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Exam level5 past questions

Superposition, Beats, and Resonance

Explore the principles of superposition, formation of beats, beat frequency, resonance, and phase difference in wave phenomena.

Why this shows up in the exam

You will encounter NEET questions on how waves interact, form beats, and resonate in different systems.

How NEET tests this

Numerical · 5 QsApplicationDirect recall

Learn the idea

Superposition lets waves add linearly; when two close frequencies interfere they produce beats whose rate equals the frequency difference, and resonance occurs when the length of a pipe matches a standing‑wave condition (closed pipe: L = (2n‑1)λ/4).

🧠 Memory hook: Closed pipe climbs only the odd rungs of a ladder – 1st, 3rd, 5th… → L = (2n‑1)λ/4

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Resultant displacement = algebraic sum of individual displacements (superposition)
  • Beat frequency f_beats = |f1‑f2|
  • For a pipe closed at one end, allowed standing‑wave lengths satisfy L = (2n‑1)λ/4 (n=1,2,3…)
  • Speed of sound v = f λ (use v≈330 m s⁻¹ in air for NEET)
  • Constructive interference when phase difference Δφ = 2πm, destructive when Δφ = (2m+1)π

How to approach it

  1. 1Read the question and note whether it asks for length, frequency or beat rate
  2. 2Write the appropriate formula: v = fλ for wavelength, L = (2n‑1)λ/4 for closed pipe, f_beats = |f1‑f2| for beats
  3. 3Calculate λ from the given frequency (or vice‑versa) using v
  4. 4Insert λ into the resonance condition to get the required length or frequency
  5. 5For beats, take the absolute difference of the two frequencies and compare with the given beat count

Worked example — watch it click

An air column in a pipe, which is closed at one end will be in resonance with a vibrating body of frequency 166 Hz, if the length of the air column is

  • A)2 m
  • B)1.5 m
  • C)1 m
  • ✅0.5 m

The concept behind this problem

The example checks that you can convert the given frequency to wavelength and then apply the closed‑pipe resonance condition to obtain the correct length.

Step by step

  1. 1For a pipe closed at one end, resonance occurs at L = (2n-1)λ/4, where n = 1,2,3...
  2. 2For first resonance (n=1), L = λ/4.
  3. 3Given f = 166 Hz, v = 332 m/s (speed of sound in air), λ = v/f = 332/166 = 2 m.
  4. 4Therefore L = 2/4 = 0.5 m.

Watch out

Students often use L = λ/2 for the first resonance, forgetting that a closed pipe’s first mode is a quarter‑wave (L = λ/4).

Common slip-ups that cost marks

  • •Mixing up the speed of sound value – NEET uses 330 m s⁻¹ unless stated otherwise
  • •Treating a closed pipe as if it supports all harmonics; only odd harmonics (1st, 3rd, 5th…) are allowed
  • •Confusing beat frequency with the sum of the two frequencies

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5

An air column in a pipe, which is closed at one end will be in resonance with a vibrating body of frequency 166 Hz, if the length of the air column is

Push further

More challenging

3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 3

A closed organ pipe of length 0.6 m is resonating in its fundamental mode. If the speed of sound in air is 330 m/s, what is the wavelength of the sound wave in the pipe?

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