NEET · Physics

Waves — Wave Motion, Sound & Doppler Effect

Comprehensive NEET Physics lesson on wave motion, sound waves, standing waves, beats, and the Doppler effect with solved examples and practice MCQs.

Introduction to Waves

A wave is a disturbance that transfers energy from one point to another without the net transfer of matter. Waves are classified as mechanical (require a medium — sound, water, string waves) and electromagnetic (do not require a medium — light, radio waves). In NEET physics, mechanical waves, particularly sound waves and waves on strings, are the primary focus.

Key parameters of a wave: Wavelength (λ) — distance between two consecutive identical points; Frequency (f) — number of oscillations per second (Hz); Time period (T) — time for one complete oscillation (T = 1/f); Amplitude (A) — maximum displacement from equilibrium; Wave speed (v) — speed of propagation (v = fλ).

Types of Waves

Transverse Waves

In transverse waves, particles oscillate perpendicular to the direction of wave propagation. Examples: waves on a string, light waves, ripples on water surface. The wave profile at any instant is a sine or cosine function. Crests are the highest points, troughs are the lowest points. Distance between successive crests or troughs equals one wavelength.

Longitudinal Waves

In longitudinal waves, particles oscillate parallel to the direction of wave propagation. Examples: sound waves in air, seismic P-waves. The wave consists of alternate compressions (high pressure/density regions) and rarefactions (low pressure/density regions). Distance between successive compressions or rarefactions equals one wavelength. Sound waves in fluids are always longitudinal; in solids, they can be both longitudinal and transverse.

Progressive Waves

A progressive (travelling) wave transfers energy continuously. The wave function for a sinusoidal progressive wave travelling in the positive x-direction: y(x,t) = A sin(ωt − kx + φ) where ω = 2πf (angular frequency), k = 2π/λ (wave number), and φ is the initial phase. For a wave travelling in the negative x-direction: y(x,t) = A sin(ωt + kx + φ).

Example 1 — Wave Parameters
A wave is described by y = 0.02 sin(100πt − 4πx). Find amplitude, frequency, wavelength, and wave speed.

a) A=0.02m, f=50Hz, λ=0.5m, v=25m/s   b) A=0.02m, f=50Hz, λ=0.5m, v=25m/s   c) A=0.04m, f=100Hz, λ=0.25m, v=25m/s   d) A=0.02m, f=100Hz, λ=0.5m, v=50m/s
Solution: Comparing with y = A sin(ωt − kx): A = 0.02 m. ω = 100π = 2πf → f = 50 Hz. k = 4π = 2π/λ → λ = 0.5 m. v = fλ = 50 × 0.5 = 25 m/s. Option (a).

Wave Equation & Speed

Speed of Transverse Waves on a String

The speed of a transverse wave on a stretched string depends on tension T and linear mass density μ (mass per unit length): v = √(T/μ). This is derived from Newton's second law applied to a small element of the string. The speed is independent of frequency and amplitude — only the medium properties matter.

Speed of Sound Waves

The speed of longitudinal waves (sound) in a medium depends on the elastic properties and density. In a gas: v = √(γP/ρ) = √(γRT/M) where γ = Cp/Cv (adiabatic index), P is pressure, ρ is density, M is molar mass, and T is temperature. In air at NTP: v ≈ 332 m/s. The speed of sound increases with temperature: vt = v0√(1 + t/273) ≈ v0 + 0.6t m/s.

In solids: v = √(Y/ρ) for longitudinal waves, v = √(G/ρ) for transverse waves. In liquids: v = √(B/ρ) where B is bulk modulus.

Example 2 — Speed of Sound
The speed of sound in air at 27°C is 340 m/s. Find the speed of sound at 127°C.

a) 350 m/s   b) 370 m/s   c) 393 m/s   d) 400 m/s
Solution: v ∝ √T. v2/v1 = √(T2/T1) = √((273+127)/(273+27)) = √(400/300) = √(4/3) = 1.155. v2 = 340 × 1.155 = 392.7 m/s ≈ 393 m/s. Option (c).
Example 3 — String Wave Speed
A 2 m long string of mass 10 g is under tension 200 N. Find the speed of transverse waves on the string.

a) 100 m/s   b) 141 m/s   c) 200 m/s   d) 50 m/s
Solution: μ = m/L = 0.01/2 = 0.005 kg/m. v = √(T/μ) = √(200/0.005) = √(40000) = 200 m/s. Option (c).

Wave Equation — Detailed Derivation

The wave function for a sinusoidal progressive wave travelling in the positive x-direction is derived from the fact that the disturbance (displacement) at any point x and time t is a function of the form y = f(t − x/v). For a periodic wave, the source executes simple harmonic motion: y(0,t) = A sin(ωt). If the wave travels with speed v, the disturbance at a point x is delayed by time x/v. Therefore:

y(x,t) = A sin[ω(t − x/v)] = A sin(ωt − ωx/v)

Since ω/v = 2πf/v = 2π/λ = k (wave number), we get the standard form: y(x,t) = A sin(ωt − kx + φ) where φ is the initial phase. This equation satisfies the one-dimensional wave equation: ∂2y/∂t2 = v22y/∂x2.

Phase Difference & Path Difference

Phase difference (Δφ) and path difference (Δx) are related by: Δφ = (2π/λ) Δx. Similarly, time difference Δt is related as: Δφ = ωΔt = 2πfΔt. Points separated by a path difference of λ have a phase difference of 2π (they are in phase). Points separated by λ/2 have a phase difference of π (they are out of phase).

For two waves of the same frequency travelling in the same direction, the resultant amplitude depends on their phase difference: if Δφ = 0, 2π, 4π,... → constructive interference (AR = A1 + A2). If Δφ = π, 3π, 5π,... → destructive interference (AR = |A1 − A2|).

Phase Velocity

The phase velocity (wave speed) is the speed at which a point of constant phase travels: vp = ω/k = fλ. For non-dispersive media (e.g., sound in air, waves on an ideal string), the phase velocity is independent of frequency. In dispersive media, different frequencies travel at different speeds, causing a wave pulse to spread out over time.

Example 15 — Phase & Path Difference
Two points on a wave are separated by a distance of 0.25 m. If the wavelength is 1 m, find the phase difference between them.

a) π/2 rad   b) π rad   c) π/4 rad   d) 2π rad
Solution: Δφ = (2π/λ) Δx = (2π/1) × 0.25 = π/2 rad. Option (a).
NEET Key — Phase & Path Difference
Memorise: Δφ = (2π/λ) Δx = (2π/T) Δt = (2πf) Δt. A path difference of λ corresponds to a phase difference of 2π. For constructive interference: Δx = nλ (Δφ = 2nπ). For destructive interference: Δx = (2n+1)λ/2 (Δφ = (2n+1)π).

Sound Waves

Characteristics of Sound

Sound is a longitudinal mechanical wave. Human ear can detect frequencies from 20 Hz to 20,000 Hz. Frequencies below 20 Hz are infrasonic (earthquakes, whales), above 20 kHz are ultrasonic (bats, medical imaging). The loudness of sound depends on intensity (power per unit area): I = ½ρvω2A2. Sound level in decibels: β = 10 log(I/I0) where I0 = 10−12 W/m2 (threshold of hearing). Threshold of pain: 120 dB (I = 1 W/m2).

Intensity & Inverse Square Law

The intensity of sound from a point source decreases with distance: I ∝ 1/r2. The amplitude decreases as 1/r. The sound level in dB decreases by approximately 6 dB for each doubling of distance from a point source.

Reflection & Echo

Sound waves obey the laws of reflection (angle of incidence = angle of reflection). An echo is heard when the reflected sound reaches the ear with a delay ≥ 0.1 s (requires reflecting surface at least 17 m away). Reverberation is the persistence of sound due to multiple reflections. Acoustics of halls is designed to optimise reverberation time.

Reflection of Waves — Phase Change

When a wave is reflected from a boundary, the phase change depends on the nature of the boundary:

Rigid boundary (fixed end): The reflected wave undergoes a phase change of π (180°). This is equivalent to a path difference of λ/2. For a transverse wave on a string fixed at the end, the displacement of the fixed point must always be zero, so the reflected wave is inverted relative to the incident wave. The incident and reflected waves interfere to produce a node at the rigid boundary.

Free boundary (free end): The reflected wave undergoes no phase change (0°). The end is free to move, and the reflected wave is not inverted. An antinode is formed at the free boundary.

Sound waves: When sound reflects from a rigid wall (closed end of a pipe), a displacement node (pressure antinode) is formed. When sound reflects from an open end, a displacement antinode (pressure node) is formed. This explains why closed pipes have a node at the closed end and an antinode at the open end.

