NEET Physics - Chapter 17
Beats – Complete Notes, Revision, Important Questions & Downloads
Beats is built around the TOC subtopic Sound Intensity Variation with Frequency Difference, where two nearly equal frequencies superimpose and produce periodic loud-soft variation at a fixed position. The page formulas are A = 2a cos pi(n1 - n2)t, beat frequency n = n1 - n2, and beat period T = 1/(n1 - n2). NEET usually tests this by asking either beat count per second from two tuning forks, or unknown frequency identification after loading or filing one fork. The scoring edge is to interpret whether beat count increases or decreases after changing one fork, then choose nB = nA +/- x correctly.
Superposition ApplicationFrequency DifferenceNCERT-Aligned
Expected QuestionsQ
1
Typically one direct or mixed question appears on beat frequency, beat period, or unknown tuning-fork frequency from loading/filing observations.
Time Required⏱
1.25 h
Roughly 30 minutes to lock formulas and sign logic, 35 minutes for beat-count numericals, and 10 minutes to revise loading and filing cases.
Difficulty⚡
Medium
Core formulas are short, but options are close and students lose marks by reversing frequency difference direction after modifying one fork.
NRI USA Curriculum GapUS
Bridge Needed
Many US classroom wave exercises emphasize qualitative interference, while NEET expects quick algebraic decisions from beat-change experiments with tuning forks.
3Subtopics
20Practice Questions
4Free Downloads
1.25 hPrep Time
Beats Weightage and Trend
Waves and Sound - Topic 23| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2020 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2023 | 1 | 4 | |
| 2024 | 1 | 4 | |
| 2025 | 0 | 0 | |
| Estimated beat-related asks in recent NEET papers | 4 | 16 |
Most beats questions are one-step if n = n1 - n2 is used with absolute frequency difference and unit in hertz is preserved.
Unknown-frequency items become two-step when the fork is loaded or filed and beat count changes from x to x'; this tests direction logic, not heavy algebra.
When two moving sources are involved, NEET still reduces final beat count to difference of apparent frequencies, so sign tracking is the main discriminator.
Unknown-frequency items become two-step when the fork is loaded or filed and beat count changes from x to x'; this tests direction logic, not heavy algebra.
When two moving sources are involved, NEET still reduces final beat count to difference of apparent frequencies, so sign tracking is the main discriminator.
0.7
Avg Questions / Year
16
Total Marks (6 yrs)
Direct
Pattern
Medium
Difficulty
5-Step Beats Solve Routine
1
Start with frequency-gap equation Write beat frequency as nbeat = |n1 - n2| before substituting values, and keep all frequencies in hertz to avoid hidden unit slips.
2
Link beat period only after beat frequency Use T = 1/nbeat only after nbeat is found from frequency difference, not by averaging frequencies or using source period directly.
3
Handle loading and filing as sign tests Remember loading decreases fork frequency and filing increases it; compare new beat count x' with old x to decide whether unknown was initially above or below known fork.
4
Map outcome to nB = nA +/- x Once direction is fixed, choose nB = nA + x or nB = nA - x, then check if the modified case behavior (x' increase/decrease/zero) is physically consistent.
5
Run a plausibility check with persistence limit Distinct beats require near frequencies; if computed difference is very large, re-check whether the question asked beats per second or time between beats.
Beats Download Kit
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete beats theory with amplitude modulation derivation, beat frequency and period formulas, and unknown-frequency determination cases.
Download PDFFormula Sheet
Quick formula map for A = 2a cos pi(n1 - n2)t, nbeat = n1 - n2, T = 1/(n1 - n2), and loading/filing inference rules.
Download PDFMCQ Practice
NEET-style application MCQs on beat counting, beat period conversion, and tuning-fork unknown frequency from changed beat observations.
Download PDFPYQ Workbook
Year-tagged and NEET-style beats questions with option-level trap notes on sign errors, loading/filing direction, and x versus x' logic.
Download PDFSubtopics in Beats
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision Cards
Concept → Trap → Example1) Sound Intensity Variation with Frequency Difference
Beat count from nearby frequenciesFor two nearly equal frequencies n1 and n2, beat frequency nbeat = n1 - n2 and beat period T = 1/(n1 - n2), with resultant amplitude A = 2a cos pi(n1 - n2)t.
