Formula
Beats
When two waves of nearly equal frequencies ν₁ and ν₂ superpose, the resultant amplitude varies periodically. Beat frequency = |ν₁ - ν₂|. Used to tune musical instruments.
-- NCERT Class 11 Physics, Ch. 14, p. 294Beats
8 MCQs1 revision card9-step worked example
Source: NCERT Oscillations and WavesPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 26 Sep 2026
Beats, explained for NEET
When two sound sources with nearly equal frequencies play simultaneously, the combined sound rises and falls in loudness at a regular rate. This periodic variation is called beats. You hear a rhythmic "wah-wah-wah" — loud when the two waves are in phase, quiet when they are out of phase.
The beat frequency — the number of loudness maxima per second — equals the absolute difference of the two source frequencies:
f_beat = |f₁ − f₂|
This follows directly from the principle of superposition. When two waves y₁ = A sin(2πf₁t) and y₂ = A sin(2πf₂t) add up, the resultant amplitude oscillates at frequency (f₁ − f₂)/2, and the loudness (proportional to amplitude squared) oscillates at |f₁ − f₂|. NCERT Class 11 Physics Chapter 14, page 294 derives this result.
Key conditions: (1) The two frequencies must be close — typically within about 10 Hz for the human ear to perceive distinct beats. Beyond that, the fluctuations blend into a rough tone. (2) Linear superposition must hold.
The high-frequency trap in NEET: a question gives you two frequencies and asks for the beat frequency. The temptation is to add them. Beat frequency is always the difference, never the sum. A second trap variant gives you the beat frequency and one source frequency, then asks for the unknown frequency — the answer has two possible values (f₁ + f_beat or f₁ − f_beat) unless additional information pins down which is higher.
Watch out: if a question says "5 beats are heard in 2 seconds," the beat frequency is 2.5 Hz, not 5 Hz. Read whether the count is total beats or beats per second.
Can you answer these Beats MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPYQ Pattern
Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. What is the beat frequency?
Show answer and why every option is right or wrong
Answer: C. Beat frequency = |f₁ − f₂| = |256 − 260| = 4 Hz (NCERT Class 11 Physics Chapter 14, page 294).
Why A is wrong: A is wrong because 516 Hz is the sum of the two frequencies, not the difference. Beats arise from superposition producing amplitude variation at the difference frequency.
Why B is wrong: B is wrong because 258 Hz is the average of the two frequencies — this is the frequency at which the resultant wave oscillates, not the beat frequency.
Why D is wrong: D is wrong because 2 Hz would result from halving the difference. The beat frequency equals the full absolute difference |f₁ − f₂|, not half of it.
MCQ 2Easy RecallPYQ Pattern
Beats are produced when two sound waves of nearly equal frequencies superpose. The phenomenon is based on which principle?
Show answer and why every option is right or wrong
Answer: A. Beats result from the algebraic addition of two wave displacements at every point — the principle of superposition (NCERT Class 11 Physics Chapter 14, page 293).
Why B is wrong: B is wrong because resonance is the enhanced response of a system driven at its natural frequency; beats involve two independent sources, not a driven system.
Why C is wrong: C is wrong because the Doppler effect involves a frequency shift due to relative motion between source and observer; beats occur even with stationary sources.
Why D is wrong: D is wrong because reflection of sound produces echoes or standing waves at boundaries; beats require two distinct frequencies arriving at the same point.
MCQ 3Easy RecallPYQ Pattern
A musician hears 6 beats per second when playing a 440 Hz tuning fork alongside a guitar string. Which of the following COULD be the frequency of the guitar string?
Show answer and why every option is right or wrong
Answer: C. C is correct. The beat frequency is |f_string − 440|, so 6 beats per second means the string is 6 Hz away from the fork on one side or the other: 434 Hz or 446 Hz. Only 434 Hz appears among the options. Note that the beat frequency alone cannot tell you WHICH side — that is why two answers are physically possible, and why a question of this type must offer just one of them.
Why A is wrong: A is wrong because |448 − 440| = 8 beats per second, not 6.
Why B is wrong: B is wrong because |436 − 440| = 4 beats per second, not 6. Being close to 440 Hz is not the test — the gap has to be exactly the beat frequency.