Phase Change on Reflection — Comparison

Table 3: Phase Change on Reflection
Boundary TypePhase ChangePath Difference EquivalentResult at BoundaryExample
Rigid (Fixed)π (180°)λ/2NodeString fixed end, closed pipe
Free (Open)0AntinodeString free end, open pipe
Rarer to denserπ (180°)λ/2Node-likeSound: air to water
Denser to rarer0Antinode-likeSound: water to air
Example 4 — Sound Intensity
The intensity of sound at a distance of 2 m from a source is 10−4 W/m2. Find the intensity at 4 m.

a) 2.5 × 10−5 W/m2   b) 5.0 × 10−5 W/m2   c) 1.0 × 10−4 W/m2   d) 0.5 × 10−4 W/m2
Solution: I ∝ 1/r2. I2/I1 = (r1/r2)2 = (2/4)2 = 1/4. I2 = 10−4/4 = 2.5 × 10−5 W/m2. Option (a).

Intensity & Amplitude Relationship

The intensity of a wave is proportional to the square of its amplitude: I ∝ A2. For a sound wave: I = ½ρvω2A2 where ρ is the density of the medium, v is the wave speed, and ω is the angular frequency. This means that doubling the amplitude quadruples the intensity, and increasing the frequency (keeping amplitude constant) also increases the intensity proportionally to f2.

In interference of two waves of equal amplitude A and phase difference Δφ, the resultant amplitude: AR = 2A cos(Δφ/2). The resultant intensity: IR = I1 + I2 + 2√(I1I2) cos(Δφ). For equal intensities I0: IR = 4I0 cos2(Δφ/2). At constructive interference (Δφ = 0): IR = 4I0. At destructive interference (Δφ = π): IR = 0.

The loudness of sound is logarithmic: β = 10 log(I/I0) dB. A 10× increase in intensity corresponds to a 10 dB increase in sound level. A 100× increase corresponds to 20 dB.

NEET Tip — Sound Level Change
For a point source, every time distance doubles, intensity drops to 1/4, which means sound level drops by 10 log(4) = 6.02 dB ≈ 6 dB. If the distance increases by a factor of 10, the sound level drops by 20 dB.
Example 16 — Superposition & Intensity
Two sound waves of equal amplitude and frequency interfere at a point. If their phase difference is 60°, find the ratio of resultant intensity to the intensity of each wave.

a) 2 : 1   b) 3 : 1   c) 4 : 1   d) 1 : 1
Solution: For equal amplitudes: IR = 4I0 cos2(Δφ/2) = 4I0 cos2(30°) = 4I0 × (0.866)2 = 4I0 × 0.75 = 3I0. Ratio IR : I0 = 3 : 1. Option (b).

Interference of Sound Waves

When two sound waves of the same frequency travel in the same medium and meet at a point, they interfere. The resultant displacement is the algebraic sum of the individual displacements (superposition principle). The nature of interference depends on the phase difference between the waves arriving at that point.

Constructive interference occurs when the waves arrive in phase (path difference = nλ). The resultant amplitude is maximum (A1 + A2). For sound, this produces a loud sound. Destructive interference occurs when the waves arrive out of phase (path difference = (2n+1)λ/2). The resultant amplitude is minimum (|A1 − A2|), producing a soft or silent region.

For two sources emitting sound in phase, the condition for constructive interference at a point is: path difference = nλ, and for destructive interference: path difference = (2n+1)λ/2. If the sources are out of phase by π, these conditions are reversed.

In Young's double-slit experiment with sound (using two loudspeakers driven by the same oscillator), alternating loud and soft regions are observed as the listener moves across the room. The angular position of maxima: sin θ = nλ/d where d is the separation between sources.

Example 23 — Interference of Sound
Two loudspeakers placed 2 m apart are driven by the same oscillator at 340 Hz. Speed of sound = 340 m/s. At what angle from the centre line does the first maximum occur?

a) 15°   b) 30°   c) 45°   d) 60°
Solution: λ = v/f = 340/340 = 1 m. For first maximum: d sin θ = λ. sin θ = λ/d = 1/2 = 0.5. θ = 30°. Option (b).

Characteristics of Musical Sound

Musical sound has three distinguishing characteristics: Pitch (determined by frequency — higher frequency = higher pitch), Loudness (determined by intensity/amplitude — greater amplitude = louder sound, measured in dB), and Quality or Timbre (determined by the waveform shape / presence of harmonics — distinguishes the same note played on different instruments).

A musical note consists of a fundamental frequency (first harmonic) along with several overtones (higher harmonics). The relative amplitudes of these harmonics determine the timbre. A pure tone has only the fundamental frequency (sine wave). Most musical instruments produce complex waves with many harmonics. The human ear can distinguish these differences, allowing us to identify different instruments playing the same note.

Consonance and Dissonance: Certain frequency combinations sound pleasant (consonance — e.g., octave 2:1, fifth 3:2, fourth 4:3). Others sound harsh (dissonance — e.g., semitone 16:15). The concept of beats explains this — consonant intervals have simple frequency ratios and produce few or no audible beats.

Standing (Stationary) Waves

Standing waves are formed by the superposition of two identical waves travelling in opposite directions. The wave function: y(x,t) = 2A sin(kx) cos(ωt). In standing waves, all particles oscillate with the same frequency but different amplitudes. Points of zero displacement are called nodes (permanent rest). Points of maximum displacement are called antinodes. Distance between successive nodes = λ/2. Distance between successive antinodes = λ/2. Distance between a node and adjacent antinode = λ/4.

Energy is not transported in standing waves — it is trapped between nodes. The medium particles oscillate between extreme positions, converting potential to kinetic energy and back. Standing waves represent the normal modes of vibration of a bounded system — only specific frequencies (eigenfrequencies) produce sustained standing waves.

Formation of Standing Waves

When a travelling wave reflects from a rigid boundary, it undergoes a phase change of π (equivalent to λ/2 path difference). From a free boundary, no phase change occurs. Standing waves are produced when a wave and its reflected counterpart interfere. In strings fixed at both ends, nodes form at the ends; in open pipes, antinodes form at the open ends.

Vibrations in Strings & Pipes

Standing Waves in Stretched Strings

For a string fixed at both ends of length L, standing waves exist only at specific frequencies (natural frequencies or harmonics). The fundamental frequency (first harmonic): f1 = v/2L. Higher harmonics: fn = nv/2L for n = 1, 2, 3... The integer n denotes the number of loops (segments). The wavelength of the nth harmonic: λn = 2L/n.

Laws of vibrating strings: (i) f ∝ 1/L (law of length), (ii) f ∝ √T (law of tension), (iii) f ∝ 1/√μ (law of mass). These are collectively known as Sonometer laws.

Example 5 — String Harmonics
A 1 m long string has mass per unit length 0.01 kg/m and tension 100 N. Find the fundamental frequency.

a) 25 Hz   b) 50 Hz   c) 75 Hz   d) 100 Hz
Solution: v = √(T/μ) = √(100/0.01) = √(10000) = 100 m/s. f1 = v/2L = 100/(2×1) = 50 Hz. Option (b).

Standing Waves in Air Columns (Pipes)

Closed pipe (closed at one end, open at the other): The closed end is always a node (displacement) / pressure antinode. The open end is always an antinode (displacement) / pressure node. Fundamental frequency: f1 = v/4L. Only odd harmonics exist: fn = (2n−1)v/4L for n = 1, 2, 3,... Wavelength: λn = 4L/(2n−1). The first three resonances correspond to n = 1 (fundamental, f1), n = 2 (third harmonic, 3f1), n = 3 (fifth harmonic, 5f1).

Open pipe (open at both ends): Both ends are displacement antinodes (pressure nodes). Fundamental frequency: f1 = v/2L. All harmonics exist: fn = nv/2L. Wavelength: λn = 2L/n. The first three resonances are the fundamental (f1), second harmonic (2f1), and third harmonic (3f1).

End correction: For real pipes, the effective length is slightly more than the geometric length because the antinode lies slightly beyond the open end: Leff = L + e where e = 0.6r (for a circular pipe of radius r) or approximately 0.3d where d is the diameter. The end correction is important in accurate determination of the speed of sound using resonance column experiments. For a closed pipe: f1 = v/4(L + e). For an open pipe: f1 = v/2(L + 2e) (both ends have end correction).

String Instruments — Laws of Vibration

The frequency of vibration of a stretched string depends on three factors:

(i) Law of length: f ∝ 1/L (for constant T and μ). Doubling the length halves the frequency. This is why bass strings in musical instruments are longer than treble strings.

(ii) Law of tension: f ∝ √T (for constant L and μ). Increasing tension increases frequency. This is how string instruments are tuned — tightening increases pitch.

(iii) Law of mass: f ∝ 1/√μ (for constant L and T). Thicker strings (higher μ) produce lower frequencies. This is why the bass strings of a guitar are thicker than the treble strings.

