Beat frequency = |f₁ - f₂| = 4 Hz. 4 beats heard per second. Why beats occur — constructive/destructive interference.

What are Beats? Two Tuning Forks at 440 Hz and 444 Hz

Question

Two tuning forks have frequencies 440 Hz and 444 Hz. When both are sounded together, what is the beat frequency? How many beats are heard in 5 seconds?

Solution — Step by Step

We have $f_1 = 440$ Hz and $f_2 = 444$ Hz. Both forks vibrate simultaneously and their sound waves superpose in the air reaching your ear.

The beat frequency is simply the difference between the two frequencies:

n_{beat} = |f_1 - f_2| = |440 - 444| = 4 \text{ Hz}

This means 4 beats are heard per second.

Since beat frequency = 4 Hz (4 beats every second):

\text{Beats in 5 s} = n_{beat} \times t = 4 \times 5 = \mathbf{20 \text{ beats}}

Beats are perceptible only when $n_{beat} \leq 10$ Hz. Here $n_{beat} = 4$ Hz, so yes — these beats are clearly audible. Beyond ~10 Hz, the variations blend into a rough tone rather than distinct pulses.

Why This Works

Sound is a pressure wave. When two waves of slightly different frequencies reach your ear together, they alternately reinforce and cancel each other — this is superposition. When the two waves are in phase, you hear a loud sound (constructive interference). When they are exactly out of phase, they cancel and you hear silence (destructive interference).

One complete cycle of loud → soft → loud counts as one beat. The time for this cycle equals the time it takes the faster fork to gain exactly one full oscillation over the slower one. Mathematically, this happens every $\frac{1}{|f_1 - f_2|}$ seconds — which is why beat frequency = $|f_1 - f_2|$ .

n_{beat} = |f_1 - f_2|

where $f_1$ and $f_2$ are the individual frequencies (in Hz).

Beat period: $T_{beat} = \dfrac{1}{n_{beat}}$

This is a scoring topic in CBSE Class 11 board exams and shows up as a 2-mark conceptual question. In JEE Main, it appears combined with the Doppler effect or string resonance.

Alternative Method — Using Superposition Directly

Take two waves: $y_1 = A\sin(2\pi f_1 t)$ and $y_2 = A\sin(2\pi f_2 t)$ .

By the sum-to-product identity:

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

The $\sin$ term gives the average frequency — what you hear as the pitch (here, 442 Hz). The $\cos$ term is the slowly varying amplitude envelope — it completes one full cycle ( $\cos$ goes from +1 to −1 and back) in time $\frac{1}{|f_1 - f_2|}$ .

Since the amplitude peaks twice per cosine cycle (once at +1, once at −1), the perceived loud-soft pattern repeats at $2 \times \frac{|f_1-f_2|}{2} = |f_1 - f_2|$ . Same result — 4 beats per second. This derivation is sometimes asked directly in JEE.

Common Mistake

Students write $n_{beat} = f_1 - f_2 = 440 - 444 = -4$ Hz and conclude the answer is “negative” or get confused. Beat frequency is always positive — it’s a count of pulses per second, never negative. Always take the absolute value: $n_{beat} = |f_1 - f_2|$ . It does not matter which fork has the higher frequency.

A classic PYQ twist: “Fork A and a piano key at 256 Hz produce 5 beats/s. After loading fork A with wax, beats reduce to 3/s. Find the original frequency of A.” If beats decrease after loading (which lowers frequency), then A was above 256 Hz — so $f_A = 261$ Hz. If beats increase, A was below. This logic appears often in both boards and JEE Main.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Superposition Directly

Common Mistake

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