, gases are poor thermal conductors. Hence, sound propagation is strictly an adiabatic process ($P V^\gamma = \text{constant}$):
$$P (\gamma V^{\gamma – 1} dV) + V^\gamma dP = 0 \implies B_{\text{adia}} = -V \frac{dP}{dV} = \gamma P$$
Therefore, the corrected speed of sound in an ideal gas is:
$$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M_{\text{mol}}}}$$
For air ($\gamma \approx 1.4$): $v_{\text{Laplace}} = \sqrt{1.4} \times 280\text{ m/s} \approx 331.3\text{ m/s}$, matching experimental data.
Factors Affecting Speed of Sound in Gas:
• Effect of Pressure: At constant temperature, $P/\rho = \frac{RT}{M_{\text{mol}}} = \text{constant}$. Thus, pressure changes at constant temperature have zero effect on the speed of sound.
• Effect of Temperature: $v \propto \sqrt{T}$. For small Celsius variations around $0^\circ\text{C}$: $v(t) \approx v_0 + 0.61 t$ (in $\text{m/s}$).
• Effect of Humidity: Moist air is less dense than dry air ($\rho_{\text{moist}} < \rho_{\text{dry}}$ because molar mass of $H_2 O$ is $18\text{ g/mol}$ vs $28.8\text{ g/mol}$ for air). Hence, $v_{\text{moist}} > v_{\text{dry}}$.
2. Displacement Wave vs. Pressure Wave
A sound wave propagating along the $+x$ direction can be expressed in terms of the longitudinal displacement $s(x, t)$ of medium particles from their equilibrium positions:
$$s(x, t) = s_0 \sin(k x – \omega t)$$
Consider a thin cylindrical slice of gas of equilibrium thickness $\Delta x$ and cross-sectional area $A$. The original volume is $V = A \Delta x$. As the wave passes, the left face displaces by $s(x)$ and the right face by $s(x + \Delta x)$, changing the volume by $\Delta V = A [s(x + \Delta x) – s(x)] \approx A \frac{\partial s}{\partial x} \Delta x$. The volumetric strain is:
$$\frac{\Delta V}{V} = \frac{\partial s}{\partial x}$$
Using the bulk modulus definition $\Delta P = -B \frac{\Delta V}{V} = -B \frac{\partial s}{\partial x}$:
$$\Delta P(x, t) = -B \frac{\partial}{\partial x}[s_0 \sin(k x – \omega t)] = -B k s_0 \cos(k x – \omega t) = \Delta P_0 \sin\left(k x – \omega t – \frac{\pi}{2}\right)$$
where the pressure amplitude $\Delta P_0$ is:
$$\Delta P_0 = B k s_0 = (\rho v^2)\left(\frac{\omega}{v}\right) s_0 = \rho v \omega s_0$$
Phase Duality: The pressure excess $\Delta P$ is out of phase with displacement $s$ by $\frac{\pi}{2}\text{ rad}$ ($90^\circ$). At points of maximum displacement ($s = \pm s_0$), the pressure excess is zero ($\Delta P = 0$). Conversely, at points of zero displacement ($s = 0$), the compression or rarefaction strain is maximal, so pressure excess reaches its extreme peaks ($\Delta P = \pm \Delta P_0$).
3. Intensity of Sound Waves & The Decibel Scale
1. Acoustic Intensity ($I$): The time-averaged acoustic energy flowing per unit time across unit area perpendicular to the direction of propagation:
$$I = \langle \Delta P \cdot v_p \rangle = \frac{1}{2} \rho v \omega^2 s_0^2 = \frac{\Delta P_0^2}{2 \rho v}$$
where $Z = \rho v$ is the characteristic acoustic impedance of the medium ($\text{kg}\cdot\text{m}^{-2}\cdot\text{s}^{-1}$).
2. Point Source & Spatial Attenuation:
For an isotropic point source emitting power $P_{\text{source}}$ in a lossless 3D medium:
$$I(r) = \frac{P_{\text{source}}}{4 \pi r^2} \implies I \propto \frac{1}{r^2}, \quad \Delta P_0 \propto \frac{1}{r}, \quad s_0 \propto \frac{1}{r}$$
3. Sound Level in Decibels ($\beta$):
Due to the logarithmic response of human auditory perception (Weber-Fechner Law), the sound intensity level $\beta$ is defined relative to the reference threshold of human hearing at $1\text{ kHz}$ ($I_0 = 10^{-12}\text{ W/m}^2$):
$$\beta = 10 \log_{10}\left(\frac{I}{I_0}\right) \quad (\text{in dB})$$
If intensity increases by a factor of $10$, $\beta$ increases by $+10\text{ dB}$. If intensity doubles ($I \to 2I$), $\beta$ increases by $10 \log_{10}(2) \approx +3.01\text{ dB}$.
