Q.A student tunes his guitar by striking a 120 Hertz with a tuning fork, and simultaneously plays the 4th string on his guitar. By keen observation, he hears the amplitude of the combined sound oscillating thrice per second. Which of the following frequencies is the most likely the frequency of the 4th string on his guitar?

The Intuition: When Two Almost-Identical Notes Clash

Imagine you're tuning a guitar. You pluck the string you want to tune, and simultaneously strike the 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 almost the same — say one is 440 Hz and the other is 442 Hz — something strange happens. The sound doesn't stay steady. Instead, it swells in loudness, fades away, swells again, fades again, in a slow, throbbing rhythm. That rhythmic pulsing is what we call beats.

Why does this happen? Because the two waves, travelling together, sometimes add up constructively (making a loud sound) and sometimes cancel each other out (making a near-silence). They are constantly shifting in and out of phase.

The Precise Physics: Superposition of Two Slightly Different Frequencies

Let two waves of equal amplitude $A$ but slightly different angular frequencies $\omega_1$ and $\omega_2$ arrive at your ear. Their displacements at a point can be written as:

By the principle of superposition, the resultant displacement is:

Using the sum-to-product identity:

we get:

This is the key result. The resultant wave has two parts:

• A fast oscillation at the average frequency $\frac{\omega_1 + \omega_2}{2}$ (which is nearly the same as the original frequencies). This is what your ear hears as the pitch.
• A slowly varying amplitude given by $2A \cos\left(\frac{\omega_1 - \omega_2}{2} t\right)$. This envelope modulates the loudness.

The beat frequency is the absolute difference of the two original frequencies. Your ear perceives one loud-soft cycle for every complete cycle of the cosine envelope. Since the cosine goes through a full cycle when its argument changes by $2\pi$, the time period of one beat is:

And since $f = \omega/2\pi$, the beat frequency in hertz is simply:

What You Actually Hear

Your ear does not follow the rapid $(\omega_1 + \omega_2)/2$ oscillations individually — that's just the pitch you perceive. What you notice is the envelope: the amplitude rises and falls at the beat frequency. So with 440 Hz and 442 Hz, you hear a note of roughly 441 Hz that grows louder and softer 2 times every second.

A Concrete Example

Suppose two tuning forks of 256 Hz and 260 Hz are sounded together.

• The average frequency is 258 Hz — that's the pitch you'll hear.
• The beat frequency is $|256 - 260| = 4$ Hz. You'll hear the sound swell and fade 4 times every second.

If you slowly adjust one fork's frequency towards the other, the beats slow down. When they vanish entirely, the two frequencies are identical — that's how musicians tune instruments.

Why This Matters

Beats are not just a curiosity. They are the principle behind:

• Tuning musical instruments (listening for beats to match pitches)
• Doppler ultrasound (beats between emitted and reflected sound reveal blood flow speed)
• Heterodyne radio receivers (mixing two signals to produce an audible beat frequency)

The same mathematics applies to any wave — light, radio, water waves — whenever two slightly different frequencies interfere.

Beats is one of the most frequently examined topics in the NCERT Class 11 Physics Waves chapter, and 'beats formula f_beat = |f1 - f2|' or 'beats important questions class 11 physics' are common exam-prep searches for both boards and JEE Main. Because instrument tuning and Doppler ultrasound both rely on this idea, it also appears often as an applied, real-world question in competitive exams.