Class 11Physics · Oscillations & WavesFull chapter

Waves

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Transverse and Longitudinal Waves

Quick answer In a transverse wave, particles vibrate perpendicular to the direction of wave travel; in a longitudinal wave, they vibrate along it, creating compressions and rarefactions.

A mechanical wave is a disturbance that transfers energy through a medium without any net transport of matter; only the disturbance (and energy) travels while individual particles oscillate about their mean positions.

Waves are broadly classified by the relationship between the direction of particle vibration and the direction of wave propagation.

  • Transverse wave: particles of the medium vibrate perpendicular to the direction of propagation of the wave. Crests (maximum positive displacement) and troughs (maximum negative displacement) are formed. Example: waves on a stretched string, waves on the surface of water, and electromagnetic waves (which need no medium). Transverse mechanical waves can travel only through media possessing shear elasticity (rigidity), i.e. solids and stretched strings, because sideways restoring forces are required.
  • Longitudinal wave: particles of the medium vibrate parallel to (along) the direction of propagation. Regions of crowded particles are called compressions and regions of spread-out particles are called rarefactions. Example: sound waves in air, waves in a spring pushed and pulled along its length. Longitudinal waves need only volume elasticity and can therefore travel through solids, liquids and gases.

For a longitudinal wave, the distance between two consecutive compressions (or two consecutive rarefactions) equals one wavelength, λ, exactly as the distance between two consecutive crests equals one wavelength for a transverse wave.

Worked example.

Given: a sound wave travels through air with speed v = 340 m/s. The distance measured between two successive compressions is 1.7 m.

Formula: wavelength λ = distance between successive compressions; wave speed v = ν λ, so frequency ν = v / λ.

Substitution: λ = 1.7 m, so ν = 340 / 1.7

Result: ν = 200 Hz.

Wave speed–wavelength–frequency relation v = ν λ v = wave speed, ν = frequency, λ = wavelength
General 1-D progressive wave y(x,t) = f(x − vt) travelling along +x; f(x + vt) travels along −x
Remember
  • Transverse waves: particle motion perpendicular to propagation direction; crests and troughs.
  • Longitudinal waves: particle motion parallel to propagation direction; compressions and rarefactions.
  • Transverse mechanical waves need shear elasticity (rigidity); longitudinal waves need only volume elasticity.
  • Sound in air/gases is always longitudinal; waves on a string are transverse.
  • Distance between two successive compressions/rarefactions = one wavelength λ.

Displacement Relation for a Progressive Wave

Quick answer A harmonic progressive wave is written as y(x,t) = a sin(kx − ωt); its amplitude, wavelength, period, frequency and speed can all be read off directly from this equation.

A one-dimensional sinusoidal (harmonic) progressive wave travelling along the positive x-direction is represented as:

y(x, t) = a sin(kx − ωt)

where each symbol has a precise meaning:

  • Amplitude (a): the maximum displacement of a particle of the medium from its equilibrium position.
  • Wavelength (λ): the distance between two consecutive points that are in the same phase (e.g. two successive crests); related to the propagation constant k by k = 2π/λ.
  • Period (T): the time taken by a particle to complete one full oscillation; related to angular frequency ω by ω = 2π/T.
  • Frequency (ν): the number of oscillations a particle completes per second, ν = 1/T.
  • Wave speed (v): the speed at which the phase (e.g. a crest) advances through the medium, v = ω/k = ν λ.

The quantity (kx − ωt) is called the phase of the wave; every point on the medium having the same phase at a given instant is displaced by the same amount.

Worked example.

Given: a transverse wave travelling on a string is described by y(x,t) = 0.005 sin(80x − 3t), with x, y in metres and t in seconds.

Formula: compare with y = a sin(kx − ωt) to identify a, k, ω; then λ = 2π/k, T = 2π/ω, ν = 1/T, v = ω/k.