Combining all three: f = (1/2L) √(T/μ) which is exactly f1 = v/2L since v = √(T/μ).

Sonometer Experiment

A sonometer (or monochord) is used to verify the laws of vibrating strings. It consists of a hollow wooden box with a long wire stretched over two bridges. One end of the wire is fixed, the other passes over a pulley carrying a load to provide tension. The length of the vibrating segment can be adjusted by moving a movable bridge. A tuning fork of known frequency is struck and held near the wire, and the length is adjusted until the wire vibrates in resonance with the fork (maximum loudness). Using the sonometer, one can verify:

(a) f ∝ 1/L: For constant T and μ, measure L for different tuning forks. fL = constant.

(b) f ∝ √T: For constant L, vary tension and find the resonant frequency using a set of tuning forks.

(c) f ∝ 1/√μ: Use wires of different materials/thicknesses (different μ) keeping L and T constant.

The sonometer is also used to determine the unknown frequency of a tuning fork and to find the AC mains frequency using a magnet and solenoid arrangement.

Example 17 — String Vibration Law
A stretched string vibrates with a fundamental frequency of 100 Hz. If the tension is increased by 44%, find the new fundamental frequency.

a) 120 Hz   b) 144 Hz   c) 110 Hz   d) 80 Hz
Solution: f ∝ √T. New tension T' = 1.44T. f'/f = √(T'/T) = √(1.44) = 1.2. f' = 100 × 1.2 = 120 Hz. Option (a).

Echo, Reverberation & Acoustics of Buildings

Echo: An echo is the reflection of sound that reaches the listener after a delay of at least 0.1 s (the persistence of hearing). Since the speed of sound in air is approximately 340 m/s, the minimum distance for a distinct echo is: d = vt/2 = 340 × 0.1/2 = 17 m. If the reflecting surface is closer than 17 m, the reflected sound overlaps with the original sound and is perceived as reverberation rather than a distinct echo.

Multiple Echoes: When sound reflects multiple times between two or more surfaces, multiple echoes are heard. This is commonly observed in hills, canyons, large empty halls, and stairwells. The time interval between successive echoes depends on the distances between reflecting surfaces.

Reverberation: Reverberation is the persistence of sound in an enclosed space after the original sound source has stopped. It is caused by multiple reflections of sound from walls, ceiling, and floor. The reverberation time (RT60) is the time required for the sound intensity to decay by 60 dB (i.e., to one-millionth of its original intensity) after the source stops. Sabine's formula for reverberation time is: RT = 0.161V/A, where V is the volume of the room (in m3) and A is the total absorption (in sabins). An optimal reverberation time is crucial for different applications: concert halls (1.8–2.2 s), theatres (1.0–1.5 s), classrooms (0.5–0.8 s), recording studios (0.2–0.4 s).

Acoustic Treatment: Sound absorption can be increased using porous materials (fibreglass, foam, carpets, curtains), resonant panels, or Helmholtz resonators. Diffusers scatter sound waves to prevent echoes and standing waves. Bass traps absorb low-frequency sound. The choice of materials and their placement determines the acoustic quality of a hall.

Sabine's Law: Wallace Sabine found empirically that RT = 0.161V/Σ(αiSi), where αi is the absorption coefficient of surface i (ranging from 0 for perfect reflection to 1 for perfect absorption) and Si is its surface area. Common absorption coefficients: brick wall 0.03, carpet 0.3–0.6, acoustic tiles 0.5–0.9, audience 0.8–0.9 per person.

Example 30 — Minimum Echo Distance
A person claps their hands in front of a large wall. If the speed of sound is 340 m/s and the minimum time delay for a distinct echo is 0.1 s, what is the minimum distance of the wall from the person?

a) 17 m   b) 34 m   c) 51 m   d) 68 m
Solution: The sound travels to the wall and back (total distance = 2d). Time = 2d/v. For a distinct echo, 2d/v ≥ 0.1 s. d ≥ v × 0.1/2 = 340 × 0.05 = 17 m. Option (a).
Example 31 — Speed of Sound by Echo Method
A person stands 85 m from a cliff and fires a gun. The echo is heard after 0.5 s. Find the speed of sound in air.

a) 320 m/s   b) 330 m/s   c) 340 m/s   d) 350 m/s
Solution: Sound travels 2 × 85 = 170 m in 0.5 s. v = 170/0.5 = 340 m/s. Option (c).
NEET Key — Echo & Reverberation
Minimum distance for echo: d = v/20 (since 2d/v ≥ 0.1 s, so d ≥ v/20). For v = 340 m/s, d ≥ 17 m. If the reflecting surface is closer, the reflected sound reinforces the original (reverberation). In NEET questions, always remember to account for the round trip: 2d = vt. If a question states "echo is heard after t seconds", the total distance travelled by sound is 2d = vt, where d is the distance to the reflector.

Human Ear & Hearing

The human ear is a remarkable transducer that converts sound waves (pressure variations) into electrical signals that the brain interprets. The audible frequency range for young humans is typically 20 Hz to 20,000 Hz (20 kHz). With age, the upper limit gradually decreases to about 12–16 kHz by middle age — this is called presbycusis (age-related hearing loss).

Structure of the Ear: The outer ear (pinna and ear canal) collects and directs sound waves to the eardrum. The middle ear (three small bones: malleus, incus, stapes) amplifies the vibrations and transmits them to the oval window of the inner ear. The inner ear (cochlea) is a fluid-filled, spiral-shaped organ lined with hair cells of different lengths, each tuned to a specific frequency. Different regions of the cochlea respond to different frequencies (tonotopy): the base responds to high frequencies, the apex to low frequencies.

Loudness Perception: The perceived loudness depends on both intensity and frequency. The threshold of hearing is about 10−12 W/m2 at 1 kHz. The threshold of pain is about 1 W/m2. The ear is most sensitive in the 2–5 kHz range (conversational speech frequencies). Equal-loudness contours (Fletcher–Munson curves) show the sound pressure level required at different frequencies to produce the same perceived loudness.

Sound Localisation: Humans can localise the direction of sound using two cues: interaural time difference (ITD) — the time delay between sound reaching the two ears (effective for frequencies below 1.5 kHz when the wavelength is larger than the head); and interaural level difference (ILD) — the difference in sound intensity at the two ears (effective for frequencies above 1.5 kHz when the head acts as a barrier creating a sound shadow).

Resonance & Tuning

Resonance occurs when a system is driven at its natural frequency, resulting in a large amplitude of vibration. The driving force is in phase with the velocity of the oscillator, maximising energy transfer. In the absence of damping, the amplitude would theoretically become infinite at resonance.

Examples of resonance:

Tacoma Narrows Bridge (1940): Wind-induced vibrations matched the bridge's natural frequency, causing large-amplitude torsional oscillations that destroyed the bridge. This is a classic cautionary example of mechanical resonance.

Musical instruments: The air column in a flute/guitar body resonates at specific frequencies, amplifying certain harmonics. The soundboard of a piano or violin resonates to amplify the string vibrations. The resonance of the air column in an organ pipe determines the pitch produced.

Other examples: Tuning a radio circuit (electrical resonance), microwave oven (water molecules resonate at ~2.45 GHz), breaking a wine glass with sound (acoustic resonance), and soldiers breaking step while crossing a bridge to avoid resonant excitation.

In the context of sound, resonance is used to determine the speed of sound (resonance column experiment), to tune instruments, and in the design of musical instruments and auditoriums.

Example 6 — Closed Pipe
A closed organ pipe of length 0.4 m produces its fundamental frequency. If the speed of sound is 320 m/s, find the frequency.

a) 100 Hz   b) 200 Hz   c) 300 Hz   d) 400 Hz
Solution: For closed pipe: f1 = v/4L = 320/(4×0.4) = 320/1.6 = 200 Hz. Option (b).
Example 7 — Open Pipe
An open organ pipe of length 0.5 m produces its fundamental frequency. If speed of sound is 340 m/s, find the frequency of the third harmonic.

a) 340 Hz   b) 510 Hz   c) 680 Hz   d) 1020 Hz
Solution: For open pipe: f1 = v/2L = 340/(2×0.5) = 340 Hz. Third harmonic: f3 = 3f1 = 1020 Hz. Option (d).
NEET Trick — Open vs Closed Pipe
Open pipe: all harmonics (n = 1,2,3,...). Closed pipe: only odd harmonics (n = 1,3,5,...). The fundamental frequency of an open pipe is twice that of a closed pipe of the same length. For the same fundamental frequency, closed pipe length is half of open pipe length.

Resonance Column Experiment

This experiment uses a closed pipe with variable length (water column) to determine the speed of sound. The first resonance occurs at L1 = λ/4 (approximately). The second resonance at L2 = 3λ/4. Therefore, λ = 2(L2 − L1) and v = fλ. This is a classic NEET experiment-based question.

