Welcome to Lecture 119 (Master Index Entry: L113) of our comprehensive JEE Physics Master Series! Deepening our study of continuous wave phenomena in Unit 10: Oscillations and Waves, we present the master comparative lecture on wave mechanics: Longitudinal and Transverse Waves — Mechanisms, Elastic Constraints, Pressure Formulations & Polarization. While all mechanical waves transport energy through elastic deformation, the orientation of particle oscillation relative to energy flow fundamentally alters the physical behavior of the wave. In this masterclass, we contrast the mechanical requirements of transverse and longitudinal propagation, derive the mathematical duality between displacement and pressure waves in fluids, evaluate the spatial phase shift of $\pi/2$, analyze seismic $P$-waves and $S$-waves, explain why polarization uniquely diagnoses transverse motion, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Transverse vs Longitudinal Wave Mechanics
1. Transverse Mechanical Waves
In a transverse wave, the individual particles of the medium oscillate along lines that are strictly perpendicular ($\perp$) to the direction of wave propagation:
$$\mathbf{\vec{v}_{\text{particle}} \perp \vec{v}_{\text{wave}}}$$
- Geometric Profile: Characterized by alternating crests (points of maximum positive displacement $+A$) and troughs (points of maximum negative displacement $-A$).
- Constitutive Elastic Requirement: Requires shear elasticity (modulus of rigidity, $\eta$). When a layer of medium is displaced sideways, the intermolecular forces in adjacent layers must exert restoring forces opposing the lateral shear deformation.
- Permissible Media:
- Solids: Possess non-zero rigidity modulus ($\eta > 0$) and can transmit transverse mechanical waves throughout their bulk volume.
- Liquid Surfaces: Transverse surface waves (ripples) can propagate along the free boundary of liquids where surface tension and gravity supply transverse restoring forces.
- Bulk Fluids (Liquids & Gases): Cannot support static shear stresses ($\eta = 0$). Hence, transverse mechanical waves cannot propagate in the interior of gases or liquids.
- Polarization: Because transverse particle vibrations occur in a two-dimensional plane perpendicular to the propagation axis, transverse waves can be polarized (linear, circular, or elliptical polarization).
2. Longitudinal Mechanical Waves
In a longitudinal wave, the individual particles of the medium oscillate along lines that are strictly parallel ($||$) to the direction of wave propagation:
$$\mathbf{\vec{v}_{\text{particle}} \parallel \vec{v}_{\text{wave}}}$$
- Geometric Profile: Characterized by alternating compressions (regions of elevated particle density and pressure) and rarefactions (regions of reduced particle density and pressure).
- Constitutive Elastic Requirement: Requires volume elasticity (bulk modulus, $B$). As layers compress and expand, the medium resists changes in volume.
- Permissible Media: Because every state of matter (solids, liquids, and gases) possesses volume elasticity ($B > 0$), longitudinal waves propagate through all states of matter. Sound in air and water is strictly longitudinal.
- Absence of Polarization: Since oscillations occur solely along the line of propagation, longitudinal waves cannot be polarized.
