JEE Main Physics · Chapter 11

Waves

Wave equation, speed of waves, standing waves in strings and pipes, beats and the Doppler effect.

NCERT Class 11JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

A wave carries energy from one place to another without carrying matter along with it. This chapter covers transverse and longitudinal waves, the wave equation, speed of waves on strings and of sound, superposition, standing waves on strings and in pipes, beats and the Doppler effect. JEE Main asks one question almost every year — favourites are pipes, beats and Doppler shift.

What JEE asks most
  • Reading amplitude, frequency, wavelength and speed from y(x, t).
  • Harmonics of strings, open pipes and closed pipes.
  • Beats and finding an unknown frequency by loading or filing a fork.
  • Doppler effect for moving source and/or observer.

2. Wave Basics

  • Transverse: particles vibrate perpendicular to the direction of travel (waves on strings, light). Longitudinal: parallel to travel (sound in air), forming compressions and rarefactions.
  • Progressive wave: y = A sin(ωt − kx + φ), moving in +x. ω = 2πf, k = 2π/λ, v = fλ = ω/k.
  • Particle velocity ∂y/∂t is different from wave velocity v; maximum particle speed = Aω.
  • Phase difference = (2π/λ) × path difference.

Speed of waves

MediumSpeed
Stretched stringv = √(T/μ) (μ = mass per length)
Solid rod (longitudinal)v = √(Y/ρ)
Fluidv = √(B/ρ)
Gas (Laplace)v = √(γP/ρ) = √(γRT/M)

The speed of sound in a gas depends on temperature (v ∝ √T) and not on pressure at constant temperature. It increases with humidity.

3. Superposition and Standing Waves

  • When waves overlap, displacements add. Two waves of the same frequency travelling in opposite directions make a standing wave: y = 2A sin(kx) cos(ωt).
  • Nodes (zero amplitude) are λ/2 apart; antinodes lie midway. Node to next antinode = λ/4. No energy is carried along a standing wave.
string: L = λ/2 open pipe: L = λ/2 closed pipe: L = λ/4 string / open: fn = nv/2L (all n) closed: fn = nv/4L (odd n)
Fundamental modes. A closed end is a displacement node; an open end is an antinode.
SystemFrequenciesHarmonics present
String fixed at both endsfn = n v/2LAll (1, 2, 3, …)
Open pipefn = n v/2LAll
Closed pipefn = n v/4LOdd only (1, 3, 5, …)

End correction: the antinode forms slightly outside an open end, about 0.6r beyond it. In a resonance tube, λ/2 = l₂ − l₁ and end correction e = (l₂ − 3l₁)/2.

4. Beats and Sound Intensity

  • Two sounds of slightly different frequencies f₁ and f₂ produce beats at a rate |f₁ − f₂| per second.
  • Loading a tuning fork with wax lowers its frequency; filing it raises its frequency. If loading fork B reduces the beat rate, B was the higher frequency.
  • Sound level β = 10 log₁₀(I/I₀) dB with I₀ = 10⁻¹² W m⁻². Every 10-fold rise in intensity adds 10 dB. Intensity ∝ A²f².

5. Doppler Effect

f′ = f (v ± vo)/(v ∓ vs)

Use the upper signs when source and observer move towards each other (frequency rises), the lower signs when they move apart. v is the speed of sound; the medium (air) is taken at rest. The Doppler effect is used in radar, speed guns and medical ultrasound.