Substitution: a = 0.005 m, k = 80 rad/m, ω = 3 rad/s.

  • λ = 2π/80 = 0.0785 m ≈ 7.85 × 10−2 m
  • T = 2π/3 = 2.09 s
  • ν = 1/2.09 = 0.48 Hz
  • v = 3/80 = 0.0375 m/s = 3.75 cm/s

Result: amplitude 5 mm, wavelength ≈ 7.85 cm, period ≈ 2.09 s, frequency ≈ 0.48 Hz, wave speed = 3.75 cm/s (all along +x).

Displacement relation y(x,t) = a sin(kx − ωt)
Propagation constant k = 2π/λ
Angular frequency ω = 2π/T = 2πν
Wave speed v = ω/k = ν λ
Remember
  • y(x,t) = a sin(kx − ωt) describes a harmonic wave moving along +x; a sin(kx + ωt) moves along −x.
  • Propagation constant k = 2π/λ; angular frequency ω = 2π/T = 2πν.
  • Wave speed v = ω/k = νλ is the speed of the phase/waveform, not the particle's own speed.
  • Amplitude, wavelength, period, frequency and speed can all be read directly from the wave equation.

Speed of a Travelling Wave

Quick answer Wave speed depends only on the elastic and inertial properties of the medium: v = √(T/μ) for a stretched string and v = √(γP/ρ) (Laplace's corrected formula) for sound in a gas.

The speed of a mechanical wave is fixed entirely by the properties of the medium (its elasticity and inertia) — it does not depend on the frequency or amplitude of the wave.

  • Transverse wave on a stretched string: v = √(T/μ), where T is the tension in the string and μ is the mass per unit length (linear mass density).
  • Longitudinal wave in a medium (general): v = √(E/ρ), where E is the relevant modulus of elasticity (Young's modulus Y for a solid rod, bulk modulus B for a liquid) and ρ is the density.
  • Longitudinal wave (sound) in a gas — Newton's formula: Newton proposed v = √(P/ρ), assuming compressions and rarefactions occur isothermally. This underestimates the measured speed of sound in air by about 15%.
  • Laplace's correction: Laplace pointed out that compressions and rarefactions occur too fast for heat exchange, so the process is adiabatic. The corrected formula is v = √(γP/ρ), where γ = Cp/Cv is the ratio of specific heats of the gas.

Worked example (Laplace's correction for sound in air).

Given: atmospheric pressure P = 1.01 × 105 Pa, density of air ρ = 1.29 kg/m3, γ (air) = 1.4.

Formula: v = √(γP/ρ)

Substitution: v = √[(1.4 × 1.01 × 105) / 1.29] = √(1.096 × 105)

Result: v ≈ 331 m/s, matching the experimentally measured speed of sound in air at 0 °C (Newton's uncorrected formula gives only about 280 m/s, about 15% lower).

Speed on a stretched string v = √(T/μ) T = tension, μ = mass per unit length
Speed in a solid rod v = √(Y/ρ)
Speed in a liquid v = √(B/ρ)
Newton's formula (isothermal, gas) v = √(P/ρ)
Laplace's corrected formula (adiabatic, gas) v = √(γP/ρ) γ = Cp/Cv
Remember
  • Wave speed depends only on the medium's elasticity and inertia, not on frequency or amplitude.
  • String: v = √(T/μ). Solid rod: v = √(Y/ρ). Liquid: v = √(B/ρ).
  • Newton's formula v = √(P/ρ) assumes isothermal changes and underestimates sound speed in gases.
  • Laplace corrected it to v = √(γP/ρ) using the adiabatic assumption, matching experiment (≈331 m/s in air, versus Newton's ≈280 m/s — about 15% lower).

Principle of Superposition of Waves

Quick answer When two or more waves overlap, the net displacement at any point is simply the algebraic sum of the displacements each wave would produce alone.

