Why Waves and Oscillations belong together
Every particle in a travelling wave performs simple harmonic motion about its mean position. The wave equation is the SHM equation with a position term added, and quantities like angular frequency and amplitude carry over unchanged. A student who has just finished Oscillations already knows perhaps half of Waves without realising it.
The genuinely new content is superposition: what happens when two waves occupy the same space. That single idea generates standing waves, beats and interference, which between them account for most of the chapter's questions.
How JEE actually asks Waves
NTA publishes no chapter-wise weightage, so any figure you see online is a coaching estimate from past papers; confirm the syllabus and pattern for your session in the official NTA information bulletin.
The observable pattern is that Waves questions cluster around a small number of standard set-ups. Organ pipes and strings under tension supply the resonance questions; beats supply short numerical ones; and the Doppler effect appears regularly because it produces clean, checkable numbers. Questions distinguishing wave speed from particle speed are also common, and they are conceptual rather than computational.
Key concepts, compressed
- ✓Wave speed is a property of the medium. Changing the source frequency changes the wavelength, not the speed.
- ✓In a transverse wave the particles move perpendicular to the propagation direction; in a longitudinal wave they move along it. Sound in air is longitudinal.
- ✓Superposition: when waves overlap, displacements add. Constructive interference needs a path difference of a whole number of wavelengths, destructive a half-integer number.
- ✓A standing wave is formed by two identical waves travelling in opposite directions. It has fixed nodes and transports no net energy.
- ✓Both ends fixed (or both open) gives all harmonics; one end fixed and one free gives only odd harmonics.
- ✓Beats occur when two close frequencies superpose, producing a loudness variation at the difference frequency.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Wave relation | v = fλ | |
| Speed on a stretched string | v = √(T/μ) | μ is mass per unit length |
| Node-to-node spacing | λ/2 | |
| Node to adjacent antinode | λ/4 | |
| String fixed at both ends | fₙ = nv/2L | all harmonics |
| Open organ pipe | fₙ = nv/2L | all harmonics |
| Closed organ pipe | fₙ = nv/4L | odd n only |
| Beat frequency | |f₁ − f₂| | |
| Doppler (general) | f′ = f(v ± v(o))/(v ∓ v(s)) | signs chosen by reasoning |
| Speed of sound and temperature | v ∝ √T | T in kelvin |
The five mistakes that cost the most marks
- ✓Memorising the Doppler signs instead of reasoning them. Ask first whether the observed frequency must rise or fall, then choose the signs that give that - the formula then checks itself.
- ✓Giving a closed pipe even harmonics. It supports only odd ones, so its overtones are three, five and seven times the fundamental.
- ✓Confusing wave speed with particle speed. The wave travels at v = fλ while a particle oscillates about its mean position at a continuously varying speed with maximum ωA.
- ✓Treating μ in v = √(T/μ) as total mass. It is mass per unit length, so a question giving total mass and length needs that division first.
- ✓Assuming a louder sound travels faster. Amplitude carries the loudness; speed is fixed by the medium alone.
Practice set 1: wave basics
1. A wave of frequency 500 Hz has a wavelength of 0.68 m. What is its speed?
340 m/s. v = fλ = 500 × 0.68 = 340 m/s, which is the speed of sound in air at room temperature - so this is a sound wave. The relation v = fλ is the definition-level equation of the chapter and holds for every kind of wave, mechanical or electromagnetic.
2. A sound wave of frequency 170 Hz travels in air at 340 m/s. What is its wavelength?
2 m. λ = v/f = 340/170 = 2 m. Note the direction of causation: the source sets the frequency and the medium sets the speed, so the wavelength is whatever the two require. Raising the frequency shortens the wavelength; it does not speed the wave up.
3. Is sound in air a transverse or a longitudinal wave, and why?
Longitudinal - the air particles oscillate back and forth along the direction the wave travels, producing compressions and rarefactions. Air is a fluid and cannot sustain the shear stress a transverse wave requires. This is why sound cannot be polarised, and it is a standard one-line question.
4. A string has tension 100 N and mass per unit length 0.01 kg/m. What is the wave speed on it?
100 m/s. v = √(T/μ) = √(100/0.01) = √10000 = 100 m/s. Note this is independent of the frequency you shake the string at - the speed is fixed by the tension and the string, and changing frequency simply changes the wavelength.
5. The tension in a string is quadrupled. What happens to the wave speed?
It doubles. Since v = √(T/μ), the speed goes as the square root of tension, so multiplying T by four multiplies v by two. This is why tightening a guitar string raises its pitch: the speed rises, and with the string length fixing the wavelength, the frequency must rise too.
6. Distinguish between wave velocity and particle velocity in a travelling wave.
Wave velocity is the constant speed v = fλ at which the disturbance propagates through the medium. Particle velocity is the speed of an individual particle oscillating about its mean position, which varies continuously between zero at the extremes and a maximum of ωA at the centre. The two are unrelated in magnitude, and a wave can travel fast while its particles move very slowly.