Example 8 — Resonance Column
In a resonance column experiment, the first resonance occurs at 16 cm and the second at 50 cm. If the tuning fork frequency is 512 Hz, find the speed of sound.

a) 328 m/s   b) 340 m/s   c) 348 m/s   d) 332 m/s
Solution: λ = 2(L2 − L1) = 2(50 − 16) = 68 cm = 0.68 m. v = fλ = 512 × 0.68 = 348.16 m/s ≈ 348 m/s. Option (c).
Example 18 — Resonance Column with End Correction
In a resonance column experiment, the first resonance occurs at 16 cm and the second at 50 cm. If the tuning fork frequency is 512 Hz, find the speed of sound and the end correction.

a) 348 m/s, 1 cm   b) 348 m/s, 0.67 cm   c) 340 m/s, 1 cm   d) 332 m/s, 0.67 cm
Solution: λ = 2(L2 − L1) = 2(50 − 16) = 68 cm = 0.68 m. v = fλ = 512 × 0.68 = 348.16 m/s ≈ 348 m/s. End correction e = (L2 − 3L1)/2 = (50 − 48)/2 = 1 cm. Option (a).
NEET Key — Organ Pipe Frequencies
Open pipe: fn = nv/2L (n = 1,2,3,...). Closed pipe: fn = (2n−1)v/4L (n = 1,2,3,...). The fundamental frequency of a closed pipe is half that of an open pipe of the same length. For a given fundamental frequency, the closed pipe is half as long as an open pipe. Remember: Closed = Odd harmonics only (1,3,5,...). Open = All harmonics (1,2,3,...).
NEET Key — Resonance Conditions
Resonance occurs when driving frequency equals natural frequency. For a string: fn = nv/2L. For a closed pipe: fn = (2n−1)v/4L. For an open pipe: fn = nv/2L. The natural frequencies depend on the boundary conditions (fixed/free ends). In a resonance column, the water level is adjusted until the air column resonates with a tuning fork — this occurs at L = (2n−1)λ/4.

Beats

When two sound waves of nearly equal frequencies (f1 and f2) travel in the same direction and superimpose, the intensity at a point varies periodically with time. This phenomenon is called beats. The beat frequency: fbeat = |f1 − f2|. The number of beats per second equals the difference in frequencies. Beat period Tbeat = 1/fbeat.

Beat Frequency — Mathematical Derivation

Consider two waves of equal amplitude A and slightly different angular frequencies ω1 and ω2: y1 = A sin(ω1t − k1x) and y2 = A sin(ω2t − k2x). At a fixed point x = 0 (for simplicity), the displacements are: y1 = A sin(ω1t) and y2 = A sin(ω2t). By the principle of superposition:

y = y1 + y2 = A[sin(ω1t) + sin(ω2t)]

Using the trigonometric identity sin α + sin β = 2 sin((α+β)/2) cos((α−β)/2):

y = 2A cos[(ω1 − ω2)t/2] × sin[(ω1 + ω2)t/2]

The resultant wave has an average angular frequency (ω1 + ω2)/2 (≈ either frequency) and its amplitude is modulated by the cos term with angular frequency (ω1 − ω2)/2. The intensity varies as cos2((ω1 − ω2)t/2). The intensity maxima occur twice per modulation cycle (when cos = ±1), so the beat frequency is twice the modulation frequency:

fbeat = 2 × [(ω1 − ω2)/4π] = |f1 − f2|

The human ear can distinguish beats up to about 10–15 beats per second. Beyond that, the beats merge into a continuous roughness or dissonance.

Graphical Representation of Beats

If we plot the displacement of the two individual waves and their resultant, we observe that the resultant amplitude varies sinusoidally between 2A (constructive interference) and 0 (destructive interference). The envelope of the resultant wave is given by 2A cos(π(f1−f2)t). The time interval between successive maxima (loud sounds) is: tmax = 1/|f1 − f2|. During one beat period, the resultant wave completes approximately (f1 + f2)/2fbeat oscillations.

Applications of beats include: tuning musical instruments (two instruments produce beats when slightly out of tune; adjusting to minimise beats brings them into unison), detecting small frequency differences in Doppler shift measurements, and in heterodyne detection used in radio receivers and ultrasonic imaging.

Beats in instrument tuning: When tuning a violin or guitar, the musician sounds a reference tuning fork alongside the string. If beats are heard, the string is slightly out of tune. The tension is adjusted until the beat frequency drops to zero (unison). Similarly, piano tuners use beat frequencies to tune intervals.

Beats with more than two waves: When multiple waves with closely spaced frequencies interfere, the beat pattern becomes more complex. For n waves with frequencies f, f+Δf, f+2Δf, ..., the resultant is a periodic pulse train — this is the principle behind mode-locked lasers and certain musical effects like chorus and phasing.

Detection of partials in musical instruments: Beats can also occur between the harmonics of a single note (e.g., between the 2nd and 3rd harmonics if they are not perfectly harmonic). This causes the sensation of "roughness" and is an important factor in the perception of musical consonance and dissonance.

Example 9 — Beats
Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. Find the beat frequency.

a) 2 Hz   b) 4 Hz   c) 6 Hz   d) 8 Hz
Solution: fbeat = |256 − 260| = 4 Hz. So 4 beats per second. Option (b).
Example 10 — Unknown Frequency Using Beats
An unknown tuning fork when sounded with a 256 Hz fork gives 4 beats/s. When loaded with a little wax, the beat frequency increases to 6 beats/s. Find the unknown frequency.

a) 250 Hz   b) 252 Hz   c) 260 Hz   d) 262 Hz
Solution: Initially funknown = 256 ± 4 = 252 Hz or 260 Hz. Adding wax decreases frequency. If frequency increases on loading, unknown was lower (i.e., 252 Hz, which would give 256−252 = 4 beats/s). After loading, f decreases further, so |256 − fnew| > 4, giving 6 beats/s. This matches: 252 − wax → lower → beat freq increases. So f = 252 Hz. Option (b).
NEET Trick — Beat Frequency Analysis
When an unknown fork produces n beats/s with a known fork of frequency fk, the unknown frequency is fk ± n. To determine which, use loading (adding wax lowers frequency) or filing (removing material raises frequency). If beat frequency increases on loading, unknown was lower than known; if beat frequency decreases, unknown was higher.
Example 19 — Beats Application
A fork of unknown frequency gives 5 beats when sounded with a 256 Hz fork. When the unknown fork is loaded with wax, the beat frequency becomes 3 beats. Find the unknown frequency.

a) 251 Hz   b) 259 Hz   c) 261 Hz   d) 253 Hz
Solution: Unknown f = 256 ± 5 = 251 Hz or 261 Hz. Loading decreases frequency. If unknown was 251 Hz, loading would decrease it further, making beat frequency > 5 (not 3). If unknown was 261 Hz, loading decreases it towards 256 Hz, reducing beat frequency to 3. So f = 261 Hz. Option (c).
Example 24 — Beats in Instrument Tuning
A guitar string and a 440 Hz tuning fork produce 4 beats/s. When the string tension is increased slightly, the beat frequency becomes 2 beats/s. Was the original frequency of the string above or below 440 Hz, and what was it?

a) 436 Hz, below   b) 444 Hz, above   c) 436 Hz, above   d) 444 Hz, below
Solution: Original f = 440 ± 4 = 436 Hz or 444 Hz. Increasing tension increases frequency. If f = 436 Hz, increasing tension moves it towards 440 Hz, decreasing beat frequency (from 4 to 2) — this matches. If f = 444 Hz, increasing tension moves it away from 440 Hz, increasing beat frequency — doesn't match. So f = 436 Hz, below 440 Hz. Option (a).
NEET Key — Beat Frequency Formula
fbeat = |f1 − f2|. The beat period T = 1/fbeat.
NEET Key — Musical Scale & Harmonics
The Western musical scale is based on the following frequency ratios relative to the tonic (sa): Sa=1:1, Re=9:8, Ga=5:4, Ma=4:3, Pa=3:2, Dha=5:3, Ni=15:8, Sa'=2:1 (octave). The 12-tone equal temperament scale divides the octave into 12 semitones, each with a frequency ratio of 21/12 ≈ 1.0595. A frequency difference of 1 semitone corresponds to about 6% change. The human ear can detect frequency changes of about 0.3% (3 Hz at 1000 Hz) under ideal conditions.

Musical Instruments — Sound Production

Different musical instruments produce sound through different mechanisms:

String instruments (violin, guitar, sitar, veena): Sound is produced by vibrating strings. The string vibrations are transmitted to the soundboard/hollow body, which resonates and amplifies the sound. The pitch is determined by the length, tension, and mass per unit length of the string. Harmonics are produced by lightly touching the string at nodal points (e.g., at L/2 for the 2nd harmonic).