- Apply this when two forks or sources are sounded together and the question asks number of loudness cycles per second at one point.
- Use absolute frequency difference for count, then convert to period only if needed by taking reciprocal in seconds.
- Trap: treating average frequency (n1 + n2)/2 as beat frequency gives options close to source frequency and is always wrong for beats.
Example (NEET-style)If two forks are 256 Hz and 262 Hz, beat frequency is |262 - 256| = 6 Hz, so six beats occur each second and beat period is 1/6 s.
Curriculum Gap: India vs USA
Two concrete preparation gaps to bridge for NEET readinessAP Physics 1 introduces superposition, but NEET emphasizes tuning-fork beat diagnostics
AP courses often stop at qualitative interference descriptions, while NEET expects immediate extraction of beat frequency and unknown source frequency from measured beat-count changes.
- Practice sets where one fork is loaded or filed and infer nB = nA +/- x from x and x' change pattern.
- Train to convert between beats per second and time between beats in under 10 seconds.
US lab writing focuses method reporting; NEET rewards option-level sign discipline
Even mathematically strong students lose marks by reversing increase/decrease interpretation after frequency modification, so objective-speed sign control is essential.
- Create a one-page loading-versus-filing decision table and rehearse with mixed random cases.
- Use elimination by physical trend: if new beat count increases after loading unknown fork, original unknown was below known frequency.
NEET-style practice questions
2 MCQs1Two tuning forks A and B produce 5 beats per second when sounded together. Fork A has known frequency 320 Hz. When B is loaded with wax, beats become 7 per second. The original frequency of B is:Sound Intensity Variation with Frequency Difference
315 Hz
325 Hz
313 Hz
327 Hz
Initially, |nB - 320| = 5 so nB could be 315 Hz or 325 Hz. Loading with wax decreases nB. After loading, beats increase from 5 to 7, so the modified nB moved farther away from 320 Hz, not closer. That can happen only if original nB was below 320 Hz and became even smaller after loading. Hence original nB = 315 Hz. If nB had been 325 Hz, loading would move it toward 320 and beats would decrease, not increase. This question tests the directional logic of beat change, not only the difference formula.
2Two sound sources of frequencies 500 Hz and 504 Hz are heard together at a fixed point. Which statement is correct?Sound Intensity Variation with Frequency Difference
Beat frequency is 1004 Hz and beat period is 1/4 s
Beat frequency is 4 Hz and beat period is 0.25 s
Beat frequency is 2 Hz and beat period is 0.5 s
Beat frequency is 4 Hz and beat period is 4 s
Beat frequency is the difference of nearby source frequencies, so nbeat = |504 - 500| = 4 Hz. Beat period is reciprocal of beat frequency, T = 1/4 s = 0.25 s. Option A incorrectly uses the sum of frequencies, which corresponds to carrier oscillation behavior, not loudness variation count. Option C halves the difference without any physical basis. Option D inverts reciprocal conversion and treats period as numerically equal to frequency. This is a classic NEET conversion trap where both formula and unit interpretation must be correct together.
Practice Questions
Click "Reveal Answer" after attempting1Two forks of frequencies 256 Hz and 260 Hz are sounded together. How many beats are heard in 15 s?
15
30
60
120
👁 Reveal Answer
Correct option: C. Beat frequency is nbeat = |260 - 256| = 4 Hz, meaning 4 beats each second. In 15 seconds, total beats = 4 x 15 = 60. Option B comes from mistakenly taking half of the frequency difference. Option D doubles by confusing maxima and minima counting with beats count. The standard definition counts one complete loudness cycle per beat.
2A known fork of 400 Hz gives 6 beats per second with an unknown fork. If filing the unknown fork reduces beats to 2 per second, the unknown frequency before filing was:
394 Hz
406 Hz
398 Hz
404 Hz
👁 Reveal Answer
Correct option: A. Initially unknown could be 400 +/- 6, so 394 Hz or 406 Hz. Filing increases unknown frequency. After filing, beat count decreases from 6 to 2, so unknown moved closer to 400 Hz after frequency increase. That is possible only if unknown was initially below 400 Hz, namely 394 Hz. If it had been 406 Hz, filing would push it higher and beats would increase, not decrease.