The principle of superposition states that when two or more waves travel through the same region of a medium simultaneously, the resultant displacement of any particle at any instant is the algebraic sum of the displacements each wave would have produced individually:

y(x,t) = y1(x,t) + y2(x,t) + …

This holds because the wave equation is linear in y, and it explains phenomena such as interference, beats, and standing waves.

Consider two waves of the same frequency and amplitude a, arriving at a point with a phase difference φ between them. Using the superposition principle, the resultant is again a simple harmonic oscillation of amplitude:

R = 2a cos(φ/2)

  • When φ = 0, 2π, 4π, … (in phase), R = 2a — constructive interference (maximum amplitude).
  • When φ = π, 3π, … (out of phase), R = 0 — destructive interference (waves cancel).

Worked example.

Given: two waves of equal amplitude a1 = a2 = 0.02 m and the same frequency arrive at a point with a phase difference φ = 60° = π/3 rad.

Formula: resultant amplitude R = √(a12 + a22 + 2a1a2 cos φ), which for a1 = a2 = a reduces to R = 2a cos(φ/2).

Substitution: R = 2 × 0.02 × cos(30°) = 0.04 × 0.866

Result: R ≈ 0.0346 m = 3.46 cm.

Principle of superposition y = y1 + y2 + … + yn
Resultant of two waves (general phase difference) R = √(a1² + a2² + 2 a1 a2 cos φ)
Resultant, equal amplitudes R = 2a cos(φ/2)
Remember
  • Superposition: net displacement = algebraic sum of individual wave displacements (valid because the wave equation is linear).
  • Resultant of two equal-amplitude waves of same frequency: R = 2a cos(φ/2), φ = phase difference.
  • φ = 0 (or a multiple of 2π) gives constructive interference; φ = π (an odd multiple) gives destructive interference.
  • Superposition underlies interference, standing waves, and beats.

Reflection of Waves and Standing Waves in Strings and Organ Pipes

Quick answer A wave reflected from a boundary combines with the incident wave to form a stationary (standing) wave with fixed nodes and antinodes; strings and organ pipes support only specific discrete (normal-mode) frequencies.

Reflection of waves. When a travelling wave meets a boundary, part or all of it is reflected. The nature of reflection depends on the boundary:

  • At a rigid (fixed) boundary — e.g. a string end tied to a wall — the reflected wave undergoes a phase reversal of π (180°); a crest returns as a trough. The fixed point remains a node (zero displacement).
  • At a free (open) boundary — e.g. the open end of an air column — the wave is reflected without any phase change, and that end becomes an antinode (maximum displacement).

Standing (stationary) waves. When two identical waves travelling in opposite directions (the incident and reflected waves) superpose, the result is a standing wave — the waveform no longer travels; certain points (nodes) never move, while points midway between them (antinodes) oscillate with maximum amplitude. Adjacent nodes (or adjacent antinodes) are separated by λ/2, and a node and its neighbouring antinode are separated by λ/4.

Normal modes of a string fixed at both ends (e.g. a sonometer wire or a guitar string of length L): both ends must be nodes, so only wavelengths satisfying L = nλ/2 (n = 1, 2, 3, …) are allowed. The normal-mode (natural) frequencies are:

νn = n v / (2L), n = 1, 2, 3, …

All harmonics (integer multiples of the fundamental) are possible.

Organ pipes. In a pipe closed at one end, the closed end must be a node and the open end an antinode, allowing only odd harmonics:

νn = (2n − 1) v / (4L), n = 1, 2, 3, …

In a pipe open at both ends, both ends are (approximately) antinodes, so, as for the stretched string, all harmonics are allowed:

νn = n v / (2L), n = 1, 2, 3, …

Worked example.

Given: a string of length L = 1.0 m, fixed at both ends, along which transverse waves travel with speed v = 200 m/s.