Wind instruments (flute, clarinet, saxophone, shehnai): Sound is produced by the vibration of an air column inside the instrument. Flutes and shehnais behave like open pipes (both ends open), producing all harmonics. Clarinets behave like closed pipes (one end closed by the reed), producing only odd harmonics. The pitch is changed by opening/closing holes along the instrument, which changes the effective length of the air column.

Percussion instruments (drums, tabla, mridangam): Sound is produced by the vibration of a stretched membrane. The vibrations are more complex than strings because membranes are 2D. The pitch depends on the tension and area of the membrane. Tabla and mridangam are unique in that the loading (syahi) on the drumhead is carefully designed to produce specific harmonic overtones.

Brass instruments (trumpet, French horn, trombone): Sound is produced by the vibration of the player's lips against the mouthpiece. The air column resonates at frequencies determined by the length of the tube. The player can select different harmonics by changing lip tension, and the length is changed using valves or a sliding tube.

In all cases, the fundamental principle is the same: a vibrating system (string, air column, membrane) resonates at its natural frequencies, producing standing waves. The superposition of these harmonics creates the characteristic timbre of each instrument.

Quality of Sound — Timbre and Waveform Analysis

The quality or timbre of a sound is determined by the relative amplitudes of the fundamental frequency and its harmonics. Fourier analysis shows that any periodic waveform can be expressed as a sum of sine waves: y(t) = A1 sin(ωt + φ1) + A2 sin(2ωt + φ2) + A3 sin(3ωt + φ3) + ... where An are the amplitudes of the harmonics and φn are their phases.

A pure sine wave (single frequency) sounds like a tuning fork or a flute (in its fundamental register). A square wave (odd harmonics with amplitudes 1/n) sounds like a clarinet. A sawtooth wave (all harmonics with amplitudes 1/n) sounds like a violin or string instrument. A triangle wave (odd harmonics with amplitudes 1/n2) has a mellower sound.

The human ear can distinguish these different waveforms even when the fundamental frequency is the same, because the inner ear (cochlea) performs a frequency analysis, separating the sound into its component frequencies. The brain interprets the relative strengths of these frequencies as timbre.

Doppler Effect — Extended Applications

Laser Doppler Velocimetry (LDV): A laser beam is split into two beams that intersect at the measurement point. Particles moving through the intersection scatter light, and the Doppler shift in the scattered light frequency is proportional to the particle velocity. This non-contact technique is used for measuring flow velocities in fluids, blood flow in biomedical research, and vibration analysis of structures.

Doppler in Satellite Communication: Low Earth orbit (LEO) satellites move at speeds of about 7.5 km/s relative to a ground station. This causes a Doppler shift of about ±3.5 kHz for a 400 MHz UHF signal and proportionally more for higher frequencies. Ground stations must continuously adjust their receiver frequency to compensate for this shift — a process called Doppler correction or frequency tracking. The Doppler shift pattern (increasing as the satellite approaches, then decreasing as it recedes) is also used to determine the satellite's orbital parameters.

Acoustic Doppler Current Profiler (ADCP): An ADCP is a sonar device that measures water current velocities by transmitting sound pulses at a fixed frequency and measuring the Doppler shift of the echoes reflected from suspended particles (sediment, plankton, bubbles) in the water column. By measuring the shift at different depths, a profile of current velocity vs. depth can be obtained. ADCPs are used in oceanography, river monitoring, and harbour navigation.

Doppler in Astrophysics — Redshift & Blueshift: The Doppler effect for light (relativistic Doppler effect) is used extensively in astronomy to measure the radial velocities of stars, galaxies, and other celestial objects. When a star or galaxy is moving away from Earth, its spectral lines are shifted to longer wavelengths (redshift); when moving toward Earth, they shift to shorter wavelengths (blueshift). The redshift z = (λobserved − λemitted)/λemitted ≈ v/c for v << c. Edwin Hubble's observation of the redshift of distant galaxies led to the discovery of the expansion of the universe (Hubble's law: v = H0d). The current value of the Hubble constant H0 ≈ 70 km/s/Mpc.

Exoplanet Detection using Doppler Spectroscopy: The radial velocity method (Doppler spectroscopy) is one of the most successful techniques for detecting exoplanets. As an exoplanet orbits its parent star, the star wobbles around the centre of mass of the star-planet system. This wobble causes periodic Doppler shifts in the star's spectral lines. By measuring the amplitude and period of these shifts, astronomers can determine the planet's orbital period, eccentricity, and minimum mass. This technique has discovered thousands of exoplanets, including 51 Pegasi b (the first exoplanet found around a Sun-like star, 1995) and the TRAPPIST-1 system of seven Earth-sized planets.

Example 32 — Doppler in Satellite Communication
A LEO satellite transmits at 400 MHz. If the maximum Doppler shift observed at a ground station is ±3.5 kHz, what is the approximate orbital speed of the satellite? (Take c = 3 × 108 m/s)

a) 2.6 km/s   b) 5.2 km/s   c) 7.8 km/s   d) 10.4 km/s
Solution: For EM waves, Δf/f = v/c. v = c × Δf/f = 3 × 108 × 3500/(400 × 106) = 3 × 108 × 8.75 × 10−6 = 2625 m/s ≈ 2.6 km/s. This is the relative speed along the line of sight. The actual orbital speed is higher because the satellite is not moving directly toward/away from the ground station. Option (a).
Example 33 — Hubble Expansion
A distant galaxy is found to have a redshift z = 0.1. If H0 = 70 km/s/Mpc, what is the approximate distance to the galaxy? (1 Mpc = 3.26 × 106 light years)

a) 143 Mpc   b) 429 Mpc   c) 714 Mpc   d) 1000 Mpc
Solution: For small redshifts, v = cz = 3 × 105 × 0.1 = 3 × 104 km/s. Hubble's law: v = H0d. d = v/H0 = 3 × 104/70 = 428.6 Mpc ≈ 429 Mpc. Option (b).

In problems where a loaded fork produces different beats, remember: loading decreases frequency, filing increases frequency. If beat frequency increases after loading, the unknown was lower than the known frequency. If beat frequency decreases after loading, the unknown was higher.

Doppler Effect

The Doppler effect is the apparent change in frequency of a wave when the source, observer, or both are in relative motion. For sound waves, the formula depends on the velocities of source (vs) and observer (vo) relative to the medium:

f' = f × (v ± vo) / (v ∓ vs)

where v is the speed of sound in the medium. Sign convention: Numerator: + if observer moves towards source, − if away. Denominator: − if source moves towards observer, + if away. The apparent frequency increases when source and observer approach each other, and decreases when they recede.

Derivation of Doppler Effect Formulas

Case 1 — Source moving towards stationary observer: The source moves with velocity vs towards the observer. In one second, the source emits f waves. The first wave travels a distance v, while the source itself moves vs. The waves are compressed into a shorter distance (v − vs). The apparent wavelength: λ' = (v − vs)/f. The apparent frequency: f' = v/λ' = vf/(v − vs).

Case 2 — Source moving away from stationary observer: The waves are stretched: λ' = (v + vs)/f. Apparent frequency: f' = vf/(v + vs).

Case 3 — Observer moving towards stationary source: The observer intercepts more waves per second. Relative speed of sound with respect to observer = v + vo. The apparent frequency: f' = (v + vo)/λ = (v + vo)f/v.

Case 4 — Observer moving away from stationary source: Relative speed = v − vo. Apparent frequency: f' = (v − vo)f/v.

Case 5 — Both source and observer moving: Combine the effects: f' = f(v ± vo)/(v ∓ vs). The signs depend on directions of motion relative to each other.

Effect of wind: If wind is blowing with speed w from source to observer, the effective speed of sound changes to veff = v ± w (depending on direction). Replace v with veff in all formulas. Note that wind affects the wave speed, not the source or observer velocities relative to the ground.

Supersonic motion: When vs > v, the source moves faster than the waves it produces. The wavefronts form a cone (Mach cone) with half-angle θ given by sin θ = v/vs. The ratio vs/v is called the Mach number. The Doppler formula breaks down when vs ≥ v as the denominator becomes zero or negative.

