1. Chapter at a Glance
A wave is a disturbance that carries energy from one place to another without the medium itself moving along. This chapter covers transverse and longitudinal waves, the equation of a travelling wave, the speed of waves on strings and of sound, the principle of superposition, reflection, standing waves in strings and air columns, and beats. NEET asks one or two questions here, often on pipes, strings or beats.
- Finding amplitude, frequency, wavelength and speed from y = A sin(kx − ωt).
- Speed of transverse waves on a string; speed of sound and its dependence on temperature.
- Harmonics of strings, open pipes and closed pipes.
- Beats and the unknown frequency of a tuning fork.
2. Types of Waves and the Wave Equation
- Mechanical waves need a medium (sound, water waves, waves on strings); electromagnetic waves do not.
- Transverse wave: particles vibrate perpendicular to the direction of travel (crests and troughs). Possible in solids and on liquid surfaces.
- Longitudinal wave: particles vibrate along the direction of travel (compressions and rarefactions). Sound in air is longitudinal.
- Travelling wave along +x: y(x, t) = A sin(kx − ωt + φ), with angular wave number k = 2π/λ and ω = 2πν.
- Wave speed v = νλ = ω/k. Particle speed ∂y/∂t is different; its maximum is Aω.
- Phase difference = (2π/λ) × path difference.
3. Speed of Waves
| Wave | Speed | NEET Note |
|---|---|---|
| Transverse wave on a string | v = √(T/μ) | μ = mass per unit length |
| Longitudinal wave in a solid rod | v = √(Y/ρ) | Sound is fastest in solids |
| Sound in a fluid | v = √(B/ρ) | B = bulk modulus |
| Sound in a gas (Laplace) | v = √(γP/ρ) = √(γRT/M) | v ∝ √T; independent of P at fixed T |
Newton assumed isothermal compressions and got v = √(P/ρ) ≈ 280 m s⁻¹, which was too low; Laplace corrected it by using adiabatic compressions (factor √γ). Sound travels faster in humid air (moist air is less dense).
4. Superposition, Reflection and Standing Waves
- Principle of superposition: when waves overlap, the resultant displacement is the algebraic sum of the individual displacements.
- Reflection from a rigid boundary reverses the phase (phase change π); from an open boundary there is no phase change.
- Two identical waves travelling in opposite directions form a standing wave: y = 2A sin(kx) cos(ωt). Points of zero amplitude are nodes; points of maximum amplitude are antinodes. Adjacent nodes are λ/2 apart; node to antinode is λ/4.
| System | Frequencies | Harmonics present | NEET Note |
|---|---|---|---|
| String fixed at both ends | νn = nv/2L | All | Nodes at both ends |
| Pipe open at both ends | νn = nv/2L | All | Antinodes at both ends |
| Pipe closed at one end | νn = nv/4L | Odd only (1, 3, 5 …) | Node at closed end |
In a resonance-tube experiment, successive resonance lengths differ by λ/2, and the end correction e = (l₂ − 3l₁)/2.
5. Beats
- Two sounds of nearly equal frequencies ν₁ and ν₂ produce a rise and fall of loudness called beats; beat frequency = |ν₁ − ν₂|.
- The human ear can detect up to about 10 beats per second.
- Loading a tuning fork with wax lowers its frequency; filing its prongs raises it. Watch how the beat rate changes to decide which fork was higher.
- Beats are used to tune musical instruments.
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Wave speed | v = νλ = ω/k |
| String | v = √(T/μ) |
| Sound in gas | v = √(γRT/M) |
| Standing wave | y = 2A sin kx cos ωt |
| String / open pipe | νn = nv/2L |
| Closed pipe | νn = nv/4L, n odd |
| Beats | νbeat = |ν₁ − ν₂| |
| End correction | e = (l₂ − 3l₁)/2 |
7. Master Summary Table
| Concept | Key point | NEET Note |
|---|---|---|
| Transverse vs longitudinal | ⊥ vs ∥ vibration | Sound in air is longitudinal |
| Wave speed | Depends on the medium | Frequency depends on the source |
| Sound speed | ∝ √T | Not affected by pressure at fixed T |
| Standing waves | Nodes λ/2 apart | No net energy transport |
| Closed pipe | Odd harmonics only | Fundamental = v/4L |
| Beats | |ν₁ − ν₂| | Wax lowers, filing raises frequency |
8. Mind Map
- Transverse
- Longitudinal
- String √(T/μ)
- Sound √(γRT/M)
- Strings
- Open, closed pipes
- Resonance tube
- |ν₁ − ν₂|
- Tuning forks
9. Tips and Tricks
- From y = A sin(kx − ωt): speed = (coefficient of t)/(coefficient of x).