Formula: νn = n v / (2L)

Substitution: fundamental, n = 1: ν1 = (1 × 200)/(2 × 1); first overtone, n = 2: ν2 = (2 × 200)/(2 × 1)

Result: fundamental frequency ν1 = 100 Hz; first overtone (second harmonic) ν2 = 200 Hz.

Standing wave (superposition of incident + reflected) y = 2a sin(kx) cos(ωt) nodes where sin(kx) = 0
Node/antinode spacing adjacent nodes (or antinodes) separated by λ/2; node–antinode separated by λ/4
String fixed both ends / pipe open both ends νn = n v / (2L), n = 1,2,3,…
Pipe closed at one end νn = (2n−1) v / (4L), n = 1,2,3,…
Remember
  • Rigid boundary: phase reversal (π) on reflection, that end is a node. Free boundary: no phase change, that end is an antinode.
  • Standing wave = superposition of incident + reflected wave; nodes are permanently at rest, antinodes oscillate maximally.
  • Adjacent nodes (or antinodes) are λ/2 apart; a node and its adjacent antinode are λ/4 apart.
  • String fixed at both ends / pipe open at both ends: νn = n v/(2L) — all harmonics.
  • Pipe closed at one end: νn = (2n−1) v/(4L) — only odd harmonics.

Beats

Quick answer Beats are periodic rises and falls in loudness heard when two sound waves of slightly different frequencies superpose; the beat frequency equals the difference of the two frequencies.

When two sound waves of nearly equal (but not identical) frequencies ν1 and ν2 and equal amplitude a travel together, their superposition produces a resultant whose amplitude rises and falls periodically. This periodic waxing and waning of loudness is called beats.

If y1 = a sin(2πν1t) and y2 = a sin(2πν2t), superposition gives:

y = y1 + y2 = [2a cos(2π Δν t)] sin(2π ν̄ t)

where Δν = (ν1 − ν2)/2 and ν̄ = (ν1 + ν2)/2. This is interpreted as a wave of frequency ν̄ (close to both original frequencies) whose amplitude is modulated by the slowly varying term 2a cos(2πΔνt).

Loudness is maximum whenever cos(2πΔνt) = ±1; since both +1 and −1 give a maximum, the number of loud beats heard per second (the beat frequency) works out to be simply the difference of the two original frequencies:

νbeat = |ν1 − ν2|

Beats are commonly used to tune musical instruments: two notes are adjusted until the beat frequency falls to zero, indicating equal frequencies.

Worked example.

Given: two tuning forks of frequencies ν1 = 256 Hz and ν2 = 260 Hz are sounded together.

Formula: νbeat = |ν1 − ν2|; beat period Tbeat = 1/νbeat.

Substitution: νbeat = |256 − 260| = 4; Tbeat = 1/4

Result: 4 beats are heard every second (beat frequency = 4 Hz), with a beat period of 0.25 s.

Beat frequency νbeat = |ν1 − ν2|
Beat period Tbeat = 1 / νbeat
Resultant of two nearly-equal-frequency waves y = [2a cos(2π Δν t)] sin(2π ν̄ t) Δν = (ν1−ν2)/2, ν̄ = (ν1+ν2)/2
Remember
  • Beats arise from superposition of two waves of slightly different, nearly equal frequencies.
  • Beat frequency νbeat = |ν1 − ν2| (independent of which frequency is larger).
  • The amplitude envelope is modulated at Δν = (ν1−ν2)/2, but audible loud beats occur at the rate |ν1−ν2|.
  • Used practically to tune musical instruments to the same frequency (zero beat).