Apparent Frequency Formulas — Quick Reference

Table 4: Doppler Effect — All Cases
ScenarioSource MotionObserver MotionApparent Frequency f'Effect
Source → stationary OTowards O (vs)Stationaryf × v/(v − vs)Increases
Source ← stationary OAway from O (vs)Stationaryf × v/(v + vs)Decreases
Stationary S → observerStationaryTowards S (vo)f × (v + vo)/vIncreases
Stationary S ← observerStationaryAway from S (vo)f × (v − vo)/vDecreases
Both towards each otherTowards O (vs)Towards S (vo)f × (v + vo)/(v − vs)Increases (strong)
Both away from each otherAway from O (vs)Away from S (vo)f × (v − vo)/(v + vs)Decreases (strong)
Source → away, O → towardsAway from O (vs)Towards S (vo)f × (v + vo)/(v + vs)Depends on magnitudes
Perpendicular motionPerpendicular to S-O lineAnyf (unchanged)No change
NEET Cheat Sheet — Doppler Effect Signs
Rule: f' = f(v ± vo)/(v ∓ vs). Words: TOWARDS = numerator +, denominator − (because waves are compressed). AWAY = numerator −, denominator + (waves stretched). The top signs go together (+ in numerator, − in denominator for approach). The bottom signs go together (− in numerator, + in denominator for recession). If the source and observer are not moving along the same line, only the component along the line joining them matters.

Special Cases of Doppler Effect

1. Source moving, observer stationary: f' = f × v/(v ∓ vs).

2. Observer moving, source stationary: f' = f × (v ± vo)/v.

3. Both source and observer moving: Use the general formula with appropriate signs.

4. Wind blowing from source to observer: Effective speed of sound changes. If wind speed is w in the direction from source to observer: veff = v + w for the forward direction.

5. Source moving perpendicular to the line joining source-observer: No Doppler shift (only radial component matters).

6. Source moving supersonically (vs > v): The source outruns its own waves, producing a conical shock wave (Mach cone). The Mach angle: sin θ = v/vs. This is the basis of sonic booms.

Example 11 — Source Moving Towards Observer
A source of sound of frequency 500 Hz is moving towards a stationary observer at 30 m/s. Speed of sound = 330 m/s. Find the apparent frequency.

a) 455 Hz   b) 500 Hz   c) 550 Hz   d) 600 Hz
Solution: f' = f × v/(v − vs) = 500 × 330/(330 − 30) = 500 × 330/300 = 500 × 1.1 = 550 Hz. Option (c).
Example 12 — Observer Moving Towards Source
An observer moves towards a stationary sound source of frequency 400 Hz at 20 m/s. Speed of sound = 340 m/s. Find the apparent frequency.

a) 376 Hz   b) 400 Hz   c) 424 Hz   d) 440 Hz
Solution: f' = f × (v + vo)/v = 400 × (340 + 20)/340 = 400 × 360/340 = 400 × 1.0588 = 423.5 Hz ≈ 424 Hz. Option (c).
Example 13 — Both Moving
A source of frequency 600 Hz moves at 20 m/s towards an observer moving at 10 m/s away from the source. Speed of sound = 340 m/s. Find the apparent frequency.

a) 566 Hz   b) 600 Hz   c) 636 Hz   d) 650 Hz
Solution: Source moving towards: denominator v − vs. Observer moving away: numerator v − vo. f' = 600 × (340 − 10)/(340 − 20) = 600 × 330/320 = 600 × 1.03125 = 618.75 Hz ≈ 619 Hz. Option (a) 566 Hz if observer moves towards, (c) 636 Hz if observer moves towards and source moves away. Check: if observer moves away from source approaching, then effective vo is away = −10 in numerator. f' = 600(340+(−10))/(340−20) = 600×330/320 = 619 Hz.

Doppler Effect in Reflection (Echo)

When sound reflects from a moving surface (e.g., a moving wall), the Doppler shift occurs twice — the wall acts first as an observer (moving), then as a source (moving at the same velocity). The final apparent frequency after reflection: f'' = f × (v + vw)/(v − vw) for a wall moving towards the source. This principle is used in radar speed guns and medical ultrasound (Doppler echocardiography).

Applications of Doppler Effect

Medical Ultrasound (Echocardiography): Ultrasonic waves (2–10 MHz) are directed at blood cells. The reflected waves undergo a Doppler shift proportional to the velocity of blood flow. By analysing the shift, doctors can measure blood flow velocity in the heart and blood vessels, detect blockages, and assess heart valve function. Colour Doppler imaging maps blood flow velocities onto a 2D image.

Radar Speed Guns: A radar gun emits microwaves that reflect off a moving vehicle. The frequency shift between the emitted and received waves is proportional to the vehicle's speed: Δf = 2f0v/c, where c is the speed of light. The factor of 2 accounts for the double Doppler shift (out and back).

Astronomy (Redshift): The Doppler effect for light waves causes a shift in spectral lines. When a star or galaxy moves away from Earth, its spectral lines shift to longer wavelengths (redshift). The amount of redshift reveals the recessional velocity, which Edwin Hubble used to discover the expansion of the universe. Blueshift occurs when objects move towards Earth.

Flow Measurement (Laser Doppler Velocimetry): A laser beam is split and directed at a flowing fluid. Light scattered by particles in the fluid is Doppler-shifted. By measuring the shift, the flow velocity can be determined without disturbing the flow.

Weather Radar: Doppler radar is used in meteorology to detect the motion of raindrops and hail. The frequency shift indicates the speed and direction of wind and precipitation, helping forecasters track storms and tornadoes.

Military Sonar: Submarines use active sonar (sound navigation and ranging) to detect objects. The Doppler shift of the reflected sound reveals whether the target is moving towards or away from the submarine, and its speed.

Example 14 — Doppler in Reflection
A sound source of frequency 1000 Hz is stationary. A wall moves towards the source at 10 m/s. Speed of sound = 330 m/s. Find the frequency of the reflected wave heard by a stationary observer near the source.

a) 1000 Hz   b) 1031 Hz   c) 1062 Hz   d) 1094 Hz
Solution: The wall acts as a moving observer: fwall = 1000 × (330 + 10)/330 = 1000 × 340/330 = 1030.3 Hz. The wall then acts as a moving source towards the observer: f'' = fwall × 330/(330 − 10) = 1030.3 × 330/320 = 1030.3 × 1.03125 = 1062 Hz. Option (c).
Example 20 — Doppler with Wind
A source of frequency 500 Hz is stationary. Wind is blowing from the source to an observer at 20 m/s. Speed of sound in still air = 340 m/s. Find the apparent frequency heard by the stationary observer.

a) 500 Hz   b) 529 Hz   c) 472 Hz   d) 544 Hz
Solution: Wind carries the sound waves, effectively increasing the speed of sound. veff = v + w = 340 + 20 = 360 m/s. Since both source and observer are stationary relative to the ground: f' = f × veff/veff = f = 500 Hz. The wind does NOT by itself cause a shift if both are stationary in the ground frame. Option (a).
Example 22 — Doppler Shift for Receding Source
A fire engine siren emitting 800 Hz sound moves away from a stationary observer at 20 m/s. Speed of sound = 340 m/s. Find the apparent frequency and the percentage change.

a) 756 Hz, 5.5%   b) 800 Hz, 0%   c) 848 Hz, 6%   d) 756 Hz, 5.9%
Solution: f' = f × v/(v + vs) = 800 × 340/(340 + 20) = 800 × 340/360 = 800 × 0.9444 = 755.6 Hz ≈ 756 Hz. % change = (800 − 756)/800 × 100 = 44/800 × 100 = 5.5%. Option (a).
Example 25 — Doppler in Medicine
Blood flowing at 0.5 m/s in an artery is examined using 5 MHz ultrasound. Speed of sound in tissue = 1540 m/s. Find the frequency shift due to the Doppler effect (assuming the beam is along the flow direction).

a) 1.6 kHz   b) 3.2 kHz   c) 4.8 kHz   d) 6.4 kHz
Solution: For reflection from a moving scatterer: Δf = 2f0v/vs. Here v = 0.5 m/s, f0 = 5 MHz, vs = 1540 m/s. Δf = 2 × 5 × 106 × 0.5 / 1540 = 3246.75 Hz ≈ 3.2 kHz. Option (b).
NEET Key — Doppler Effect Summary
f' = f(v ± vo)/(v ∓ vs). Rule: Use + vo if observer moves TOWARDS source (catches more waves). Use − vs if source moves TOWARDS observer (waves compressed). Remember: Towards = numerator +, denominator −. Away = numerator −, denominator +.

Ultrasonic Waves & Applications

Ultrasonic waves are sound waves with frequencies above 20 kHz (the upper limit of human hearing). They have important applications due to their high frequency (short wavelength), which allows better resolution and directional beams. Common frequencies used in applications range from 20 kHz to several GHz.

Properties of ultrasonic waves: (i) They are highly directional and can be focused into narrow beams. (ii) They can penetrate solids and liquids. (iii) They produce cavitation in liquids (formation and collapse of tiny bubbles, releasing energy). (iv) They are reflected at interfaces between different media (used in echolocation). (v) They cause heating when absorbed by materials.