6. Formula Master Table

QuantityFormula
Wave speedv = fλ = ω/k
Stringv = √(T/μ)
Sound in gasv = √(γRT/M)
String / open pipefn = nv/2L
Closed pipefn = nv/4L (n odd)
Beatsfbeat = |f₁ − f₂|
Sound levelβ = 10 log(I/I₀)
Dopplerf′ = f(v ± vo)/(v ∓ vs)

7. Mind Map

Waves
Basics
  • Transverse, longitudinal
  • y = A sin(ωt − kx)
  • Wave speeds
Standing waves
  • Nodes, antinodes
  • Strings, pipes
  • End correction
Sound
  • Beats
  • Intensity, dB
Doppler
  • Moving source
  • Moving observer

8. Tips and Tricks

Shortcuts
  • From y = A sin(ωt − kx): speed = coefficient of t ÷ coefficient of x.
  • A closed pipe's fundamental equals that of an open pipe twice its length.
  • For a string, f ∝ √T/L: quadrupling tension doubles the frequency.
  • Doppler sign rule: anything moving "towards" makes the frequency go up.
Mistakes students make
  • Including even harmonics for a closed pipe.
  • Thinking the speed of sound in air changes with pressure at fixed temperature.
  • Swapping the roles of source and observer speeds in the Doppler formula.

9. Important One-liners

  • Sound cannot travel through vacuum; light can.
  • Sound travels fastest in solids and slowest in gases.
  • A closed pipe produces only odd harmonics.
  • Standing waves do not transport energy.
  • Beat frequency equals the difference of the two frequencies.
  • The pitch of a sound depends on its frequency; loudness on its intensity.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take the speed of sound in air as 340 m s⁻¹ unless stated.

Average level

N1Average

A wave is given by y = 0.05 sin(20πt − 0.5πx) (SI units). Find its amplitude, frequency, wavelength and speed.

Show answer & solution

Concept: compare with y = A sin(ωt − kx).

  1. A = 0.05 m.
  2. ω = 20π ⇒ f = ω/2π = 10 Hz.
  3. k = 0.5π ⇒ λ = 2π/k = 4 m.
  4. v = fλ = 40 m s⁻¹ (or ω/k = 20π/0.5π).

Answer: 0.05 m; 10 Hz; 4 m; 40 m s⁻¹

N2Average

A string with linear mass density 0.01 kg m⁻¹ is under a tension of 100 N. Find the speed of transverse waves on it.

Show answer & solution

Concept: v = √(T/μ).

  1. v = √(100/0.01) = √10⁴.
  2. = 100 m s⁻¹.

Answer: 100 m s⁻¹

N3Average

Find the fundamental frequency of a 0.5 m pipe when it is (a) open at both ends, (b) closed at one end.

Show answer & solution

Concept: open f₁ = v/2L; closed f₁ = v/4L.

  1. (a) f₁ = 340/(2 × 0.5) = 340 Hz.
  2. (b) f₁ = 340/(4 × 0.5) = 170 Hz.

Answer: 340 Hz; 170 Hz

N4Average

The speed of sound in air is 332 m s⁻¹ at 0 °C. Find it at 27 °C.

Show answer & solution

Concept: v ∝ √T (T in kelvin).

  1. v₂ = 332 × √(300/273).
  2. = 332 × 1.048 ≈ 348 m s⁻¹.

Answer: about 348 m s⁻¹

N5Average

By how many decibels does the sound level rise when the intensity becomes 100 times?

Show answer & solution

Concept: Δβ = 10 log₁₀(I₂/I₁).

  1. Δβ = 10 log₁₀ 100 = 10 × 2.
  2. = 20 dB.

Answer: 20 dB

Advanced level

N6Advanced

A siren of frequency 900 Hz moves at 34 m s⁻¹. What frequency does a stationary listener hear when the siren (a) approaches, (b) moves away?

Show answer & solution

Concept: moving source: f′ = f v/(v ∓ vs).

  1. (a) f′ = 900 × 340/(340 − 34) = 900 × 340/306 = 1000 Hz.
  2. (b) f′ = 900 × 340/(340 + 34) = 900 × 340/374 ≈ 818 Hz.

Answer: 1000 Hz; about 818 Hz

N7Advanced

Fork A (256 Hz) produces 4 beats per second with fork B. When B is loaded with a little wax, the beat rate falls to 2 per second. Find the original frequency of B.

Show answer & solution

Concept: B is 256 ± 4 Hz; loading lowers B's frequency.

