Beats and Superposition of Waves

What Beats Are, Physically

Strike two tuning forks of slightly different frequencies — say $256$ Hz and $260$ Hz — close to each other, and you hear a sound that periodically swells and fades, almost throbbing. That throbbing is beats, and it happens because the two waves alternately add up (constructive interference) and cancel (destructive interference) over time.

The frequency of this throbbing — how many loud-soft cycles per second we hear — equals the difference between the two source frequencies. So $256$ Hz and $260$ Hz produce $4$ beats per second. This is the simplest, cleanest demonstration of wave superposition that we encounter in school physics.

For JEE and NEET, beats appear in two forms: numerical questions where we compute beat frequency from given source frequencies, and conceptual questions about identifying unknown frequencies using a tuning fork of known frequency. Both rely on the same superposition principle.

Key Terms & Definitions

Superposition principle: When two or more waves overlap in space, the resultant displacement at any point is the algebraic sum of the individual displacements. This is the foundation for interference, beats, and standing waves.

Beats: The periodic variation in amplitude (and hence loudness) when two waves of slightly different frequencies superpose. Audible only when the frequency difference is small — roughly less than $10$ Hz for the human ear.

Beat frequency: The number of beats per second, $f_{\text{beat}} = |f_1 - f_2|$ . The absolute value matters because we can’t hear a “negative” beat rate.

Constructive interference: When two waves arrive in phase (path difference $= n\lambda$ ), amplitudes add. Loud moment.

Destructive interference: When two waves arrive out of phase by $\pi$ (path difference $= (n + 1/2)\lambda$ ), amplitudes subtract. Soft (or silent) moment.

Methods & Concepts

Mathematical Derivation of Beats

Consider two waves of equal amplitude $A$ but different frequencies $f_1$ and $f_2$ at a fixed point in space:

$y_1 = A \sin(2\pi f_1 t), \quad y_2 = A \sin(2\pi f_2 t)$

By superposition:

$y = y_1 + y_2 = 2 A \cos\left(2\pi \frac{f_1 - f_2}{2} t\right) \sin\left(2\pi \frac{f_1 + f_2}{2} t\right)$

The $\sin$ term oscillates at the average frequency $(f_1 + f_2)/2$ — that’s the pitch we hear. The $\cos$ term modulates the amplitude slowly at frequency $(f_1 - f_2)/2$ . But because amplitude squared is what determines loudness, and $\cos^2$ goes through one full cycle when $\cos$ goes from peak to peak, the loudness cycle happens at $|f_1 - f_2|$ — the beat frequency.

$f_{\text{beat}} = |f_1 - f_2|$

Beats are audible only when this difference is small, typically under 10 Hz.

Standing Waves vs Beats

Both are superposition phenomena, but they’re different. Standing waves form when two waves of the SAME frequency travel in opposite directions — fixed nodes and antinodes in space. Beats form when two waves of DIFFERENT frequencies travel in the same direction — amplitude varies in TIME at a fixed point.

For string and pipe problems, standing waves dominate. For tuning fork problems, beats dominate.

Worked Example: Identifying Unknown Frequency

A tuning fork of unknown frequency produces $5$ beats per second with a standard $256$ Hz fork. When the unknown fork is loaded with a small piece of wax (which lowers its frequency), the beat frequency increases to $7$ Hz. What is the original frequency?

The unknown frequency is either $251$ Hz or $261$ Hz. Wax lowers the frequency. If the original was $251$ , loading drops it further, and the beat frequency $|f - 256|$ increases — consistent. If the original was $261$ , loading lowers it toward $256$ , and beat frequency would decrease — inconsistent. So the original frequency is $\mathbf{261}$ Hz.

This kind of problem appears almost every year in NEET. The trick is the wax test — it tells you which side of the standard fork the unknown sits on.

Solved Examples

Easy (CBSE)

Two sources of sound emit at $300$ Hz and $304$ Hz. Find the beat frequency.

Beat frequency $= |304 - 300| = 4$ Hz.

Medium (JEE Main)

A pipe open at both ends has fundamental frequency $200$ Hz. When held next to a tuning fork, $5$ beats per second are heard. The tuning fork could be $195$ Hz or $205$ Hz. Heating the pipe (which raises the speed of sound, hence the pipe’s frequency) causes the beat frequency to drop. What is the tuning fork’s frequency?