Applications: (1) Medical imaging (sonography) — ultrasonic waves are reflected from different tissue layers; the echoes are processed to create images of internal organs, foetus, etc. (2) Therapeutic ultrasound — used in physiotherapy for deep heating of tissues, and in surgery (high-intensity focused ultrasound, HIFU) to destroy tumours. (3) Sonar (Sound Navigation and Ranging) — ships and submarines use ultrasonic waves to detect underwater objects and measure depth. (4) Non-destructive testing (NDT) — ultrasonic waves are used to detect cracks and flaws in metal structures, welds, and pipelines. (5) Industrial cleaning — ultrasonic baths use cavitation to remove dirt from instruments, jewellery, and electronic components. (6) Insect and rodent repellers — devices that emit high-frequency sound to deter pests. (7) Welding plastics — ultrasonic vibrations generate heat at the interface of plastic parts, fusing them together.

Detection of ultrasonic waves: Since they cannot be heard by the human ear, special detectors are used: piezoelectric transducers (quartz or ceramic crystals that generate electric signals when subjected to ultrasonic pressure), magnetostriction oscillators, and microphones designed for high-frequency response. The same transducer often serves as both transmitter and receiver (pulse-echo principle).

In NEET, questions on ultrasonics typically focus on their properties (directional, high energy, penetrative) and the principle of echo-based distance measurement: distance = (v × t)/2, where t is the time between emission and reception of the reflected pulse.

Example 26 — Ultrasonic Distance Measurement
A ship sends an ultrasonic pulse towards the seabed and receives the echo after 0.6 s. Speed of sound in water = 1500 m/s. Find the depth of the sea.

a) 450 m   b) 500 m   c) 900 m   d) 300 m
Solution: Distance = (v × t)/2 = (1500 × 0.6)/2 = 900/2 = 450 m. The factor of 2 accounts for the round trip. Option (a).

Shock Waves & Sonic Boom

When a source moves through a medium at a speed greater than the wave speed (vs > v), it creates a shock wave. The wavefronts form a cone (Mach cone) with the source at the apex. The half-angle of the cone (Mach angle) is given by: sin θ = v/vs = 1/M, where M = vs/v is the Mach number.

When an aircraft breaks the sound barrier (M > 1), the shock wave produces a sonic boom — a sudden, loud sound heard on the ground as the cone of compressed air passes by. The sonic boom is not a single event at the moment of breaking the barrier; rather, it is continuously generated as long as the aircraft travels supersonically. The boom is heard as two distinct thumps (from the nose and tail shock waves).

Key points for NEET: (1) Subsonic: vs < v (M < 1). (2) Supersonic: vs > v (M > 1). (3) Mach angle decreases as speed increases: sin θ = v/vs. (4) The Doppler formula is not applicable for vs ≥ v because the denominator becomes zero or negative — the source outruns its own waves. (5) A bullet fired from a rifle produces a mini sonic boom (crack sound) as it travels supersonically. (6) The crack of a whip is a mini sonic boom — the tip of the whip moves faster than sound.

Example 27 — Mach Angle
An aircraft flies at Mach 2. Find the Mach angle.

a) 15°   b) 30°   c) 45°   d) 60°
Solution: M = vs/v = 2. sin θ = v/vs = 1/M = 1/2 = 0.5. θ = 30°. Option (b).
NEET Key — Shock Waves & Sonic Boom
sin θ = v/vs = 1/M. As the speed increases, the Mach angle decreases — the cone becomes narrower. For M = 1 (sound speed), θ = 90°. For M → ∞, θ → 0°. The sonic boom is heard on the ground as a double thump when the Mach cone intersects the ground. Remember: Doppler formulas break down for vs ≥ v.

Numerical Problem Solving — Waves

Solving numerical problems in waves requires a systematic approach. Here are step-by-step strategies for each type of problem commonly asked in NEET:

Strategy 1 — Wave Parameters from Equation

Given y = A sin(ωt ± kx + φ), identify: Amplitude A (coefficient of sin), ω (coefficient of t), k (coefficient of x). Then: f = ω/2π, T = 2π/ω, λ = 2π/k, v = ω/k = fλ. The sign determines direction: − for +x direction, + for −x direction.

Strategy 2 — Standing Wave Problems

Identify boundary conditions: (a) String fixed at both ends — nodes at ends, fn = nv/2L. (b) Open pipe — antinodes at both ends, fn = nv/2L. (c) Closed pipe — node at closed end, antinode at open end, fn = (2n−1)v/4L. For string, n = number of loops. For pipes, n = 1,2,3,... For closed pipe, remember only odd harmonics exist.

Strategy 3 — Doppler Effect Problems

Step 1: Draw a diagram showing source, observer, and their velocities. Step 2: Determine direction of motion relative to each other. Step 3: Apply formula f' = f(v ± vo)/(v ∓ vs). Step 4: For reflection, treat the reflector as first an observer, then a source. Step 5: For wind, adjust the effective speed of sound. Remember: only the component of velocity along the line joining source and observer matters.

Strategy 4 — Beat Frequency Problems

Step 1: fbeat = |f1 − f2|. Step 2: For unknown fork, f = fknown ± fbeat. Step 3: To determine sign, use loading (decreases f) or filing (increases f). Step 4: If beat frequency increases after loading, unknown < known. If beat frequency decreases after loading, unknown > known.

Strategy 5 — Speed of Sound Problems

Use v = √(γRT/M) for gases. Remember v ∝ √T (at constant composition). vt = v0√(1 + t/273) ≈ v0 + 0.6t. For resonance column: λ = 2(L2 − L1), v = fλ, end correction e = (L2 − 3L1)/2.

Common NEET Pitfalls to Avoid

(1) Confusing open and closed pipe harmonics — open: all harmonics, closed: only odd. (2) Forgetting the factor of 2 in round-trip problems (echo, sonar). (3) Using the wrong sign in Doppler effect — always check "towards = increases frequency". (4) Confusing ω (angular frequency) with f (frequency) — ω = 2πf. (5) Mixing up node and antinode positions — node: zero displacement, antinode: maximum displacement. (6) Forgetting that wave speed on a string depends on T and μ only, not on f. (7) Applying v = fλ incorrectly — this holds for all waves but the values of f and λ change according to the medium.

NEET Exam Strategy — Waves
In NEET, expect 3–5 questions from Waves. Topics most frequently asked: (1) Doppler effect formula application, (2) Standing waves in strings and pipes (harmonics), (3) Beat frequency problems, (4) Speed of sound dependence on temperature, (5) Phase/path difference relationship. Practice with the formula sheet and remember that numerical accuracy is critical — carry calculations to 2–3 decimal places.
NEET Key — Doppler Effect Quick Reference
For approach: f' > f. For recession: f' < f. Formula: f' = f(v ± vo)/(v ∓ vs). Sign rule: +vo when observer moves toward source (catches more waves per second). −vs when source moves toward observer (compresses wavelength). Mnemonic: NATO = Numerator +ve for observer Approaching, TOward; Denominator −ve for source Toward Observer.

Advanced Topics in Wave Motion

Group Velocity: In dispersive media where wave speed depends on frequency (e.g., light in glass, deep water waves), a wave pulse travels at the group velocity vg = dω/dk, which may differ from the phase velocity vp = ω/k. For non-dispersive media (sound in air, string waves), vg = vp. The group velocity represents the speed of energy propagation, while phase velocity is the speed of individual wave crests.

Fourier Analysis of Waveforms: Any periodic wave can be decomposed into a sum of sinusoidal components (harmonics) using Fourier series. A square wave contains odd harmonics (1, 3, 5,...) with amplitudes inversely proportional to the harmonic number. A sawtooth wave contains all harmonics (1, 2, 3,...). The Fourier transform extends this to non-periodic waves, representing them as a continuous spectrum of frequencies.

Non-linear Waves & Solitons: When the amplitude of a wave is very large, the medium response becomes non-linear, and the wave speed depends on amplitude. This can lead to the formation of shock waves (in sound) or solitons (solitary waves that maintain their shape during propagation). Solitons were first observed by Scott Russell in 1834 on the Edinburgh-Glasgow canal and are important in fibre optics communications, plasma physics, and tsunami propagation.

Water Waves (Surface Waves): Water waves are a combination of transverse and longitudinal motion. Water particles move in circular or elliptical paths. The wave speed depends on water depth: for deep water (depth > λ/2), v = √(gλ/2π); for shallow water (depth << λ), v = √(gh). Tsunamis are shallow-water waves even in the deep ocean because their wavelength (100–500 km) is much larger than the ocean depth (4–5 km). This is why tsunamis travel at speeds up to 700 km/h in the deep ocean.

Seismic Waves: Earthquakes produce two main types of waves: P-waves (primary, longitudinal, faster, v ≈ 5–8 km/s in crust) and S-waves (secondary, transverse, slower, v ≈ 3–5 km/s in crust, cannot travel through liquids). The time difference between the arrival of P and S waves at a seismograph station is used to determine the distance to the earthquake epicentre. Surface waves (Love and Rayleigh waves) travel along the Earth's surface and are responsible for most of the damage during earthquakes.