3. Mathematical Duality: Displacement Wave vs Pressure Wave
Consider a longitudinal sound wave propagating along the $+x$-direction in a fluid of ambient density $\rho$ and bulk modulus $B$. Let the longitudinal displacement of a thin fluid layer from its equilibrium position be:
$$\mathbf{s(x, t) = s_0 \cos(kx – \omega t)}$$
A fluid element initially between $x$ and $x + \Delta x$ undergoes a fractional change in volume (volumetric strain):
$$\frac{\Delta V}{V} = \frac{A \Delta s}{A \Delta x} = \frac{\partial s}{\partial x} = -k s_0 \sin(kx – \omega t)$$
From the physical definition of bulk modulus, $B = -\frac{\Delta P}{\Delta V / V}$, the excess gauge pressure $\Delta P(x, t)$ is:
$$\Delta P(x, t) = -B \frac{\partial s}{\partial x} = -B \left[-k s_0 \sin(kx – \omega t)\right] = \mathbf{B k s_0 \sin(kx – \omega t)}$$
Defining the pressure amplitude $\mathbf{\Delta P_0 = B k s_0}$:
$$\mathbf{\Delta P(x, t) = \Delta P_0 \sin(kx – \omega t) = \Delta P_0 \cos\left(kx – \omega t – \frac{\pi}{2}\right)}$$
| Physical Location in Wave | Displacement State ($s$) | Displacement Gradient ($\partial s/\partial x$) | Gauge Pressure State ($\Delta P$) |
|---|---|---|---|
| Center of Compression | $s = 0$ (Passing mean position) | Maximum negative ($-\text{max}$) | $$\Delta P = +\Delta P_0 \quad (\text{Maximum Overpressure})$$ |
| Center of Rarefaction | $s = 0$ (Passing mean position) | Maximum positive ($+\text{max}$) | $$\Delta P = -\Delta P_0 \quad (\text{Maximum Underpressure})$$ |
| Displacement Extremes | $$s = \pm s_0 \quad (\text{Turning Points})$$ | $\frac{\partial s}{\partial x} = 0$ | $$\Delta P = 0 \quad (\text{Normal Ambient Pressure } P_0)$$ |
- Crucial Phase Difference: The pressure variation wave leads/lags the displacement wave by a phase of $\mathbf{\frac{\pi}{2}\text{ radians} \ (90^\circ)}$.
A displacement node is a pressure antinode, and a displacement antinode is a pressure node! - Alternative Forms for Pressure Amplitude: Using $v = \sqrt{B/\rho} \implies B = \rho v^2$ and $k = \omega/v$: $$\mathbf{\Delta P_0 = B k s_0 = (\rho v^2)\left(\frac{\omega}{v}\right) s_0 = \rho v \omega s_0}$$
4. Acoustic Energy, Intensity & Characteristic Impedance
The time-averaged energy transported per unit cross-sectional area per unit time is the wave intensity ($I$):
$$\mathbf{I = \frac{1}{2} \rho v \omega^2 s_0^2 = \frac{(\Delta P_0)^2}{2\rho v} = \frac{(\Delta P_0)^2}{2 Z}}$$
where $\mathbf{Z = \rho v}$ is the characteristic acoustic impedance of the transmitting medium (measured in $\text{kg}/(\text{m}^2\cdot\text{s})$ or Rayls). Higher acoustic impedance means that a given pressure amplitude produces smaller displacement amplitudes.
5. Seismic Waves: Primary ($P$) vs Secondary ($S$) Waves
| Parameter | Primary ($P$) Waves | Secondary ($S$) Waves |
|---|---|---|
| Wave Type | Longitudinal (Compressional) | Transverse (Shear) |
| Propagation Speed | $$v_P = \sqrt{\frac{B + \frac{4}{3}\eta}{\rho}}$$ | $$v_S = \sqrt{\frac{\eta}{\rho}}$$ |
| Relative Arrival Time | Fastest seismic wave; arrives first ($t_P < t_S$). | Slower wave; arrives after delay $\Delta t$. |
| Transmission Through Earth’s Core | Transmitted through both solid mantle and liquid outer core. | Blocked completely by the liquid outer core ($\eta = 0$), creating the S-wave shadow zone. |
| Epicenter Distance ($d$) | $$d = \Delta t \left(\frac{v_P v_S}{v_P – v_S}\right)$$ | |
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Evaluating Pressure Amplitude & Particle Kinematics
Problem Statement: A longitudinal harmonic sound wave propagates through air of ambient density $\rho = 1.29\text{ kg/m}^3$ at speed $v = 340\text{ m/s}$. The wave has frequency $f = 1000\text{ Hz}$ and displacement amplitude $s_0 = 1.0 \times 10^{-5}\text{ m} = 10\text{ }\mu\text{m}$.