  1. Possible values: 252 Hz or 260 Hz.
  2. If B = 252, loading lowers it further, so beats would increase — not observed.
  3. If B = 260, loading brings it closer to 256, so beats decrease ✓.

Answer: 260 Hz

N8Advanced

In a resonance-tube experiment with a 500 Hz fork, the first and second resonances occur at 16 cm and 50 cm. Find the speed of sound and the end correction.

Show answer & solution

Concept: l₁ + e = λ/4, l₂ + e = 3λ/4.

  1. l₂ − l₁ = λ/2 ⇒ λ = 2 × 34 = 68 cm.
  2. v = fλ = 500 × 0.68 = 340 m s⁻¹.
  3. e = (l₂ − 3l₁)/2 = (50 − 48)/2 = 1 cm.

Answer: 340 m s⁻¹; 1 cm

N9Advanced

A 1 m string fixed at both ends has μ = 0.01 kg m⁻¹ and tension 100 N. Find the fundamental frequency, the third harmonic, and the number of nodes and antinodes in the third harmonic.

Show answer & solution

Concept: fn = nv/2L; the nth mode has n loops.

  1. v = √(100/0.01) = 100 m s⁻¹.
  2. f₁ = 100/(2 × 1) = 50 Hz; f₃ = 150 Hz.
  3. Third harmonic: 3 loops ⇒ 3 antinodes and 4 nodes (including both ends).

Answer: 50 Hz; 150 Hz; 4 nodes and 3 antinodes

N10Advanced

A car horn (500 Hz) moves at 20 m s⁻¹ towards a cyclist who rides towards the car at 10 m s⁻¹. What frequency does the cyclist hear?

Show answer & solution

Concept: both approach: f′ = f(v + vo)/(v − vs).

  1. Numerator: 340 + 10 = 350.
  2. Denominator: 340 − 20 = 320.
  3. f′ = 500 × 350/320 ≈ 547 Hz.

Answer: about 547 Hz

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

A pipe closed at one end can produce:

  1. All harmonics
  2. Only even harmonics
  3. Only odd harmonics
  4. Only the fundamental
Show solution

A closed end must be a node and the open end an antinode, so L = (2n − 1)λ/4: only odd harmonics.

Answer: (C)

Q2Medium

At constant temperature, if the pressure of air is doubled, the speed of sound in it:

  1. Doubles
  2. Halves
  3. Remains the same
  4. Becomes √2 times
Show solution

v = √(γP/ρ); at constant T, P/ρ is constant, so v is unchanged.

Answer: (C)

Q3Easy

In a standing wave, the distance between a node and the nearest antinode is:

  1. λ
  2. λ/2
  3. λ/4
  4. 2λ
Show solution

Nodes are λ/2 apart and antinodes lie midway, so λ/4.

Answer: (C)

Q4Hard

The tension in a sonometer wire is made four times. Its fundamental frequency becomes:

  1. Four times
  2. Twice
  3. Half
  4. Unchanged
Show solution

f = (1/2L)√(T/μ) ∝ √T, so f doubles.

Answer: (B)

Q5Numerical

Two tuning forks of 256 Hz and 260 Hz are sounded together. How many beats are heard per second?

Show solution

260 − 256 = 4.

Answer: 4

Q6Numerical

The intensity of a sound increases 1000 times. By how many decibels does the sound level increase?

Show solution

10 log₁₀ 1000 = 30 dB.

Answer: 30

12. Quick Revision Summary

Remember before the exam
  • v = fλ = ω/k; string √(T/μ); gas √(γRT/M).
  • String/open pipe: all harmonics nv/2L; closed pipe: odd harmonics nv/4L.
  • Node–antinode λ/4; resonance tube λ/2 = l₂ − l₁.
  • Beats = |f₁ − f₂|; loading lowers, filing raises frequency.
  • Doppler: f′ = f(v ± vo)/(v ∓ vs).
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