4. Solved Examples
Example 1 (Foundational – JEE Main):
Calculate the speed of sound in hydrogen gas ($H_2$, rigid diatomic with $\gamma = 7/5$, $M_{\text{mol}} = 2.0 \times 10^{-3}\text{ kg/mol}$) at a temperature of $27^\circ\text{C}$. Take $R = 8.314\text{ J/(mol}\cdot\text{K)}$.
Solution:
Temperature in Kelvin: $T = 27 + 273.15 = 300.15\text{ K} \approx 300\text{ K}$.
Applying Laplace’s formula:
$$v = \sqrt{\frac{\gamma R T}{M_{\text{mol}}}} = \sqrt{\frac{(1.4)(8.314)(300)}{2.0 \times 10^{-3}}} = \sqrt{\frac{3491.88}{2.0 \times 10^{-3}}} = \sqrt{1.74594 \times 10^6} \approx \mathbf{1321.3\text{ m/s}}$$
Example 2 (Pressure Amplitude to Displacement Amplitude – JEE Main/Advanced):
A sound wave in air at STP ($\rho = 1.2\text{ kg/m}^3, v = 340\text{ m/s}$) has a frequency of $1000\text{ Hz}$ and a sound intensity level of $80\text{ dB}$. Find: (a) the acoustic intensity $I$, (b) the pressure amplitude $\Delta P_0$, and (c) the displacement amplitude $s_0$.
Solution:
(a) From the decibel definition:
$$\beta = 10 \log_{10}\left(\frac{I}{10^{-12}}\right) = 80 \implies \log_{10}\left(\frac{I}{10^{-12}}\right) = 8 \implies \frac{I}{10^{-12}} = 10^8 \implies I = \mathbf{10^{-4}\text{ W/m}^2}$$
(b) Using $I = \frac{\Delta P_0^2}{2 \rho v}$:
$$\Delta P_0 = \sqrt{2 \rho v I} = \sqrt{2 \times 1.2 \times 340 \times 10^{-4}} = \sqrt{816 \times 10^{-4}} = \sqrt{0.0816} \approx \mathbf{0.2857\text{ N/m}^2}$$
(c) Using $\Delta P_0 = \rho v \omega s_0$, where $\omega = 2\pi(1000) = 2000\pi\text{ rad/s}$:
$$s_0 = \frac{\Delta P_0}{\rho v \omega} = \frac{0.2857}{(1.2 \times 340)(2000\pi)} = \frac{0.2857}{408 \times 6283.18} = \frac{0.2857}{2.5635 \times 10^6} \approx \mathbf{1.11 \times 10^{-7}\text{ m} = 0.111\text{ }\mu\text{m}}$$
Example 3 (Decibel Superposition of Incoherent Sources – JEE Advanced):
Ten identical uncoordinated musical instruments are playing simultaneously. At a listener’s position, each individual instrument alone produces a sound level of $60\text{ dB}$. (a) Calculate the combined sound level in decibels when all 10 instruments play together. (b) How many additional identical instruments must join so that the total sound level reaches $73\text{ dB}$?
Solution:
(a) Because the musical instruments are mutually incoherent, their intensities add directly without phase interference:
$$I_1 = I_0 \times 10^{\beta_1 / 10} = 10^{-12} \times 10^{60/10} = 10^{-6}\text{ W/m}^2$$
For $N = 10$ instruments:
$$I_{\text{total}} = 10 \times I_1 = 10 \times 10^{-6} = 10^{-5}\text{ W/m}^2$$
$$\beta_{\text{total}} = 10 \log_{10}\left(\frac{I_{\text{total}}}{I_0}\right) = 10 \log_{10}\left(\frac{10^{-5}}{10^{-12}}\right) = 10 \log_{10}(10^7) = \mathbf{70\text{ dB}}$$
(b) For a total sound level of $\beta’ = 73\text{ dB}$:
$$\beta’ – \beta_1 = 10 \log_{10}(N’) \implies 73 – 60 = 13 \implies 10 \log_{10}(N’) = 13 \implies \log_{10}(N’) = 1.3$$
$$N’ = 10^{1.3} = 10^1 \times 10^{0.301} \approx 10 \times 2 = 20\text{ instruments}$$
Therefore, the number of additional instruments required is $20 – 10 = \mathbf{10\text{ instruments}}$.