Example 28 — Group vs Phase Velocity
For a particular medium, the dispersion relation is ω = ak2. Find the group velocity in terms of the phase velocity.

a) vg = vp   b) vg = 2vp   c) vg = vp/2   d) vg = 4vp
Solution: vp = ω/k = ak2/k = ak. vg = dω/dk = d(ak2)/dk = 2ak = 2vp. Option (b).
Example 29 — Seismic Wave Timing
P-waves travel at 6 km/s and S-waves at 4 km/s. At a seismograph station, the S-wave arrives 10 s after the P-wave. Find the distance to the epicentre.

a) 60 km   b) 100 km   c) 120 km   d) 240 km
Solution: Let d be the distance. Time for P-wave: tP = d/6. Time for S-wave: tS = d/4. Difference: tS − tP = d/4 − d/6 = d(3−2)/12 = d/12 = 10 s. d = 120 km. Option (c).
NEET Key — Seismic Waves
P-waves (longitudinal) arrive first, S-waves (transverse) arrive second. S-waves cannot travel through liquids (outer core). The time delay Δt = d(1/vS − 1/vP) gives the epicentre distance. In solids, both P and S waves can propagate. In fluids, only P waves (sound) can propagate. This is important for understanding Earth's internal structure.

Quick Comparison Tables

Table 1: Transverse vs Longitudinal Waves
PropertyTransverse WavesLongitudinal Waves
Particle vibrationPerpendicular to propagationParallel to propagation
Requires medium?Mechanical ones do; EM do notAlways requires a material medium
Can travel in solids?Yes (and on surfaces)Yes
Can travel in fluids?No (except surface waves)Yes
Can travel in vacuum?EM waves can; mechanical cannotNo
Wave speed formula (solids)v = √(G/ρ)v = √(Y/ρ)
Wave speed formula (fluids)N/Av = √(B/ρ)
Waveform featuresCrests & troughsCompressions & rarefactions
Periodic shapeCan be polarizedCannot be polarized
ExamplesString waves, light, water ripples, seismic S-wavesSound in air/water, seismic P-waves, spring slinky
Energy transferTransports energy perpendicular to vibrationTransports energy parallel to vibration
Table 2: Open vs Closed Pipe Comparison
PropertyOpen PipeClosed Pipe
End conditionsBoth ends open (A at ends)One closed (N), one open (A)
Displacement at endsAntinode at both endsNode at closed end, antinode at open end
Pressure at endsNode at both endsAntinode at closed end, node at open end
Fundamental frequencyf1 = v/2Lf1 = v/4L
Harmonics presentAll harmonics (n = 1,2,3,...)Only odd harmonics (n = 1,3,5,...)
General frequencyfn = nv/2Lfn = (2n−1)v/4L
Fundamental wavelengthλ1 = 2Lλ1 = 4L
Second resonance2nd harmonic (2f1)3rd harmonic (3f1)
Third resonance3rd harmonic (3f1)5th harmonic (5f1)
End correctionLeff = L + 2e (both ends)Leff = L + e (one end)

Key NEET Formulas — Waves

Basic Wave Parameters:
v = fλ   ω = 2πf   T = 1/f   k = 2π/λ
y = A sin(ωt − kx + φ)   (progressive wave, +x direction)

Phase & Path Difference:
Δφ = (2π/λ) Δx = ωΔt   vp = ω/k = fλ

Wave Speed:
vstring = √(T/μ)   vgas = √(γP/ρ) = √(γRT/M)
vsolid = √(Y/ρ) (longitudinal)   vsolid = √(G/ρ) (transverse)
vliquid = √(B/ρ)   vt = v0√(1 + t/273) ≈ v0 + 0.6t m/s

Standing Waves — Strings (both ends fixed):
fn = nv/2L   (n = 1,2,3,...)   λn = 2L/n
f ∝ 1/L   f ∝ √T   f ∝ 1/√μ

Standing Waves — Open Pipe:
fn = nv/2L (n = 1,2,3,...)   λn = 2L/n

Standing Waves — Closed Pipe:
fn = (2n−1)v/4L (n = 1,2,3,...)   λn = 4L/(2n−1)

End Correction:
Leff = L + e   (e = 0.6r for closed pipe)
Leff = L + 2e   (for open pipe, both ends)
e = (L2 − 3L1)/2   (resonance column)

Beats:
fbeat = |f1 − f2|   Tbeat = 1/fbeat
y = 2A cos(π(f1−f2)t) sin(2π(f1+f2)t/2)

Doppler Effect:
f' = f(v ± vo)/(v ∓ vs)
Source towards stationary O: f' = f × v/(v − vs)
Source away from stationary O: f' = f × v/(v + vs)
O towards stationary source: f' = f × (v + vo)/v
O away from stationary source: f' = f × (v − vo)/v
Moving wall (reflection): f'' = f × (v + vw)/(v − vw)

Intensity & Sound Level:
I = ½ρvω2A2   I ∝ A2   I ∝ 1/r2
β = 10 log(I/I0)   (I0 = 10−12 W/m2)
Interference: IR = I1 + I2 + 2√(I1I2) cos Δφ

Resonance Column:
λ = 2(L2 − L1)   e = (L2 − 3L1)/2

Mach Number:
sin θ = v/vs   M = vs/v
Standing wave patterns for first three harmonics in a string fixed at both ends
n=1 (Fundamental)NNAn=2 (First Overtone)Nn=3 (Second Overtone)NN
Closed and open pipe standing wave patterns for fundamental mode
Closed PipeOpenNAλ/4Open PipeOpenOpenAAλ/2
Doppler effect showing wavefront compression as source moves right
SvsStationary source: concentric circular wavefrontsSvsMoving source: wavefronts compressed ahead, stretched behind
NEET Preparation Tips for Waves
(1) String and pipe problems are the most common — practice identifying nodes and antinodes. (2) In Doppler effect, always draw a diagram showing the direction of motion before applying signs. (3) Beat frequency problems often involve unknown frequency determination — use loading/filing logic carefully. (4) Remember that wave speed on a string depends only on T and μ, not on f or A. (5) In resonance column, the end correction e = (L2 − 3L1)/2. (6) For closed pipes, only odd harmonics are present — an important NEET distinction. (7) The intensity of sound is proportional to the square of amplitude and frequency.

⚠️ NEET Exam Tips & Common Mistakes

▶ Frequently Made Mistakes

Frequency vs velocity: Frequency of a wave depends ONLY on the source, not on the medium. Wavelength and velocity change with medium, but frequency stays constant.

Phase difference vs path difference: Δφ = (2π/λ)Δx. A path difference of λ corresponds to a phase difference of 2π (360°). Don't confuse the two.

Wave equation sign convention: y = A sin(ωt − kx) → wave travelling in +x direction. y = A sin(ωt + kx) → wave travelling in −x direction. The sign before kx determines direction.

Beats formula: Beat frequency = |f₁ − f₂|. The number of beats per second equals the difference in frequencies, NOT the sum.

Doppler effect sign: When source moves TOWARDS observer: apparent f increases (higher pitch). When source moves AWAY: f decreases. Use the correct sign convention: f' = f(v±v₀)/(v∓vs).

Open vs closed pipe overtones: Open pipe: all harmonics present (f₀, 2f₀, 3f₀...). Closed pipe: only odd harmonics (f₀, 3f₀, 5f₀...). NEET frequently tests this difference.

Node vs antinode: At nodes, displacement is always zero (destructive interference). At antinodes, displacement is maximum (constructive interference). Distance between adjacent nodes = λ/2.

Transverse vs longitudinal: Light = transverse (particles vibrate ⟂ to wave direction). Sound = longitudinal (particles vibrate ∥ to wave direction). Remember: sound cannot travel in vacuum.

▶ Shortcut Formulas for NEET

⚡ Wave speed on a string: v = √(T/μ). T = tension, μ = mass per unit length.

⚡ Speed of sound in air: v = √(γP/ρ) = √(γRT/M). At STP, v ≈ 332 m/s. Increases by 0.6 m/s per °C rise.

⚡ Fundamental frequency of stretched string: f₀ = (1/2L)√(T/μ).

⚡ For open pipe: f₀ = v/2L, for closed pipe: f₀ = v/4L.

⚡ Beat frequency = |f₁ − f₂|. If you hear x beats/sec and one frequency is known, the unknown is f±x.

⚡ Doppler effect for sound: f' = f(v±v₀)/(v∓vs). Numerator: + if observer moves toward source. Denominator: − if source moves toward observer.

⚡ In resonance tube experiment: first resonance = λ/4, second = 3λ/4. End correction e = (l₂ − 3l₁)/2.

Practice Questions

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