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

v = ν λ
Wave speed–wavelength–frequency relation
y(x,t) = f(x − vt)
General 1-D progressive wave
y(x,t) = a sin(kx − ωt)
Displacement relation
k = 2π/λ
Propagation constant
ω = 2π/T = 2πν
Angular frequency
v = ω/k = ν λ
Wave speed
v = √(T/μ)
Speed on a stretched string
v = √(Y/ρ)
Speed in a solid rod
v = √(B/ρ)
Speed in a liquid
v = √(P/ρ)
Newton's formula (isothermal, gas)
v = √(γP/ρ)
Laplace's corrected formula (adiabatic, gas)
y = y1 + y2 + … + yn
Principle of superposition
R = √(a1² + a2² + 2 a1 a2 cos φ)
Resultant of two waves (general phase difference)
R = 2a cos(φ/2)
Resultant, equal amplitudes
y = 2a sin(kx) cos(ωt)
Standing wave (superposition of incident + reflected)
adjacent nodes (or antinodes) separated by λ/2; node–antinode separated by λ/4
Node/antinode spacing
νn = n v / (2L), n = 1,2,3,…
String fixed both ends / pipe open both ends
νn = (2n−1) v / (4L), n = 1,2,3,…
Pipe closed at one end
νbeat = |ν1 − ν2|
Beat frequency
Tbeat = 1 / νbeat
Beat period
y = [2a cos(2π Δν t)] sin(2π ν̄ t)
Resultant of two nearly-equal-frequency waves

Test yourself

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0 correct · 0/12 answered
Q1 Wave types easy

In a transverse wave, the particles of the medium vibrate:

Q2 Wave types easy

Sound waves travelling through air are:

Q3 Displacement relation medium

A progressive wave is given by y(x,t) = 0.02 sin(4x − 20t), with x, y in metres and t in seconds. The speed of this wave is:

Q4 Standing waves medium

In a stationary wave, the distance between a node and the adjacent antinode is:

Q5 Speed of a wave medium

The speed of a transverse wave on a string of linear mass density 0.01 kg/m under a tension of 100 N is:

Q6 Reflection of waves medium

When a wave travelling on a string reflects from a rigid (fixed) end, the reflected wave undergoes:

Q7 Standing waves in strings medium

A string of length 1.5 m, fixed at both ends, supports transverse waves of speed 300 m/s. Its fundamental frequency is:

Q8 Organ pipes easy

A pipe closed at one end (a closed organ pipe) can resonate at:

Q9 Beats easy

Two tuning forks of frequencies 256 Hz and 262 Hz are sounded together. The number of beats heard per second is:

Q10 Superposition easy

According to the principle of superposition of waves, the resultant displacement at a point where two waves overlap is:

Q11 Organ pipes medium

An organ pipe open at both ends, of length 0.5 m, resonates in its fundamental mode. If the speed of sound in air is 340 m/s, the fundamental frequency is:

Q12 Speed of a wave (Laplace correction) hard

Using Newton's (uncorrected, isothermal) formula v = √(P/ρ) with P = 1.01 × 10⁵ Pa and ρ = 1.29 kg/m³, the calculated speed of sound in air is approximately:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If a transverse jerk is struck at one end, how long does the disturbance take to reach the other end?Speed of a wave

Given: mass of string M = 2.50 kg, length L = 20.0 m, tension T = 200 N.

Formula: linear mass density μ = M/L; wave speed v = √(T/μ); time to travel the length t = L/v.

Substitution:

  • μ = 2.50/20.0 = 0.125 kg/m
  • v = √(200/0.125) = √1600 = 40 m/s
  • t = 20.0/40 = 0.5 s

Result: the disturbance takes 0.5 s to reach the other end.

2 A pipe, 30.0 cm long, is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1.1 kHz source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as 330 m/s.Organ pipes

Given: L = 0.300 m, source frequency ν = 1.1 kHz = 1100 Hz, v = 330 m/s.

Formula (open at both ends): νn = n v / (2L)

Substitution: νn = n × 330/(2×0.300) = n × 550 Hz. Setting νn = 1100 Hz gives n = 1100/550 = 2.

Result: the second harmonic (first overtone) of the open pipe is excited.