(a) Calculate the angular frequency $\omega$ and angular wave number $k$.
(b) Determine the adiabatic bulk modulus $B$ of the air.
(c) Find the pressure amplitude $\Delta P_0$ of the sound wave.
(d) Calculate the maximum particle speed $v_{p,\text{max}}$ and compare it with the wave propagation speed $v$.
Solution:
(a) Angular Frequency & Wave Number:
$$\omega = 2\pi f = 2\pi(1000\text{ Hz}) = \mathbf{2000\pi\text{ rad/s} \approx 6283\text{ rad/s}}$$
$$k = \frac{\omega}{v} = \frac{2000\pi\text{ rad/s}}{340\text{ m/s}} = \mathbf{\frac{100\pi}{17}\text{ rad/m} \approx 18.48\text{ rad/m}}$$
(b) Bulk Modulus of Air:
$$v = \sqrt{\frac{B}{\rho}} \implies B = \rho v^2 = (1.29\text{ kg/m}^3)(340\text{ m/s})^2 = 1.29 \times 115600 = \mathbf{1.491 \times 10^5\text{ Pa}}$$
(c) Pressure Amplitude:
Using $\Delta P_0 = \rho v \omega s_0$:
$$\Delta P_0 = (1.29\text{ kg/m}^3)(340\text{ m/s})(2000\pi\text{ rad/s})(1.0 \times 10^{-5}\text{ m})$$
$$\Delta P_0 = (438.6)(2\pi \times 10^{-2}) = 8.772\pi\text{ Pa} \approx \mathbf{27.56\text{ Pa}}$$
(d) Maximum Particle Speed:
$$v_{p,\text{max}} = \omega s_0 = (2000\pi\text{ rad/s})(1.0 \times 10^{-5}\text{ m}) = 0.020\pi\text{ m/s} \approx \mathbf{0.0628\text{ m/s} = 6.28\text{ cm/s}}$$
Comparison with wave speed:
$$\frac{v_{p,\text{max}}}{v} = \frac{0.0628\text{ m/s}}{340\text{ m/s}} \approx \mathbf{1.85 \times 10^{-4}}$$
The oscillating air particles move with speeds less than $0.02\%$ of the speed at which the acoustic disturbance propagates.
Example 2: Intermediate Application — Spatial Phase Distribution of Displacement & Pressure
Problem Statement: A longitudinal sound wave propagating in a gas is described by the displacement wave equation: $$s(x, t) = s_0 \sin\left(\omega t – kx + \frac{\pi}{4}\right)$$
(a) Derive the corresponding mathematical equation for the gauge pressure variation wave $\Delta P(x, t)$.
(b) Determine the phase difference between the displacement wave and the pressure variation wave.
(c) At $t = 0$, find all locations $x \in [0, \lambda)$ where:
(i) The medium displacement is zero while pressure variation is an extremum.
(ii) The medium displacement is an extremum while pressure variation is zero.
Solution:
(a) Pressure Wave Derivation:
The excess pressure is related to the spatial derivative of displacement by $\Delta P = -B \frac{\partial s}{\partial x}$.
$$\frac{\partial s}{\partial x} = \frac{\partial}{\partial x}\left[s_0 \sin\left(\omega t – kx + \frac{\pi}{4}\right)\right] = -k s_0 \cos\left(\omega t – kx + \frac{\pi}{4}\right)$$
Substituting into the pressure relation:
$$\Delta P(x, t) = -B \left[-k s_0 \cos\left(\omega t – kx + \frac{\pi}{4}\right)\right] = B k s_0 \cos\left(\omega t – kx + \frac{\pi}{4}\right)$$
Using $\Delta P_0 = B k s_0$ and $\cos\theta = \sin(\theta + \pi/2)$:
$$\mathbf{\Delta P(x, t) = \Delta P_0 \sin\left(\omega t – kx + \frac{3\pi}{4}\right)}$$
(b) Phase Difference:
Phase of pressure wave: $\Phi_P = \omega t – kx + \frac{3\pi}{4}$.