3Beat period between two forks is measured as 0.2 s. If one fork is 512 Hz, the other can be:
507 Hz
510 Hz
517 Hz
Both 507 Hz and 517 Hz
👁 Reveal Answer
Correct option: D. Beat period T = 0.2 s gives nbeat = 1/T = 5 Hz. So the other frequency must satisfy |n - 512| = 5, giving n = 517 Hz or 507 Hz. Option B corresponds to difference 2 Hz, not 5 Hz. Option C alone is incomplete because beats only fix absolute difference, not whether unknown is higher or lower. This is a direct reciprocal and absolute-difference application.
4Two sources have equal amplitude a and frequencies n1, n2 close to each other. The time-dependent resultant amplitude is:
A = a cos pi(n1 + n2)t
A = 2a cos pi(n1 - n2)t
A = 2a sin 2pi(n1 - n2)t
A = a sin pi(n1 - n2)t
👁 Reveal Answer
Correct option: B. For superposition of two equal-amplitude close-frequency waves, the resultant displacement can be written as y = 2a cos pi(n1 - n2)t sin pi(n1 + n2)t, so the envelope amplitude is A = 2a cos pi(n1 - n2)t. Option A misses factor 2 and uses sum frequency in the envelope term. Option C uses wrong functional form and argument, and option D again misses factor 2 and cosine envelope structure central to beats.
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Frequently Asked Questions
Notes · Downloads · Revision · Important QuestionsWhy are beats heard only when frequencies are close to each other?
The page notes persistence of hearing around one-tenth of a second. Distinct loud-soft cycles are perceived only when the frequency difference is small enough that intensity modulation is slow compared with auditory persistence. If the difference becomes large, separate tones dominate and the ear does not register clean beat cycles in the usual textbook sense.
What exactly is counted as one beat in NEET numericals?
One beat is one complete intensity variation cycle in which loudness returns to maximum after passing through a minimum once in between. In objective questions, beat frequency means count of these complete cycles per second, not count of maxima plus minima separately. Keeping this definition avoids common factor-of-two mistakes in final options.
Is beat frequency n1 - n2 or absolute difference |n1 - n2|?
For number of beats per second, use the absolute difference so the result is positive. The signed form helps in algebraic direction reasoning for loading and filing cases, but the physically heard beat count is always non-negative. In final NEET answers, report beat frequency as a positive value in hertz unless a symbolic sign argument is explicitly asked.
How does loading a tuning fork change the frequency?
Loading a prong with wax increases effective mass and reduces the natural frequency of that fork. In beats problems this is used as a controlled downward shift, and comparing old and new beat counts reveals whether unknown frequency was originally above or below the known fork. This direction logic is central in unknown-frequency determination questions.
How does filing a tuning fork change the frequency?
Filing removes material from the prong, reducing effective mass and increasing natural frequency. In question sets with known fork nA and unknown nB, filing provides an upward shift to the modified fork and the resulting beat-count change identifies original ordering. Students should always write this physical effect first before trying numerical elimination.
Can two candidate frequencies satisfy the initial beat count?
Yes. If beat count with known frequency nA is x, unknown can initially be nA + x or nA - x. The second experiment (loading or filing) is needed to remove this ambiguity. This is why textbook problems include a changed condition and a new beat count x', and why one cannot finalize unknown frequency from the first observation alone.
Why is resultant amplitude written as A = 2a cos pi(n1 - n2)t?
From superposition of two equal-amplitude waves, trigonometric transformation gives displacement as product of a fast oscillation term and a slowly varying envelope term. The envelope coefficient 2a cos pi(n1 - n2)t controls intensity variation with time and generates beats. This expression links mathematical form directly to the loud-soft pattern heard experimentally.
What is the fastest way to avoid mistakes in beats problems under timed conditions?
Use a fixed sequence: write nbeat = |n1 - n2|, convert to period only if asked, then for unknown-frequency cases write both candidates and test them against change trend after loading or filing. This structured order prevents sign reversals and reciprocal errors, which are the main causes of wrong options in otherwise easy beats questions.
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