Heating raises pipe frequency from $200$ Hz upward. If tuning fork is $205$ , beat frequency drops as pipe approaches $205$ — consistent. If $195$ , beat frequency rises as pipe moves away from $195$ — inconsistent. Tuning fork frequency $= \mathbf{205}$ Hz.

Hard (JEE Advanced)

Two strings of identical material, length, and tension differ in cross-sectional area by $1\%$ . They produce $4$ beats per second when plucked simultaneously. Find the higher frequency. (Recall: fundamental frequency of a string $\propto 1/\sqrt{\text{linear mass density}} \propto 1/\sqrt{\text{area}}$ .)

If areas differ by $1\%$ , frequencies differ by approximately $0.5\%$ (square root halves the percentage difference). Let the higher frequency be $f$ . Then $f - 0.995 f = 4$ , giving $0.005 f = 4$ , so $f = 800$ Hz.

Exam-Specific Tips

For NEET: Beats appear in 1-2 questions per year, almost always involving tuning forks and wax/file loading. Memorise: wax LOWERS frequency, filing RAISES frequency. The beat frequency change tells you which side of the standard fork the unknown was on.

For JEE Main: Often combined with Doppler effect or pipe resonance. Standard format: “When a source of frequency $f$ approaches an observer at speed $v$ , the apparent frequency creates beats with a stationary source. Find $v$ .”

For CBSE Boards: Derivations of the beat frequency formula are common — practice writing the trigonometric identity step.

Common Mistakes to Avoid

Mistake 1: Writing $f_{\text{beat}} = (f_1 + f_2)/2$ . That’s the average — the pitch we hear, not the beats.

Mistake 2: Forgetting the absolute value. The beat frequency is always positive: $|f_1 - f_2|$ .

Mistake 3: Assuming wax raises frequency. Wax adds mass, lowering the natural frequency of the tuning fork. Filing removes mass and raises frequency.

Mistake 4: Confusing beats with interference fringes. Beats are time-based, fringes are space-based.

Mistake 5: Thinking beats are audible at any frequency difference. Above ~10 Hz, our ears stop perceiving the throbbing as beats and instead hear two distinct tones.

Practice Questions

Q1. Two strings produce $6$ beats per second when sounded together. One is at $300$ Hz. List possible frequencies for the other.

$294$ Hz or $306$ Hz.

Q2. A $480$ Hz fork and an unknown fork produce $4$ beats per second. After loading the unknown with wax, beats reduce to $2$ per second. Find the unknown’s original frequency.

Since wax lowers frequency and beats reduce, the unknown was originally above $480$ Hz. So unknown $= 484$ Hz.

Q3. Two waves of equal amplitude superpose: $f_1 = 100$ Hz, $f_2 = 104$ Hz. State the average frequency heard and the beat frequency.

Average $= 102$ Hz, beat frequency $= 4$ Hz.

Q4. Why don’t we hear beats when frequencies differ by $50$ Hz?

The amplitude modulation happens too fast for the ear to perceive as throbbing; instead, we hear two separate tones (or even a chord).

Q5. A piano string and a tuning fork at $440$ Hz produce $3$ beats per second. The string is then tightened slightly, and the beat frequency drops to $1$ . Was the string originally above or below $440$ Hz?

Tightening raises string frequency. Beats dropped, meaning string approached $440$ — so it was below $440$ originally. Original frequency $= 437$ Hz.

FAQs

Why must the frequency difference be small for beats to be audible? The human ear integrates loudness over short windows (~100 ms). Beat frequencies above $\sim 10$ Hz happen too fast to register as separate loud-soft pulses.

What is the difference between beats and resonance? Resonance happens when an external frequency matches a natural frequency, leading to large amplitude growth. Beats happen between two close frequencies and produce amplitude modulation.

Can three waves create a beat pattern? Yes, but the result is more complex — typically you get a primary beat at the closest pair’s difference, with secondary modulations.

Do beats occur for light waves? Yes, in principle — two laser beams of slightly different frequencies produce optical beats. We use this in heterodyne detection and laser interferometry.

Why are tuning forks used for tuning instruments? Their frequency is highly stable. Using beats with a tuning fork, a tuner can match any string within a fraction of a hertz by reducing beats to zero.