Phase of displacement wave: $\Phi_s = \omega t – kx + \frac{\pi}{4}$.
$$\mathbf{\Delta\Phi = \Phi_P – \Phi_s = \frac{3\pi}{4} – \frac{\pi}{4} = \frac{\pi}{2}\text{ rad} \ (90^\circ)}$$
The pressure variation wave leads the displacement wave by $\pi/2$ radians.
(c) Specific Spatial Coordinates at $t = 0$:
At $t = 0$, the spatial profiles are:
$$s(x, 0) = s_0 \sin\left(-kx + \frac{\pi}{4}\right) = -s_0 \sin\left(kx – \frac{\pi}{4}\right)$$
$$\Delta P(x, 0) = \Delta P_0 \cos\left(kx – \frac{\pi}{4}\right)$$
(i) Locations where $s = 0$ and $|\Delta P| = \Delta P_0$:
$$s = 0 \implies \sin\left(kx – \frac{\pi}{4}\right) = 0 \implies kx – \frac{\pi}{4} = 0, \pi$$
For $kx – \frac{\pi}{4} = 0 \implies \left(\frac{2\pi}{\lambda}\right)x = \frac{\pi}{4} \implies \mathbf{x = \frac{\lambda}{8}}$
For $kx – \frac{\pi}{4} = \pi \implies \left(\frac{2\pi}{\lambda}\right)x = \frac{5\pi}{4} \implies \mathbf{x = \frac{5\lambda}{8}}$
(ii) Locations where $|s| = s_0$ and $\Delta P = 0$:
$$\Delta P = 0 \implies \cos\left(kx – \frac{\pi}{4}\right) = 0 \implies kx – \frac{\pi}{4} = \frac{\pi}{2}, \frac{3\pi}{2}$$
For $kx – \frac{\pi}{4} = \frac{\pi}{2} \implies \left(\frac{2\pi}{\lambda}\right)x = \frac{3\pi}{4} \implies \mathbf{x = \frac{3\lambda}{8}}$
For $kx – \frac{\pi}{4} = \frac{3\pi}{2} \implies \left(\frac{2\pi}{\lambda}\right)x = \frac{7\pi}{4} \implies \mathbf{x = \frac{7\lambda}{8}}$
Example 3: Advanced Application — Earthquake Epicenter Localization via Seismic Wave Separation
Problem Statement: An earthquake fault rupture generates both longitudinal $P$-waves and transverse $S$-waves. A recording seismograph station registers the first arrival of the $P$-wave and, after a time lag $\Delta t = 20.0\text{ s}$, registers the arrival of the slower $S$-wave. The average velocities through the crust are $v_P = 8.0\text{ km/s}$ and $v_S = 5.0\text{ km/s}$.
(a) Explain physically why $P$-waves propagate faster than $S$-waves in rock.
(b) Calculate the distance $d$ from the recording station to the earthquake epicenter.
(c) Given rock density $\rho = 2800\text{ kg/m}^3$ and bulk modulus $B = 7.0 \times 10^{10}\text{ Pa}$, calculate the shear modulus $\eta$ of the crustal rock using $v_S = \sqrt{\eta/\rho}$.
Solution:
(a) Physical Basis of Speed Disparity:
In an isotropic elastic solid:
$$v_P = \sqrt{\frac{B + \frac{4}{3}\eta}{\rho}}, \quad v_S = \sqrt{\frac{\eta}{\rho}}$$
Because longitudinal deformation involves both volume compression (governed by $B$) and lateral shear distortion (governed by $\eta$), the effective elastic modulus is $B + \frac{4}{3}\eta$. For transverse waves, only shear deformation occurs. Since $B > 0$, $B + \frac{4}{3}\eta > \eta$, guaranteeing that $v_P > v_S$ in all solid media.