Why D is wrong: D is wrong because |442 − 440| = 2 beats per second, not 6.
MCQ 4Direct ApplicationPYQ Pattern
Two tuning forks A and B produce 4 beats per second. Fork A has frequency 512 Hz. When a small piece of wax is attached to fork B, the beat frequency decreases to 2 beats per second. What is the original frequency of fork B?
Show answer and why every option is right or wrong
Answer: B. Without wax: |512 − f_B| = 4, so f_B = 508 or 516 Hz. Adding wax lowers f_B. If f_B were 508 Hz, lowering it increases the difference — beats would increase, not decrease. Since beats decrease, f_B must originally be 516 Hz; wax lowers it toward 512 Hz, reducing beats to 2. (NCERT Class 11 Physics Chapter 14, page 294.)
Why A is wrong: A is wrong because if f_B = 508 Hz, attaching wax would lower f_B further below 508 Hz, increasing the gap from 512 Hz and increasing beat frequency — contradicting the observed decrease to 2 beats/s.
Why C is wrong: C is wrong because |512 − 510| = 2 Hz, not 4 Hz. The original beat frequency is 4 beats/s, so 510 Hz does not satisfy the initial condition.
Why D is wrong: D is wrong because |512 − 514| = 2 Hz, not 4 Hz. The original beat frequency is 4 beats/s, so 514 Hz does not satisfy the initial condition.
MCQ 5Direct ApplicationPYQ Pattern
A tuning fork of unknown frequency produces 5 beats per second with a fork of 384 Hz. When the unknown fork is loaded with wax, the beat frequency becomes 3 beats per second. The unknown frequency is:
Show answer and why every option is right or wrong
Answer: A. |f − 384| = 5 gives f = 379 or 389 Hz. Wax loading lowers f. If f = 389 Hz, loading it reduces f toward 384, decreasing beats to 3 — consistent. If f = 379 Hz, loading would push f further from 384, increasing beats — contradicts the observation. So f = 389 Hz.
Why B is wrong: B is wrong because |381 − 384| = 3 Hz. This equals the final beat frequency after wax loading, but does not satisfy the initial condition of 5 beats/s.
Why C is wrong: C is wrong because |387 − 384| = 3 Hz, which matches the post-wax condition, not the original 5 beats/s condition.
Why D is wrong: D is wrong because if f = 379 Hz, adding wax lowers f further below 379 Hz, increasing the gap from 384 Hz and increasing beat frequency — this contradicts the observed decrease to 3 beats/s.
MCQ 6Direct ApplicationPYQ Pattern
10 beats are heard in 5 seconds when two tuning forks are sounded together. If one fork has frequency 200 Hz, a possible frequency of the other fork is:
Show answer and why every option is right or wrong
Answer: C. Beat frequency = 10 beats / 5 s = 2 Hz. So the other frequency = 200 ± 2 = 202 Hz or 198 Hz. Option C (198 Hz) matches.
Why A is wrong: A is wrong because |210 − 200| = 10 Hz. The beat frequency is 2 Hz (10 beats in 5 seconds), not 10 Hz. This is the trap of using total beat count as beat frequency.
Why B is wrong: B is wrong because |205 − 200| = 5 Hz, not 2 Hz.
Why D is wrong: D is wrong because |195 − 200| = 5 Hz, not 2 Hz.
MCQ 7Concept TrapPractice
Two sound sources produce beats. As the frequency difference between them increases from 2 Hz to 15 Hz, what happens to the perception of beats?
Show answer and why every option is right or wrong
Answer: D. The human ear can distinguish individual beats up to about 10 Hz. Beyond that, the amplitude fluctuations are too rapid to resolve, and the listener perceives a rough or "harsh" continuous tone rather than distinct wah-wah beats. The beats do not stop physically — superposition still produces amplitude variation — but the perceptual phenomenon ceases.
Why A is wrong: A is wrong because increasing the frequency difference does not make beats louder. Loudness depends on amplitude of the individual sources, not on how fast the amplitude fluctuations occur.
Why B is wrong: B is wrong because while the beat rate does increase, beyond about 10 Hz the ear cannot resolve individual beats. The beats blur into roughness rather than remaining 'clearly distinguishable.'
Why C is wrong: C is wrong because beats do not stop — the superposition still produces amplitude variation. What stops is the human ability to perceive individual beats; the sound becomes rough, not pure.