If one end is closed: only odd harmonics are allowed, νn = (2n−1) v/(4L) = (2n−1) × 330/1.2 = (2n−1) × 275 Hz, giving allowed frequencies 275 Hz, 825 Hz, 1375 Hz, … Since 1100 Hz is not an odd multiple of 275 Hz (1100/275 = 4, an even multiple), this source will not produce resonance in the closed pipe.

3 Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce 5 beats per second. When the tension of string B is slightly increased, the beat frequency decreases to 3 beats per second. If the frequency of string A is 427 Hz, what was the original frequency of string B?Beats

Given: νA = 427 Hz, initial beat frequency = 5 Hz, so νB = 427 ± 5 = 422 Hz or 432 Hz.

Formula: for a string, v ∝ √T and ν ∝ v, so increasing tension T always increases the frequency of B.

Reasoning: if νB were 432 Hz (greater than νA), raising T would raise νB further above 427 Hz, increasing the beat frequency, not decreasing it — contradicting the observation. So νB must originally have been 422 Hz (less than νA); raising T brings 422 Hz closer to 427 Hz, reducing the beat frequency to 3 Hz, exactly as observed.

Result: the original frequency of string B was 422 Hz.

4 Given below are functions representing the displacement of an elastic wave: (a) y = 5 cos(4x) sin(20t); (b) y = 4 sin(5x − t/2) + 3 cos(5x − t/2); (c) y = 10 cos[(252 − 250)πt] cos[(252 + 250)πt]. State, with reasons, which of these represents (i) a travelling wave, (ii) a stationary wave, (iii) beats.Classification of wave equations

(a) y = 5 cos(4x) sin(20t): this is a product of a function of x alone, cos(4x), and a function of t alone, sin(20t) — it cannot be written as a single function of (x − vt) or (x + vt). Points where cos(4x) = 0 remain permanently at rest (nodes). Hence this represents a stationary (standing) wave.

(b) y = 4 sin(5x − t/2) + 3 cos(5x − t/2): both terms depend on x and t only through the same combination (5x − t/2), so the sum can be written as R sin(5x − t/2 + φ) with R = √(4² + 3²) = 5 — a single function of (x − vt) with v = ω/k = (1/2)/5 = 0.1 in the given units. Hence this represents a travelling wave moving in the +x direction, of amplitude 5 and speed 0.1 (in the assumed units).

(c) y = 10 cos[(252−250)πt] cos[(252+250)πt] = 10 cos(2πt) cos(502πt): using 2cosA cosB = cos(A−B) + cos(A+B), this expands to 5 cos(500πt) + 5 cos(504πt), i.e. the sum of two waves of frequencies 250 Hz and 252 Hz (since ω/2π = 500π/2π = 250 and 504π/2π = 252). This is exactly the amplitude-modulated form obtained on superposing two waves of nearly equal frequency. Hence this represents beats, with beat frequency |252 − 250| = 2 Hz.

5 A steel wire 0.72 m long has a mass of 5.0 × 10⁻³ kg. If the wire is under a tension of 60 N, what is the speed of transverse waves on the wire?Speed of a wave

Given: length L = 0.72 m, mass M = 5.0 × 10⁻³ kg, tension T = 60 N.

Formula: μ = M/L; v = √(T/μ) = √(TL/M).

Substitution: μ = 5.0×10⁻³/0.72 = 6.94×10⁻³ kg/m; v = √(60 × 0.72 / 5.0×10⁻³) = √(8640)

Result: v ≈ 93 m/s.

6 Explain why a transverse (mechanical) wave cannot travel through a gas, whereas a longitudinal wave can travel through solids, liquids, and gases.Wave types

A transverse wave requires the medium to resist a change of shape — that is, it requires shear elasticity (rigidity), so that a layer of the medium displaced sideways experiences a restoring force from its neighbours. Gases (and, to a good approximation, liquids) offer no resistance to shear — they simply flow — so they cannot sustain a sideways restoring force and hence cannot support transverse waves.