(b) Epicenter Distance:
Let $d$ be the straight-line distance to the fault. The transit times are:
$$t_P = \frac{d}{v_P}, \quad t_S = \frac{d}{v_S}$$
The time difference is:
$$\Delta t = t_S – t_P = d\left(\frac{1}{v_S} – \frac{1}{v_P}\right) = d\left(\frac{v_P – v_S}{v_P v_S}\right)$$
Solving for $d$:
$$\mathbf{d = \Delta t \left(\frac{v_P v_S}{v_P – v_S}\right) = (20.0\text{ s}) \left[\frac{(8.0\text{ km/s})(5.0\text{ km/s})}{8.0\text{ km/s} – 5.0\text{ km/s}}\right] = (20.0)\left(\frac{40.0}{3.0}\right) = \mathbf{\frac{800}{3}\text{ km} \approx 266.7\text{ km}}}$$
(c) Shear Modulus Calculation:
Using $v_S = 5.0\text{ km/s} = 5000\text{ m/s}$:
$$\eta = \rho v_S^2 = (2800\text{ kg/m}^3)(5000\text{ m/s})^2 = (2800)(2.5 \times 10^7) = \mathbf{7.0 \times 10^{10}\text{ Pa} = 70\text{ GPa}}$$
Example 4: JEE Advanced Scenario — Sound Wave Intensity, Overpressure & Hearing Threshold
Problem Statement: An isotropic point source radiates harmonic sound waves uniformly into open air ($\rho = 1.20\text{ kg/m}^3, v = 340\text{ m/s}$) with acoustic power $P = 60\pi\text{ W} \approx 188.5\text{ W}$ at frequency $f = 1000\text{ Hz}$.
(a) Determine the acoustic intensity $I$ at a radial distance $r = 5.0\text{ m}$ from the source.
(b) Calculate the pressure amplitude $\Delta P_0$ and displacement amplitude $s_0$ of air molecules at this location.
(c) Given the reference threshold of hearing $I_0 = 1.0 \times 10^{-12}\text{ W/m}^2$, calculate the sound intensity level $\beta$ in decibels ($\text{dB}$) at $r = 5.0\text{ m}$.
Solution:
(a) Acoustic Intensity:
For a spherical wave, power spreads over the spherical area $A = 4\pi r^2 = 4\pi (5.0\text{ m})^2 = 100\pi\text{ m}^2$:
$$\mathbf{I = \frac{P}{A} = \frac{60\pi\text{ W}}{100\pi\text{ m}^2} = 0.60\text{ W/m}^2}$$
(b) Pressure & Displacement Amplitudes:
Using $I = \frac{(\Delta P_0)^2}{2\rho v}$:
$$(\Delta P_0)^2 = 2\rho v I = 2(1.20\text{ kg/m}^3)(340\text{ m/s})(0.60\text{ W/m}^2) = 489.6\text{ Pa}^2$$
$$\mathbf{\Delta P_0 = \sqrt{489.6} \approx 22.13\text{ Pa}}$$
To find displacement amplitude $s_0$, use $\Delta P_0 = \rho v \omega s_0$:
$$\omega = 2\pi f = 2000\pi\text{ rad/s}$$
$$s_0 = \frac{\Delta P_0}{\rho v \omega} = \frac{22.13}{(1.20)(340)(2000\pi)} = \frac{22.13}{408 \times 6283.18} = \frac{22.13}{2.5635 \times 10^6} \approx \mathbf{8.63 \times 10^{-6}\text{ m} = 8.63\text{ }\mu\text{m}}$$
(c) Sound Intensity Level ($\beta$):
$$\beta = 10 \log_{10}\left(\frac{I}{I_0}\right) = 10 \log_{10}\left(\frac{0.60\text{ W/m}^2}{1.0 \times 10^{-12}\text{ W/m}^2}\right) = 10 \log_{10}(6.0 \times 10^{11})$$
$$\beta = 10 \left[\log_{10}(6.0) + 11\right] = 10 [0.77815 + 11] = 10(11.77815) \approx \mathbf{117.8\text{ dB}}$$
This level borders on the human threshold of physical pain ($\approx 120\text{ dB}$).
