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Waves closes the mechanics half of JEE Main and it is almost entirely formula work. The paper likes to hand you a wave written as y = A sin(ωt − kx) and ask for its speed or wavelength, so be able to read ω and k straight off it. After that come organ pipes, stationary waves and beats, each with one relation. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 2 questions every year, which is 8 marks.
- Reading a wave equation and the string and pipe frequencies are the regular ones.
- Beats and the harmonics of a stretched string fill the rest.
What the syllabus says
In the syllabus (Unit 10, Oscillations and Waves): Wave motion, longitudinal and transverse waves, speed of a travelling wave, the displacement relation for a progressive wave, the principle of superposition, reflection of waves, standing waves in strings and organ pipes, the fundamental mode and harmonics, and beats.
Taken out in 2024: the Doppler effect in sound.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Progressive wave | y = A sin(ωt − kx), with k = 2π/λ |
| Wave speed | v = ω/k = fλ |
| Speed on a stretched string | v = √(T/μ) |
| Fundamental of a string | f = v/2L |
| Open pipe | fn = nv/2L, every harmonic |
| Closed pipe | fn = (2n − 1)v/4L, odd harmonics only |
| Stationary wave | node to node λ/2, node to antinode λ/4 |
| Beat frequency | f = |f1 − f2| |
| Speed of sound and temperature | v ∝ √T in kelvin |
| Intensity | I ∝ A2 |
For a closed pipe the resonances go 1, 3, 5 times the fundamental, so the gap between two successive resonances is twice the fundamental, not once. That single fact solves a whole family of JEE Main questions.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. A wave is given by y = 0.02 sin(100t − 2x) in SI units. Its speed is
- (a) 25 m/s
- (b) 50 m/s
- (c) 100 m/s
- (d) 2 m/s
Q2. For the same wave, the wavelength is
- (a) 2 m
- (b) π m
- (c) 2π m
- (d) 1 m
Q3. A string 1 m long of mass 10 g is stretched by a tension of 100 N. Its fundamental frequency is
- (a) 25 Hz
- (b) 50 Hz
- (c) 100 Hz
- (d) 200 Hz
Q4. An open pipe and a closed pipe have the same length. The fundamental frequency of the open pipe is
- (a) half that of the closed pipe
- (b) twice that of the closed pipe
- (c) the same
- (d) four times
Q5. A tuning fork of 200 Hz gives 5 beats per second with another fork. Loading the second fork with wax reduces the number of beats. The second fork’s frequency is
- (a) 195 Hz
- (b) 205 Hz
- (c) 200 Hz
- (d) 210 Hz
Q6. A source of 500 Hz moves towards a stationary listener at 30 m/s. Taking the speed of sound as 330 m/s, the frequency heard is
- (a) 500 Hz
- (b) 550 Hz
- (c) 455 Hz
- (d) 600 Hz
Q7. In a stationary wave, the distance between a node and the next antinode is
- (a) λ
- (b) λ/2
- (c) λ/4
- (d) 2λ
Q8. The speed of sound in air is 330 m/s at 27°C. At 127°C it becomes about
- (a) 330 m/s
- (b) 381 m/s
- (c) 440 m/s
- (d) 285 m/s
Q9. A pipe closed at one end resonates at 300 Hz and again at 500 Hz, with no resonance in between. Its fundamental frequency is
- (a) 100 Hz
- (b) 200 Hz
- (c) 150 Hz
- (d) 50 Hz
Q10. The intensity of a wave is proportional to
- (a) the amplitude
- (b) the square of the amplitude
- (c) the reciprocal of the amplitude
- (d) the frequency alone
Answer key, with a hint for each
| Q1 | (b) | v = ω/k = 100/2. |
| Q2 | (b) | λ = 2π/k = 2π/2 = π metre. |
| Q3 | (b) | μ = 0.01 kg/m, so v = √(100/0.01) = 100 m/s, and f = v/2L. |
| Q4 | (b) | The open pipe holds half a wavelength in its fundamental, the closed one only a quarter. |
| Q5 | (b) | Wax lowers the frequency. If fewer beats follow, it must have been above 200 Hz and has moved closer to it. |
| Q6 | (b) | f′ = 500 × 330/(330 − 30). |
| Q7 | (c) | Node to node is λ/2, so node to the antinode between them is half of that. |
| Q8 | (b) | v ∝ √T, so 330 × √(400/300). |
| Q9 | (a) | Two successive odd harmonics differ by 2f1, so 2f1 = 200 Hz. |
| Q10 | (b) | Energy goes as the square of the amplitude, and intensity is energy per unit area per second. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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