- Frequency of a string ∝ √T/L: doubling tension raises it by √2; halving length doubles it.
- A closed pipe has the same fundamental as an open pipe of twice its length.
- Speed of sound doubles when the absolute temperature becomes four times.
- Including even harmonics for a closed pipe.
- Thinking frequency changes when a wave enters another medium — speed and wavelength change, frequency does not.
- Confusing wave velocity with particle velocity.
10. Important One-liners
- Sound cannot travel through a vacuum.
- Sound travels fastest in solids and slowest in gases.
- The frequency of a wave is decided by its source, not by the medium.
- A closed organ pipe produces only odd harmonics.
- The distance between two consecutive nodes is half a wavelength.
- Beats are heard only when two frequencies differ slightly.
- Sound travels faster in moist air than in dry air.
11. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take the speed of sound in air as 340 m s⁻¹ unless stated. Click to reveal each answer and working.
Average level
A wave is described by y = 0.03 sin(8πt − 0.2πx) in SI units. Find its amplitude, frequency, wavelength and speed.
Show answer & solution
Concept: compare with y = A sin(ωt − kx).
- A = 0.03 m; ω = 8π ⇒ ν = 4 Hz.
- k = 0.2π ⇒ λ = 2π/k = 10 m.
- v = νλ = 40 m s⁻¹.
Answer: 0.03 m; 4 Hz; 10 m; 40 m s⁻¹
A 512 Hz tuning fork produces sound of wavelength 0.66 m in air. Find the speed of sound.
Show answer & solution
Concept: v = νλ.
- v = 512 × 0.66.
- ≈ 338 m s⁻¹.
Answer: about 338 m s⁻¹
A string with mass per unit length 0.04 kg m⁻¹ is under a tension of 64 N. Find the speed of transverse waves on it.
Show answer & solution
Concept: v = √(T/μ).
- v = √(64/0.04) = √1600.
- v = 40 m s⁻¹.
Answer: 40 m s⁻¹
Find the fundamental frequency of a 17 cm pipe that is (a) closed at one end, (b) open at both ends.
Show answer & solution
Concept: closed ν₁ = v/4L; open ν₁ = v/2L.
- (a) ν₁ = 340/(4 × 0.17) = 500 Hz.
- (b) ν₁ = 340/(2 × 0.17) = 1000 Hz.
Answer: 500 Hz; 1000 Hz
Two tuning forks of frequencies 440 Hz and 444 Hz are sounded together. How many beats are heard per second, and what is the time between two loud sounds?
Show answer & solution
Concept: beat frequency = |ν₁ − ν₂|.
- Beats = 444 − 440 = 4 per second.
- Time between loud sounds = 1/4 = 0.25 s.
Answer: 4 beats s⁻¹; 0.25 s
Advanced level
A 0.5 m string of mass 5 g is fixed at both ends under a tension of 100 N. Find its fundamental frequency and third harmonic.
Show answer & solution
Concept: μ = m/L, v = √(T/μ), νn = nv/2L.
- μ = 0.005/0.5 = 0.01 kg m⁻¹.
- v = √(100/0.01) = 100 m s⁻¹.
- ν₁ = 100/(2 × 0.5) = 100 Hz; ν₃ = 300 Hz.
Answer: 100 Hz; 300 Hz
A pipe closed at one end has a fundamental frequency of 250 Hz. Find its length and the frequencies of its first two overtones.
Show answer & solution
Concept: closed pipe: ν₁ = v/4L; overtones are odd multiples.
- L = v/(4ν₁) = 340/1000 = 0.34 m.
- First overtone = 3 × 250 = 750 Hz.
- Second overtone = 5 × 250 = 1250 Hz.
Answer: 34 cm; 750 Hz and 1250 Hz
At what temperature will the speed of sound in air be double its value at 0 °C?
Show answer & solution
Concept: v ∝ √T (kelvin).
- Doubling v needs 4 times T: T = 4 × 273 = 1092 K.