Practice Worksheet: 10 Target JEE Problems
Master wave classifications, pressure-displacement phase shifts, and acoustic impedance. Attempt all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. Which of the following wave phenomena is exhibited exclusively by transverse waves and can never be observed in longitudinal waves?
(A) Reflection
(B) Refraction
(C) Interference
(D) Polarization
Q2. Sound waves in an ideal gas cannot be polarized because:
(A) Their propagation speed is much smaller than light
(B) Medium particles oscillate strictly parallel to the direction of wave velocity
(C) They are mechanical waves requiring a material medium
(D) Their wavelength is significantly larger than atomic dimensions
Q3. In a sinusoidal longitudinal sound wave propagating through a gas, the phase difference between the particle displacement wave and the excess pressure variation wave is:
(A) $0$
(B) $\pi/4$
(C) $\pi/2$
(D) $\pi$
Q4. At a displacement node in a standing longitudinal sound wave formed inside an air column, which of the following physical conditions holds true?
(A) The gauge pressure variation is zero at all times
(B) The gauge pressure variation is an antinode (maximum variation)
(C) The particle density variation is identically zero
(D) The instantaneous particle velocity is maximum
Q5. An underwater explosion occurs in a deep lake. A detector stationed on the water surface receives two distinct acoustic signals: one transmitted through the water and one transmitted through the air, separated by time delay $\Delta t$. This observation proves that:
(A) Sound travels faster in water than in air because the bulk modulus of water is substantially higher
(B) Sound travels slower in water than in air because water has higher density
(C) Longitudinal waves cannot propagate through liquids
(D) Water transmits transverse waves while air transmits longitudinal waves
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. Which of the following statements is/are correct regarding transverse mechanical waves?
(A) They can propagate through a solid steel bar.
(B) They can propagate through the deep interior bulk of a freshwater lake.
(C) They can propagate along the free surface boundary of water under gravity and surface tension.
(D) Electromagnetic waves are transverse waves that do not require an elastic material medium.
Q7. For a harmonic longitudinal sound wave travelling through a fluid of density $\rho$ and bulk modulus $B$ with displacement amplitude $s_0$ and wave speed $v$:
(A) The pressure amplitude can be expressed as $\Delta P_0 = B k s_0$.
(B) The pressure amplitude can be expressed as $\Delta P_0 = \rho v \omega s_0$.
(C) The acoustic intensity is $I = \frac{(\Delta P_0)^2}{2\rho v}$.
(D) Regions of compression correspond to maximum negative displacement gradient ($\partial s/\partial x < 0$).
Q8. Which of the following statements is/are correct regarding seismic waves produced by an earthquake?
(A) Primary ($P$) waves are longitudinal compressional waves and arrive first at seismic monitoring stations.
(B) Secondary ($S$) waves are transverse shear waves and travel slower than $P$-waves.
(C) Secondary ($S$) waves cannot propagate through the liquid outer core of the Earth.
(D) The time interval $\Delta t$ between the arrivals of $P$ and $S$ waves is directly proportional to the distance to the earthquake epicenter.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. A longitudinal sound wave propagating in a gas is described by the displacement function $s(x, t) = 1.0 \times 10^{-5} \cos(2000 t – 4.0 x)$, where $s$ and $x$ are in meters and $t$ is in seconds. If the bulk modulus of the gas is $B = 2.0 \times 10^5\text{ Pa}$, calculate the pressure amplitude $\Delta P_0$ of the sound wave in $\text{Pa}$.
