Beats

8 MCQs1 revision card9-step worked example

Source: NCERT Oscillations and WavesPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 15 derives this result.

Key conditions: (1) The two frequencies must be close — typically within about 6–7 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.

Practice 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?

MCQ 2Easy RecallPYQ Pattern

Beats are produced when two sound waves of nearly equal frequencies superpose. The phenomenon is based on which principle?

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 CANNOT be the frequency of the guitar string?

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?

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:

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:

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?

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:

Quick recall before you leave

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. ---

Before solving, remember these

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, p. 15

Formulas

10 formulas — click to collapse

Beat frequency

f_{beat} = ∣ f_{1} - f_{2} ∣

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

Period of simple pendulum (small angle)

T = 2 π \frac{L}{g}

Period of simple pendulum of length L. Holds for small amplitudes (sin theta ~ theta).

Symbol	Quantity	SI Unit
T	period	s
L	pendulum length	m
g	gravity	m/s^2

Valid when

Small angular amplitude (typically <15°)
Massless string
Point bob

SHM displacement

x (t) = A cos (ω t + ϕ); ω = 2 π f

Displacement in simple harmonic motion. Velocity = -A*omega*sin(omega*t+phi); a = -omega^2 * x.

Symbol	Quantity	SI Unit
A	amplitude	m
omega	angular frequency	rad/s
phi	phase	rad
T	period	s
f	frequency	Hz

Valid when

Restoring force linear (F = -kx)
No damping

Total energy in SHM

E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}

Total mechanical energy is constant. Oscillates between KE (max at x=0) and PE (max at x=±A).

Symbol	Quantity	SI Unit
E	total energy	J
k	spring constant	N/m
A	amplitude	m
m	mass	kg
omega	angular frequency	rad/s

Valid when

Conservative SHM (no damping)
Elastic regime

Period of mass-spring oscillator

T = 2 π \frac{m}{k}

Period of horizontal spring with mass m, spring constant k. Independent of amplitude.

Symbol	Quantity	SI Unit
T	period	s
m	mass	kg
k	spring constant	N/m

Valid when

Hooke's law spring
No damping
Small enough amplitude to stay in elastic regime

Standing wave in closed-end pipe

f_{n} = \frac{( 2 n - 1 ) v}{4 L}

Pipe closed at one end has only odd harmonics: f, 3f, 5f, ...

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	sound speed	m/s
L	pipe length	m

Valid when

Closed at one end (open at other)
End correction neglected

Standing wave in open-open pipe

f_{n} = \frac{n v}{2 L}

Pipe open at both ends has all harmonics. Same formula as string.

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	sound speed	m/s
L	pipe length	m

Valid when

Open at both ends
End correction neglected

Standing wave frequencies on fixed-fixed string

f_{n} = \frac{n v}{2 L}

Allowed frequencies on string fixed at both ends. n=1 fundamental; harmonics 2f, 3f, ...

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	wave speed on string	m/s
L	string length	m
n	harmonic number	-

Valid when

String fixed at both ends
Wave speed v as defined above

Speed of sound in gas (Newton-Laplace)

v = \frac{γ P}{ρ}

Speed of sound in gas. Adiabatic index gamma, pressure P, density rho. Increases with sqrt(T).

Symbol	Quantity	SI Unit
v	speed of sound	m/s
gamma	adiabatic index	-
P	pressure	Pa
rho	density	kg/m^3

Valid when

Ideal gas
Adiabatic compression/expansion of sound waves

Wave speed on string

v = \frac{T}{μ}

Speed of transverse wave on string under tension T, linear mass density mu.

Symbol	Quantity	SI Unit
v	wave speed	m/s
T	tension	N
mu	linear mass density	kg/m

Valid when

Stretched uniform string
Small amplitude

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

6 items — click to collapse

Trap

Pendulum: mass-dependence claim

Category: Overthinking

Student writes T as depending on bob mass. Simple pendulum T = 2π√(L/g); independent of m.

When it triggers

Question changes pendulum bob mass and asks for new period.

How to avoid

Mass cancels in derivation (gravitational mass = inertial mass). Mass changes the bob's KE and PE proportionally; period unaffected.

Trap

SHM period: amplitude dependence claim

Category: Overthinking

Student claims SHM period depends on amplitude. For ideal SHM (Hooke's law spring or simple pendulum at small angle), period is INDEPENDENT of amplitude.

When it triggers

Question gives changes in amplitude and asks for new period.

How to avoid

T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — neither depends on A. Only at large pendulum angles does T pick up a small amplitude correction.

Trap

Pipe harmonics: open vs closed end

Category: Similar Terms

Student includes even harmonics in a closed-end pipe. Closed pipe has only ODD harmonics (f, 3f, 5f, ...).

When it triggers

Question describes pipe closed at one end (e.g. resonance tube).

How to avoid

Open both ends: all harmonics, f_n = nv/(2L). Closed one end: odd only, f_n = (2n-1)v/(4L). Fundamental of closed pipe is HALF that of open pipe of same L.

Mistake

Includes bob mass in simple pendulum period formula.

Root cause: concept gap

Correction

Simple pendulum T = 2π√(L/g) — independent of mass. Equivalence of inertial and gravitational mass cancels m.

Mistake

Claims SHM period depends on amplitude.

Root cause: concept gap

Correction

Ideal SHM: T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — no amplitude dependence. Doubling amplitude does not change period.

Mistake

Includes even harmonics in a closed-end pipe.

Root cause: concept gap

Correction

Closed-end pipe has only ODD harmonics (f, 3f, 5f, ...). Open-both-ends pipe has all (f, 2f, 3f, ...). Reason: closed end has displacement node and pressure antinode.