- In Celsius: 1092 − 273 = 819 °C.
Answer: 1092 K (819 °C)
A standing wave on a string is y = 4 sin(πx/5) cos(100πt), with x and y in cm and t in s. Find the wavelength, frequency, distance between adjacent nodes, amplitude of the component waves and their speed.
Show answer & solution
Concept: compare with y = 2A sin kx cos ωt.
- k = π/5 cm⁻¹ ⇒ λ = 2π/k = 10 cm.
- ω = 100π ⇒ ν = 50 Hz.
- Node spacing = λ/2 = 5 cm.
- 2A = 4 cm ⇒ A = 2 cm; v = νλ = 500 cm s⁻¹ = 5 m s⁻¹.
Answer: 10 cm; 50 Hz; 5 cm; 2 cm; 5 m s⁻¹
In a resonance-tube experiment with a 400 Hz fork, the first two resonance lengths are 20 cm and 62 cm. Find the speed of sound and the end correction.
Show answer & solution
Concept: l₂ − l₁ = λ/2; e = (l₂ − 3l₁)/2.
- λ = 2 × (62 − 20) = 84 cm = 0.84 m.
- v = 400 × 0.84 = 336 m s⁻¹.
- e = (62 − 60)/2 = 1 cm.
Answer: 336 m s⁻¹; 1 cm
12. NEET-pattern Questions with Detailed Answers
These questions are original and modelled on concepts repeatedly asked in NEET UG.
Sound waves in air are:
- Transverse
- Longitudinal
- Electromagnetic
- Both transverse and longitudinal
Show solution
Air cannot support shear, so only compressions and rarefactions (longitudinal waves) travel through it.
Answer: (B)
When a wave passes from one medium into another, which quantity stays the same?
- Speed
- Wavelength
- Frequency
- Amplitude
Show solution
Frequency is set by the source; speed and wavelength change with the medium, and amplitude may change due to partial reflection.
Answer: (C)
The ratio of the fundamental frequencies of an open pipe and a closed pipe of the same length is:
- 1 : 2
- 2 : 1
- 1 : 1
- 4 : 1
Show solution
Open: v/2L; closed: v/4L. Ratio = (1/2)/(1/4) = 2 : 1.
Answer: (B)
If the tension in a stretched string is made four times, the speed of transverse waves on it becomes:
- Four times
- Twice
- Half
- Unchanged
Show solution
v = √(T/μ) ∝ √T, so v doubles.
Answer: (B)
Fork A (256 Hz) gives 5 beats per second with fork B. When B is filed, the beats increase. The original frequency of B was:
- 251 Hz
- 261 Hz
- 256 Hz
- 266 Hz
Show solution
B is 251 or 261 Hz. Filing raises B's frequency. If B were 251 Hz, raising it would reduce beats; since beats increase, B must be 261 Hz.
Answer: (B)
In a standing wave, the distance between a node and the nearest antinode is:
- λ
- λ/2
- λ/4
- 2λ
Show solution
Nodes are λ/2 apart and an antinode lies halfway, λ/4 from each node.
Answer: (C)
The speed of sound in a gas at 27 °C is 330 m s⁻¹. At 127 °C it is about:
- 381 m s⁻¹
- 440 m s⁻¹
- 1552 m s⁻¹
- 286 m s⁻¹
Show solution
v ∝ √T: 330 × √(400/300) = 330 × 1.155 ≈ 381 m s⁻¹. Using the Celsius ratio would wrongly give a huge value.
Answer: (A)
The fundamental frequency of a closed pipe is 300 Hz. Which of the following frequencies can it NOT produce?
- 900 Hz
- 1500 Hz
- 600 Hz
- 2100 Hz
Show solution
A closed pipe gives only odd multiples of 300 Hz: 900 (3×), 1500 (5×), 2100 (7×). 600 Hz (2×) is an even harmonic and is absent.
Answer: (C)
13. Quick Revision Summary
- v = νλ = ω/k; string v = √(T/μ); sound v = √(γRT/M) ∝ √T.
- Frequency is fixed by the source; speed and λ by the medium.
- Standing waves: nodes λ/2 apart; node–antinode λ/4.
- String and open pipe: nv/2L (all n); closed pipe: nv/4L (odd n).
- Beats = |ν₁ − ν₂|; wax lowers, filing raises frequency.