MCQ 8CalculationPractice
Two wires of the same material, same length, and same tension have diameters in the ratio 1 : 2. Their fundamental frequencies are f₁ and f₂. When sounded together, the beat frequency is:
Show answer and why every option is right or wrong
Answer: B. For a wire, fundamental frequency f = v/(2L) where v = √(T/μ). Linear mass density μ = ρ·A = ρ·π(d/2)², so μ ∝ d². Since v = √(T/μ), v ∝ 1/d. Therefore f ∝ 1/d. With d₁ : d₂ = 1 : 2, we get f₁ : f₂ = 2 : 1, so f₂ = f₁/2. Beat frequency = |f₁ − f₂| = |f₁ − f₁/2| = f₁/2. Note: this uses the standing-wave-on-string formula for fundamental frequency combined with the wave-speed-on-string formula.
Why A is wrong: A is wrong because a beat frequency of f₁/3 needs f₂ = 2f₁/3, i.e. a diameter ratio of 2 : 3 (since f ∝ 1/d). With the given 1 : 2 ratio, f₂ = f₁/2, so the beat frequency is f₁/2.
Why C is wrong: C is wrong because |f₁ − f₂| = 3f₂ would require f₁ = 4f₂, implying a diameter ratio of 1 : 4, not 1 : 2.
Why D is wrong: D is wrong because the expression CAN be simplified. Since both wires share the same material, length, and tension, the frequency ratio depends only on the diameter ratio, which is given.
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Beats: quick recall before you leave
How do you solve a Beats question? A worked example
- 1
Given
Two tuning forks produce 4 beats per second when sounded together. Fork A has a known frequency of 256 Hz. When fork B is filed (material removed), the beat frequency increases to 6 beats per second.
- 2
Required
Find the original frequency of fork B.
- 3
Concept
Beat frequency = |f_A − f_B|. Filing a fork removes material and raises its frequency (shorter prongs vibrate faster). The direction of change in beat frequency after filing resolves the ambiguity in f_B.
- 4
Formula
f_beat = |f₁ − f₂|
- 5
Substitution
Original condition: |256 − f_B| = 4 → f_B = 252 Hz or 260 Hz.
After filing: f_B increases. Test both candidates. - 6
Calculation
Case 1: f_B = 252 Hz. Filing raises f_B → f_B moves toward 256 → difference decreases → beats should decrease. But beats increased to 6. Contradiction.
Case 2: f_B = 260 Hz. Filing raises f_B → f_B moves away from 256 → difference increases → beats increase from 4 to 6. Consistent. - 7
Final answer
f_B = 260 Hz
- 8
Common trap
The trap is choosing 252 Hz without checking which direction the beat frequency changes after the physical modification (filing/waxing). Always test both candidate frequencies against the stated change.
(This pattern — resolving the ± ambiguity using a physical modification — appears in NEET 2020 and similar sessions.) - 9
Similar NEET-style question
"Two tuning forks produce 3 beats per second. One fork has frequency 340 Hz. When wax is added to the other fork, beats decrease to 1 per second. Find the frequency of the second fork."
Strategy: |340 − f| = 3 → f = 337 or 343. Wax lowers f. If f = 343, lowering it toward 340 decreases beats — consistent. Answer: 343 Hz.
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What to remember before solving Beats questions
Which Beats formulas do you need for NEET?
1 formula — click to collapse
Beat frequency
When two waves of nearly equal frequencies superpose, amplitude oscillates at the difference frequency.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_beat | beat frequency | Hz |
| f1, f2 | superposed frequencies | Hz |
Valid when
- Linear superposition
- f1, f2 close in value
More in Oscillations and Waves: 6 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Beats questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys. — click to collapse
NEET 2020
1537 Hz
2523 Hz
3524 Hz
4536 Hz
NTA Answer: Option 3(final)
How does NEET ask about Beats?
1 recurring pattern from past papers — click to collapse
Two strings/forks slightly out of tune produce beats; find one frequency given the other and beat frequency.
Direct ApplicationEasy
Common distractors
treats beat as sum
Adds frequencies instead of subtracting
Sources
NCERT refs: Class 11 Physics Chapter 14, p.294
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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