Example 4 (Point Source Spherical Inverse-Square Attenuation – JEE Advanced):
An isotropic point source of sound radiates uniformly in all directions in an open field with zero ground reflection. At distance $r_1 = 5.0\text{ m}$, the measured sound level is $\beta_1 = 90\text{ dB}$. (a) Find the total acoustic power output of the source. (b) At what distance $r_2$ will the sound level drop to the threshold of conversational comfort ($\beta_2 = 60\text{ dB}$)?
Solution:
(a) At $r_1 = 5.0\text{ m}$, $\beta_1 = 90\text{ dB} \implies I_1 = 10^{-12} \times 10^{90/10} = 10^{-3}\text{ W/m}^2$.
Total acoustic power emitted over the spherical wavefront ($A = 4\pi r_1^2$):
$$P_{\text{source}} = I_1 (4\pi r_1^2) = (10^{-3}\text{ W/m}^2)(4\pi \times 5^2\text{ m}^2) = 10^{-3} \times 100\pi = 0.1\pi \approx \mathbf{0.314\text{ W}}$$
(b) Change in sound level between $r_1$ and $r_2$:
$$\beta_1 – \beta_2 = 10 \log_{10}\left(\frac{I_1}{I_2}\right) = 10 \log_{10}\left(\frac{r_2^2}{r_1^2}\right) = 20 \log_{10}\left(\frac{r_2}{r_1}\right)$$
Given $\beta_1 – \beta_2 = 90 – 60 = 30\text{ dB}$:
$$30 = 20 \log_{10}\left(\frac{r_2}{r_1}\right) \implies \log_{10}\left(\frac{r_2}{r_1}\right) = 1.5 \implies \frac{r_2}{r_1} = 10^{1.5} = 10\sqrt{10} \approx 31.62$$
$$r_2 = 31.62 \times 5.0\text{ m} \approx \mathbf{158.1\text{ m}}$$
5. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): The ratio of the speed of sound in oxygen gas ($O_2$) to that in hydrogen gas ($H_2$) at the same temperature is:
(A) $1 : 4$
(B) $4 : 1$
(C) $1 : 16$
(D) $16 : 1$
Q2 (Single Choice): In a progressive sound wave, at a displacement node, the pressure variation is:
(A) Zero
(B) Maximum
(C) Minimum (negative maximum only)
(D) Half of the maximum value
Q3 (Single Choice): If the pressure of an ideal gas at constant temperature is doubled, the speed of sound in it:
(A) Becomes double
(B) Becomes half
(C) Becomes $\sqrt{2}$ times
(D) Remains unchanged
Q4 (Single Choice): When the intensity of sound is increased by a factor of $100$, the sound intensity level increases by:
(A) $10\text{ dB}$
(B) $20\text{ dB}$
(C) $100\text{ dB}$
(D) $2\text{ dB}$
Q5 (Single Choice): The phase difference between the displacement wave $s(x, t)$ and the pressure wave $\Delta P(x, t)$ in a sound wave is:
(A) $0$
(B) $\pi/4$
(C) $\pi/2$
(D) $\pi$
Q6 (Single Choice): A point isotropic source emits sound waves uniformly. If you double your distance from the source, the sound level in decibels drops by approximately:
(A) $3\text{ dB}$
(B) $6\text{ dB}$
(C) $9\text{ dB}$
(D) $12\text{ dB}$
Q7 (One or More than One Correct): Which of the following statements regarding the propagation of sound in gases are strictly TRUE?
(A) The speed of sound is higher in humid air than in dry air at the same temperature and pressure.
(B) Newton’s theoretical value for the speed of sound in air was lower than the actual speed because he assumed isothermal conditions.
(C) Sound can propagate through a perfect vacuum if the frequency is ultra-high.
(D) Pressure amplitude is proportional to the frequency of the sound wave for a given displacement amplitude.
Q8 (One or More than One Correct): For a plane harmonic sound wave propagating along the $+x$ direction in air:
(A) Where particles have zero displacement, the density of the gas is either maximum or minimum.
(B) Where particles have maximum displacement, the local pressure equals ambient atmospheric pressure.
(C) Particle velocity is in phase with pressure variation $\Delta P(x, t)$.
(D) Acoustic intensity is proportional to the square of the displacement amplitude.
Q9 (Numerical Value): The sound level at a distance of $3\text{ m}$ from a point source is $70\text{ dB}$. Assuming spherical spreading without absorption, the sound level in $\text{dB}$ at a distance of $30\text{ m}$ is ________.