Q10. A seismograph station records the arrival of the primary ($P$) wave and secondary ($S$) wave from an earthquake with a time delay of $\Delta t = 25\text{ s}$. The velocities of the $P$ and $S$ waves through the crust are $v_P = 8.0\text{ km/s}$ and $v_S = 4.8\text{ km/s}$, respectively. Determine the distance $d$ (in kilometers) from the station to the earthquake epicenter.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (D) | Polarization occurs only when vibration is perpendicular to propagation |
| Q2 | (B) | Longitudinal oscillations are parallel to propagation $\implies$ no polarization |
| Q3 | (C) | $\Delta P = -B(\partial s/\partial x) \implies$ phase lead of $\pi/2$ radians |
| Q4 | (B) | Displacement node ($s=0$) is a pressure antinode ($|\Delta P|=\Delta P_0$) |
| Q5 | (A) | $v = \sqrt{B/\rho}$; high bulk modulus of water yields $v \approx 1480\text{ m/s} > 340\text{ m/s}$ |
| Q6 | (A, C, D) | Transverse mechanical waves require shear modulus; liquids lack bulk shear |
| Q7 | (A, B, C, D) | Acoustic formulations: $\Delta P_0 = Bk s_0 = \rho v \omega s_0$, intensity formula |
| Q8 | (A, B, C, D) | Seismic body waves: $P$-wave speed, $S$-wave shear barrier in liquid core |
| Q9 | 8 | $\Delta P_0 = B k s_0 = (2.0 \times 10^5)(4.0)(1.0 \times 10^{-5}) = 8\text{ Pa}$ |
| Q10 | 300 | $d = \Delta t [v_P v_S / (v_P – v_S)] = 25 [38.4 / 3.2] = 25 \times 12 = 300\text{ km}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
Polarization refers to the confinement of the oscillation vector to a single plane containing the propagation axis. Since longitudinal waves oscillate strictly along the line of propagation, there is no spatial asymmetry to polarize. Only transverse waves possess vibration vectors perpendicular to the line of propagation that can be restricted into specific orientations. Reflection, refraction, and interference are common to all wave types.
Hence, the correct choice is (D).
Solution to Q2:
In longitudinal sound waves, medium particles oscillate parallel to the direction of energy propagation. Because there are no components of motion in the plane transverse to the direction of propagation, the concept of polarization is physically inapplicable.
Hence, the correct choice is (B).
Solution to Q3:
Let displacement be $s = s_0 \cos(kx – \omega t)$. The pressure variation is:
$$\Delta P = -B \frac{\partial s}{\partial x} = -B[-k s_0 \sin(kx – \omega t)] = B k s_0 \sin(kx – \omega t) = \Delta P_0 \cos\left(kx – \omega t – \frac{\pi}{2}\right)$$
The phase angle difference between $\Delta P$ and $s$ is:
$$\Delta\Phi = \left(kx – \omega t – \frac{\pi}{2}\right) – (kx – \omega t) = -\frac{\pi}{2}\text{ rad}$$
The magnitude of the phase difference is $\mathbf{\pi/2\text{ rad} \ (90^\circ)}$. Hence, the correct choice is (C).
Solution to Q4:
In standing longitudinal sound waves, displacement and pressure variations have complementary spatial distributions:
– At a displacement node, particles are permanently at rest ($s = 0$). Adjacent layers move toward or away from this node, producing maximum compression and rarefaction. Therefore, a displacement node corresponds to a pressure antinode.
– At a displacement antinode, particle displacement is maximum, but adjacent layers move in unison without relative crowding, yielding zero pressure variation ($\Delta P = 0$).
Hence, the correct choice is (B).