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Practice set 2: superposition and standing waves
7. Two identical waves meet with zero phase difference. What is the resulting amplitude?
Twice the individual amplitude - fully constructive interference. Displacements add, so identical waves in phase reinforce completely. Note that the intensity, which goes as amplitude squared, becomes four times that of one wave, not twice, which is a frequently asked follow-up.
8. Two identical waves meet with a phase difference of π. What happens?
They cancel completely - fully destructive interference, giving zero amplitude. A phase difference of π corresponds to a path difference of half a wavelength. Energy is not destroyed; it is redistributed to the points of constructive interference elsewhere in the pattern.
9. In a standing wave of wavelength 0.4 m, what is the distance between adjacent nodes?
0.2 m. Adjacent nodes are separated by half a wavelength, so 0.4/2 = 0.2 m. This is the single most useful fact for standing-wave questions: measuring node spacing gives you the wavelength immediately, and combining it with v = fλ gives everything else.
10. For the same standing wave, what is the distance from a node to the nearest antinode?
0.1 m. Antinodes lie midway between nodes, so the node-to-antinode distance is λ/4 = 0.4/4 = 0.1 m. Keeping λ/2 and λ/4 straight is what separates a quick answer from a wrong one - the two appear in almost every standing-wave question.
11. A 1 m string fixed at both ends carries waves at 200 m/s. What is its fundamental frequency?
100 Hz. For a string fixed at both ends the fundamental has a node at each end and one antinode in the middle, so the length is half a wavelength: λ = 2L = 2 m. Then f = v/λ = 200/2 = 100 Hz. Equivalently f₁ = v/2L, which is worth remembering directly.
12. Which harmonics does a string fixed at both ends support?
All integer multiples of the fundamental: f, 2f, 3f and so on. Both ends must be nodes, and any whole number of half-wavelengths fits that condition. This is why a plucked string sounds richer than a closed pipe - it has both odd and even overtones present.
13. Does a standing wave transport energy along the medium?
No net energy is transported. It is formed by two identical waves travelling in opposite directions, and their energy flows cancel. Energy does oscillate between kinetic and potential within each segment, but none crosses a node - which is exactly why the nodes stay permanently at rest.
Practice set 3: organ pipes and resonance
14. An open organ pipe is 0.5 m long. What is its fundamental frequency? (v = 340 m/s)
340 Hz. For an open pipe, f₁ = v/2L = 340/(2 × 0.5) = 340/1 = 340 Hz. Both ends are open, so both are antinodes, and the fundamental fits half a wavelength into the pipe - the same geometry as a string fixed at both ends, with nodes and antinodes swapped.
15. A closed organ pipe of the same 0.5 m length - what is its fundamental frequency?
170 Hz. For a pipe closed at one end, f₁ = v/4L = 340/(4 × 0.5) = 340/2 = 170 Hz. The closed end must be a node and the open end an antinode, which fits only a quarter of a wavelength - so the pipe is acoustically twice as long as its physical length suggests.
16. Which harmonics does a closed organ pipe support?
Only odd harmonics: f, 3f, 5f and so on. The node-at-one-end, antinode-at-the-other condition can only be met by an odd number of quarter-wavelengths. Even harmonics are physically impossible in a closed pipe, and offering 2f as an option is the standard trap.
17. How do the fundamentals of an open and a closed pipe of the same length compare?
The open pipe's fundamental is exactly twice the closed pipe's - 340 Hz versus 170 Hz for our 0.5 m pipes. Closing one end halves the fundamental frequency and simultaneously removes all the even harmonics, which changes both the pitch and the timbre.
18. What is the frequency of the third harmonic of that 0.5 m closed pipe?
510 Hz. For a closed pipe the harmonics are odd multiples of the fundamental, so the third harmonic is 3 × 170 = 510 Hz. Be careful with terminology: this is the third harmonic but the first overtone, since the second harmonic does not exist here. Questions exploit that mismatch deliberately.
19. What happens to the frequency of an organ pipe when the air temperature rises?
It increases. The speed of sound goes as the square root of absolute temperature, and the pipe's length fixes the wavelength, so f = v/λ rises with v. This is why wind instruments go sharp as they warm up and must be retuned during a performance.
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Practice set 4: beats
20. Two tuning forks of 256 Hz and 260 Hz are sounded together. How many beats are heard per second?
4 beats per second. The beat frequency is the difference, |260 − 256| = 4 Hz. The listener hears a single tone at the average frequency of 258 Hz whose loudness rises and falls four times each second, because the two waves drift in and out of phase at that rate.
21. Why are beats not heard when the two frequencies are far apart?
Because the beat frequency becomes too high for the ear to resolve as a fluctuation in loudness - above roughly 10 to 15 Hz the variation blurs and the two tones are simply heard separately. The physics is unchanged; the limitation is perceptual, and questions sometimes test that distinction.