A longitudinal wave, in contrast, only requires the medium to resist a change of volume (compression), i.e. only volume (bulk) elasticity. Solids, liquids, and gases all resist compression to some degree (all have a finite bulk modulus), so all three states of matter can support longitudinal waves — which is why sound, a longitudinal wave, can travel through solids, liquids, and gases alike.

Previous-year board questions 4

Q1 Define beats. State the expression for beat frequency produced when two sound waves of slightly different frequencies superpose. CBSE 2 marks

Beats are the periodic waxing and waning (rise and fall) in the loudness of sound heard when two waves of the same amplitude but slightly different frequencies (ν1 and ν2) superpose at a point.

On superposition, y = y1 + y2 = [2a cos(2πΔνt)] sin(2πν̄t), where Δν = (ν1−ν2)/2 and ν̄ = (ν1+ν2)/2. The amplitude term 2a cos(2πΔνt) varies slowly, causing the loudness to rise to a maximum and fall to a minimum periodically.

The number of beats (maxima of loudness) heard per second, the beat frequency, is given by:

νbeat = |ν1 − ν2|

Q2 A string of length L is stretched and fixed at both ends. Derive an expression for the frequencies of the normal modes of vibration of the string. CBSE 3 marks

Since both ends of the string are rigidly fixed, they must remain at rest at all times — i.e. both ends must be nodes of the resulting standing wave.

A standing wave pattern with nodes only at x = 0 and x = L can exist only if the length L accommodates an exact whole number of half-wavelengths:

L = n(λn/2), n = 1, 2, 3, …

so that λn = 2L/n.

Since the wave speed v on the string is fixed by v = √(T/μ) (T = tension, μ = mass per unit length), the corresponding allowed frequencies are:

νn = v/λn = n v / (2L), n = 1, 2, 3, …

n = 1 gives the fundamental (first harmonic); n = 2, 3, … give the first, second, … overtones. All integer harmonics of the fundamental are allowed.

Q3 (a) Distinguish between transverse and longitudinal waves, giving one example of each. (b) Derive the relation v = νλ for a progressive wave. (c) A pipe closed at one end resonates with air columns. Write the expression for its normal-mode frequencies and state which harmonics are present. CBSE 5 marks

(a) In a transverse wave, particles of the medium vibrate perpendicular to the direction of wave propagation (example: waves on a stretched string). In a longitudinal wave, particles vibrate parallel to the direction of propagation, producing compressions and rarefactions (example: sound waves in air).

(b) Consider a wave travelling with speed v. In one time period T, the wave advances by exactly one wavelength λ (since the waveform repeats after every λ, and this repetition, moving at speed v, takes time T). Therefore:

v = distance travelled / time taken = λ/T

Since frequency ν = 1/T, this gives:

v = ν λ

(c) For a pipe closed at one end, the closed end is always a node and the open end an antinode. This boundary condition permits only odd harmonics:

νn = (2n − 1) v / (4L), n = 1, 2, 3, …

where L is the length of the air column and v is the speed of sound in air. Only the odd harmonics (1st, 3rd, 5th, …) of the fundamental are present; even harmonics are absent.

Q4 A progressive wave travelling along a string is represented by y(x,t) = 0.02 sin(2πx − 100πt), where x and y are in metres and t is in seconds. Calculate (i) the wavelength, (ii) the frequency, and (iii) the speed of the wave. CBSE 3 marks

Given: comparing with y = a sin(kx − ωt): a = 0.02 m, k = 2π rad/m, ω = 100π rad/s.

Formula: λ = 2π/k; ν = ω/2π; v = ω/k (equivalently v = νλ).

Substitution:

  • λ = 2π/(2π) = 1 m
  • ν = 100π/(2π) = 50 Hz
  • v = ω/k = 100π/2π = 50 m/s (check: v = νλ = 50 × 1 = 50 m/s ✓)

Result: wavelength = 1 m, frequency = 50 Hz, wave speed = 50 m/s.

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