Q10 (Numerical Value): A sound wave in air ($\rho = 1.25\text{ kg/m}^3, v = 320\text{ m/s}$) has a pressure amplitude of $40\text{ N/m}^2$. The acoustic intensity of this wave in $\text{W/m}^2$ is ________.
6. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (A) | Q2: (B) | Q3: (D) | Q4: (B) | Q5: (C) | Q6: (B) | Q7: (A, B, D) | Q8: (A, B, C, D) | Q9: 50 | Q10: 2
Detailed Solutions
Solution 1:
For both gases, $\gamma = 7/5$ (both are diatomic). Speed of sound $v = \sqrt{\frac{\gamma R T}{M_{\text{mol}}}} \propto \frac{1}{\sqrt{M_{\text{mol}}}}$.
$$\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 1 : 4$$
Hence, (A) is correct.
Solution 2:
Displacement and pressure waves are $\pi/2$ out of phase. A displacement node occurs where $s(x, t) = 0$, which corresponds to the maximum rate of spatial deformation $\left|\frac{\partial s}{\partial x}\right| = \text{max}$. Since $\Delta P = -B \frac{\partial s}{\partial x}$, a displacement node is a pressure antinode where pressure variation is maximal. Hence, (B) is correct.
Solution 3:
By the ideal gas law, $P/\rho = \frac{RT}{M_{\text{mol}}}$. At constant temperature $T$, doubling pressure doubles density, leaving the ratio $P/\rho$ invariant. Hence, $v = \sqrt{\gamma P/\rho}$ is completely unchanged. Hence, (D) is correct.
Solution 4:
Change in sound level: $\Delta \beta = 10 \log_{10}\left(\frac{I_2}{I_1}\right) = 10 \log_{10}(100) = 10 \times 2 = \mathbf{20\text{ dB}}$. Hence, (B) is correct.
Solution 5:
From $\Delta P = -B \frac{\partial s}{\partial x}$, if $s = s_0 \sin(kx – \omega t)$, then $\Delta P = \Delta P_0 \sin\left(kx – \omega t – \frac{\pi}{2}\right)$. The phase difference is $\pi/2\text{ rad}$. Hence, (C) is correct.
Solution 6:
For a point source, $I \propto 1/r^2$. When $r \to 2r$, $I \to I/4$.
$$\Delta \beta = 10 \log_{10}\left(\frac{1}{4}\right) = -10 \log_{10}(4) = -20 \log_{10}(2) \approx -20(0.301) \approx \mathbf{-6.02\text{ dB}}$$
The level drops by approximately $6\text{ dB}$. Hence, (B) is correct.
Solution 7:
(A) True: Water vapor has smaller molar mass ($18\text{ g/mol}$) than dry air ($28.8\text{ g/mol}$), making humid air less dense, so sound travels faster.
(B) True: Newton’s isothermal assumption neglected adiabatic heating in compressions.
(C) False: Sound is a mechanical wave requiring a material medium.
(D) True: $\Delta P_0 = \rho v \omega s_0 = 2\pi f \rho v s_0 \propto f$.
Hence, (A, B, D) are correct.
Solution 8:
(A) True: Displacement nodes are pressure/density antinodes.
(B) True: At displacement antinodes, $\partial s/\partial x = 0 \implies \Delta P = 0 \implies P = P_{\text{atm}}$.
(C) True: $v_p = \frac{\partial s}{\partial t} = -\omega s_0 \cos(kx – \omega t)$ and $\Delta P = -B k s_0 \cos(kx – \omega t)$. Both share the identical $-\cos(kx – \omega t)$ time-space dependence, so they oscillate strictly in phase.
(D) True: $I = \frac{1}{2}\rho v \omega^2 s_0^2 \propto s_0^2$.
Hence, all statements (A, B, C, D) are correct.
Solution 9:
Using distance attenuation for a point source:
$$\beta_2 = \beta_1 – 20 \log_{10}\left(\frac{r_2}{r_1}\right) = 70 – 20 \log_{10}\left(\frac{30}{3}\right) = 70 – 20 \log_{10}(10) = 70 – 20(1) = \mathbf{50\text{ dB}}$$
Answer is 50.
Solution 10:
Using acoustic intensity formula in terms of pressure amplitude:
$$I = \frac{\Delta P_0^2}{2 \rho v} = \frac{40^2}{2 \times 1.25 \times 320} = \frac{1600}{2.5 \times 320} = \frac{1600}{800} = \mathbf{2.0\text{ W/m}^2}$$
Answer is 2.