22. A tuning fork is loaded with a small amount of wax. What happens to its frequency?
It decreases. Adding mass to the prongs increases the effective inertia without changing the stiffness, and since frequency goes as √(k/m), a larger m lowers f. Filing the prongs has the opposite effect. This fact is the key to almost every 'identify the unknown fork' question.
23. A fork of unknown frequency gives 4 beats per second with a 256 Hz fork. On loading the unknown fork with wax, the beat rate rises to 6 per second. What was its original frequency?
252 Hz. Four beats means the unknown is either 252 Hz or 260 Hz - both differ from 256 by 4. Wax always lowers a fork's frequency, so test each candidate. If it were 260 Hz, loading moves it toward 256, and the beat rate would fall. If it is 252 Hz, loading moves it away from 256 - down to about 250 Hz - and the beat rate rises to 6, which is what was observed. So the original frequency was 252 Hz. Always test both candidates explicitly against the loading result; guessing which side of 256 the fork sits on is a coin flip.
Practice set 5: the Doppler effect
Do not memorise the sign convention. For each question, first decide whether the observed frequency must go up or down, then choose the signs that deliver that. The formula then verifies itself.
24. A source of 340 Hz approaches a stationary observer at 34 m/s. What frequency is heard? (v = 340 m/s)
About 377.8 Hz. The source is approaching, so the frequency must rise, which means the denominator must shrink: f′ = fv/(v − v(s)) = 340 × 340/(340 − 34) = 115600/306 ≈ 377.8 Hz. Check the direction before trusting the arithmetic - a result below 340 Hz would mean the signs were wrong.
25. The same source now recedes from the observer at 34 m/s. What frequency is heard?
About 309.1 Hz. Receding means the frequency must fall, so the denominator grows: f′ = 340 × 340/(340 + 34) = 115600/374 ≈ 309.1 Hz. Note the asymmetry - the rise on approach (about 38 Hz) is larger than the fall on recession (about 31 Hz), which is a genuine feature of source motion rather than an arithmetic slip.
26. An observer moves at 34 m/s toward a stationary 340 Hz source. What frequency is heard?
374 Hz. Approaching means the frequency rises, and for observer motion the change appears in the numerator: f′ = f(v + v(o))/v = 340 × (340 + 34)/340 = 374 Hz. Compare with question 24: the same relative speed gives 377.8 Hz for a moving source but 374 Hz for a moving observer. The two cases are genuinely different, which is why the formula has separate terms.
27. A source and observer are both stationary, but a steady wind blows from source to observer. Is there a Doppler shift?
No shift in frequency. The wind changes the effective speed of sound relative to the ground, so both v and λ change together, leaving f = v/λ unaltered. This is a favourite conceptual question because the intuition that 'something is moving, so there must be a shift' is wrong.
28. Under what condition is the Doppler shift zero even though the source is moving?
When the motion is entirely perpendicular to the line joining source and observer - at the instant of closest approach. Only the component of velocity along that line produces a shift. This is why a passing siren's pitch is momentarily equal to its true frequency exactly as it draws level with you.
29. Why does the pitch of a passing ambulance siren drop as it goes by?
Because the component of the source's velocity along the line of sight reverses. While approaching, successive wavefronts are emitted from closer positions and arrive more frequently, raising the pitch; after passing, they are emitted from further away and arrive less frequently. The siren itself emits an unchanging frequency throughout.
30. What happens when a source moves faster than the speed of sound?
The wavefronts pile up into a cone-shaped shock wave, heard as a sonic boom, and the ordinary Doppler formula no longer applies - it would give a negative or infinite denominator. The source outruns its own sound, so an observer hears nothing until the shock front arrives, then hears everything at once.
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How to study this chapter efficiently
- ✓Study Waves immediately after Oscillations. The overlap is large, and doing them together roughly halves the total effort.
- ✓For every resonance question, sketch the pipe or string and mark the nodes and antinodes. The wavelength then falls out of the geometry without any formula.
- ✓For Doppler, always predict the direction of the shift before calculating. It converts a memorised formula into a self-checking one.
- ✓Learn the closed-pipe harmonic sequence as odd-only. It is the most-tested single fact in the chapter.
- ✓Keep λ/2 and λ/4 explicitly separate in your notes - node-to-node and node-to-antinode respectively.
- ✓For beats questions with an unknown fork, always test both candidate frequencies against the loading result rather than guessing which one it is.
Turn this into active practice
Waves rewards recognition more than calculation. The arithmetic in a Doppler or organ-pipe question is trivial once the set-up is right, and almost all lost marks come from misidentifying the situation - treating a closed pipe as open, or applying the observer formula to a moving source.
The JEE Waves quiz on QUFF generates fresh questions across wave basics, standing waves, pipes, beats and Doppler, marks them instantly and explains each answer. Do a mixed set so the situation is not signposted, and note whether each error was a set-up error or an arithmetic one - they need different fixes.
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