Solution to Q5:
The speed of longitudinal sound in a fluid is given by $v = \sqrt{B/\rho}$. While water has a density nearly $800$ times that of air ($\rho_{\text{water}} \approx 1000\text{ kg/m}^3$ vs $\rho_{\text{air}} \approx 1.2\text{ kg/m}^3$), its bulk modulus is more than $15000$ times greater ($B_{\text{water}} \approx 2.2 \times 10^9\text{ Pa}$ vs $B_{\text{air}} \approx 1.4 \times 10^5\text{ Pa}$). Consequently, sound travels at $v \approx 1480\text{ m/s}$ in water compared to $v \approx 340\text{ m/s}$ in air. The signal through water arrives much earlier.
Hence, the correct choice is (A).
Solution to Q6:
Let us evaluate each statement:
(A) Steel possesses shear rigidity ($\eta \approx 80\text{ GPa}$) and can transmit transverse mechanical waves throughout its bulk volume. (A) is correct.
(B) Bulk liquid water has zero shear modulus and cannot support transverse waves in its interior. Thus (B) is incorrect.
(C) Water surface waves exhibit transverse motion supported by surface tension and gravity restoring forces. (C) is correct.
(D) Electromagnetic waves consist of mutually perpendicular oscillating electric and magnetic fields perpendicular to wave propagation, and require no material medium. (D) is correct.
Hence, correct choices are (A, C, D).
Solution to Q7:
(A) By definition: $\Delta P_0 = B k s_0$. (A) is correct.
(B) Using $B = \rho v^2$ and $k = \omega/v$: $\Delta P_0 = (\rho v^2)(\omega/v)s_0 = \rho v \omega s_0$. (B) is correct.
(C) Acoustic intensity is $I = \frac{1}{2}\rho v \omega^2 s_0^2 = \frac{(\rho v \omega s_0)^2}{2\rho v} = \frac{(\Delta P_0)^2}{2\rho v}$. (C) is correct.
(D) $\Delta P = -B \frac{\partial s}{\partial x}$. For compression, $\Delta P > 0 \implies \frac{\partial s}{\partial x} < 0$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
(A) Primary ($P$) waves are compressional longitudinal body waves that travel fastest through rock and arrive first. (A) is correct.
(B) Secondary ($S$) waves are shear transverse body waves whose velocity $v_S = \sqrt{\eta/\rho}$ is lower than $v_P$. (B) is correct.
(C) The Earth’s outer core is molten iron-nickel liquid with $\eta = 0$. $S$-waves cannot propagate through it, creating the global $S$-wave shadow zone. (C) is correct.
(D) The time lag is $\Delta t = d\left(\frac{1}{v_S} – \frac{1}{v_P}\right)$, which is directly proportional to distance $d$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
From the given displacement wave equation $s(x, t) = 1.0 \times 10^{-5} \cos(2000 t – 4.0 x)$:
$$s_0 = 1.0 \times 10^{-5}\text{ m}, \quad k = 4.0\text{ rad/m}$$
Given bulk modulus $B = 2.0 \times 10^5\text{ Pa}$:
The pressure amplitude is:
$$\mathbf{\Delta P_0 = B k s_0 = (2.0 \times 10^5\text{ Pa})(4.0\text{ rad/m})(1.0 \times 10^{-5}\text{ m}) = 8.0\text{ Pa} = 8\text{ Pa}}$$
Answer: 8
Solution to Q10:
Using the seismic transit time lag formula:
$$\Delta t = d\left(\frac{1}{v_S} – \frac{1}{v_P}\right) = d\left(\frac{v_P – v_S}{v_P v_S}\right)$$
Substituting $v_P = 8.0\text{ km/s}, v_S = 4.8\text{ km/s},$ and $\Delta t = 25\text{ s}$:
$$d = \Delta t \left(\frac{v_P v_S}{v_P – v_S}\right) = (25\text{ s}) \left[\frac{(8.0)(4.8)}{8.0 – 4.8}\right]\text{ km}$$
$$d = 25 \left[\frac{38.4}{3.2}\right] = 25 \times 12 = \mathbf{300\text{ km}}$$
Answer: 300