Q.Two sitar strings $A$ and $B$ playing the note 'Dha' are slightly out of tune and produce beats of frequency $5\ \text{Hz}$. The tension of the string $B$ is slightly increased and the beat frequency is found to decrease to $3\ \text{Hz}$. What is the original frequency of $B$ if the frequency of $A$ is $427\ \text{Hz}$?

Beats Frequency Analysis

The Intuition: When Two Tones "Wobble"

Imagine you're tuning a guitar. You pluck the string you're tuning, and at the same time, you play a reference note from a tuning fork. If the two notes are exactly the same pitch, you hear a single steady tone. But if they are slightly different — say one is 440 Hz and the other is 442 Hz — you don't hear two separate notes. Instead, you hear a single tone that wobbles in loudness: it gets louder, then softer, then louder again, in a slow, rhythmic pulse.

That pulse is called a beat. The phenomenon is beats.

Why does this happen? Because the two sound waves are constantly going in and out of sync. When their crests align, they add up to a louder sound (constructive interference). When a crest meets a trough, they cancel partially (destructive interference). The result is a wave whose amplitude rises and falls at a rate equal to the difference between the two original frequencies.

The Precise Statement

Let two sound waves of slightly different frequencies $f_1$ and $f_2$ (with $f_1 > f_2$) travel through the same medium. Their displacements at a point can be written as:

By the principle of superposition, the resultant displacement is:

Using the trigonometric identity $\sin P + \sin Q = 2 \sin\left(\frac{P+Q}{2}\right) \cos\left(\frac{P-Q}{2}\right)$, we get:

This is the key equation. It describes a wave with two parts:

• The carrier wave: $\sin\left( 2\pi \frac{f_1 + f_2}{2} t \right)$ — this oscillates at the average frequency $f_{\text{avg}} = \frac{f_1 + f_2}{2}$. This is the pitch you actually hear.
• The envelope: $2A \cos\left( 2\pi \frac{f_1 - f_2}{2} t \right)$ — this modulates the amplitude of the carrier. The envelope oscillates at half the difference frequency.

Why $|f_1 - f_2|$ and not half of it? Because the cosine term goes through a full cycle (from maximum to minimum and back to maximum) when its argument changes by $2\pi$. That happens when $\frac{f_1 - f_2}{2} t = 1$, i.e., $t = \frac{2}{f_1 - f_2}$. So the time period of the envelope is $T_{\text{envelope}} = \frac{2}{|f_1 - f_2|}$. But the loudness (intensity) goes through two maxima per envelope cycle — one at each positive peak of the cosine and one at each negative peak (since squaring the amplitude gives intensity). So the period of the beat (the time between successive loudness maxima) is half of that: $T_{\text{beat}} = \frac{1}{|f_1 - f_2|}$. Hence, the beat frequency is $f_{\text{beat}} = \frac{1}{T_{\text{beat}}} = |f_1 - f_2|$.

Key Conditions for Beats

1. Small difference: The two frequencies must be close (typically less than about 10–15 Hz apart). If the difference is too large, the ear perceives two separate tones instead of beats.
2. Comparable amplitudes: The amplitudes should be roughly equal for maximum contrast in loudness. If one is much louder, beats are still present but less noticeable.
3. Same medium: The waves must overlap in the same region of space.

Why This Matters for Exams …