A student tunes his guitar by striking a 120 Hertz with a tuning fork, and simultaneously plays the 4th string

I. Multiple Choice Questions · Q1

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?

(a) 130

(b) 117

(d) 120

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Concept understanding — Beats

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:

y_1 = A \sin(\omega_1 t)

y_2 = A \sin(\omega_2 t)

By the principle of superposition, the resultant displacement is:

y = y_1 + y_2 = A[\sin(\omega_1 t) + \sin(\omega_2 t)]

Using the sum-to-product identity:

\sin P + \sin Q = 2 \sin\left(\frac{P+Q}{2}\right) \cos\left(\frac{P-Q}{2}\right)

we get:

y = 2A \cos\left(\frac{\omega_1 - \omega_2}{2} t\right) \sin\left(\frac{\omega_1 + \omega_2}{2} t\right)

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.

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

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:

T_{\text{beat}} = \frac{2\pi}{|\omega_1 - \omega_2|/2} = \frac{2\pi}{|\omega_1 - \omega_2|}

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

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

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.

Watch out

A common mistake is to think the beat frequency is $\frac{f_1 + f_2}{2}$ or $\frac{|f_1 - f_2|}{2}$ . It is not. The cosine term has frequency $\frac{|f_1 - f_2|}{2}$ , but the loudness (which depends on the square of the amplitude) goes through two maxima per cycle of the cosine — one at the positive peak and one at the negative peak. So the perceived beat frequency is $|f_1 - f_2|$ , not half of it.

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.

Step 1. The beat frequency equals the magnitude of the difference between the two source frequencies: $n=|f_{fork}-f_{string}|$ .

Step 2. Here $f_{fork}=120\ \text{Hz}$ and $n=3$ beats/sec, so $f_{string}=120\pm3$ , i.e. either 123 Hz or 117 Hz.

Step 3. Comparing with the given options (130, 117, 110, 120), only 117 Hz appears, so the string frequency must be 117 Hz.

✓Final answer

(b) 117

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANN1 mark
$Q. What are beats?$
›Reveal solutionSolution

Beats are the periodic waxing and waning of sound loudness produced by the superposition of two nearly-equal-frequency sound waves.

When two sound waves of slightly different frequencies (f₁ and f₂, close to each other) but nearly equal amplitude travel in the same direction and superpose, the resultant amplitude at any point does not stay constant — it periodically rises to a maximum and falls to a minimum. This periodic variation in the intensity (loudness) of the resultant sound, heard as a regular waxing and waning, is called beats.

The number of beats produced per second (the beat frequency) equals the difference between the two individual frequencies:

fbeat = |f₁ − f₂|

Beats arise because the two waves alternately go into phase (constructive interference, giving loud sound) and out of phase (destructive interference, giving quiet/no sound) at a regular rate, since their phase difference keeps changing due to the small frequency difference.

✓Final answer
$Beats are the periodic variation (waxing and waning) in loudness heard when two sound waves of slightly different frequencies superpose; beat frequency = |f₁ - f₂|.$
CBSE 2024Set ANNUAL1 markMCQ
$Q.Which of the following phenomena in sound waves is responsible for the production of beats? (A) Reflection (B) Refraction (C) Interference (D) Resonance$
›Reveal solutionSolution

Beats are a periodic interference effect between two close-frequency waves.

When two sound waves of slightly different frequencies $f_1$ and $f_2$ travel through the same medium simultaneously, they superpose (interfere). At some instants their crests align (constructive interference, loud sound) and at others crest meets trough (destructive interference, quiet), producing a periodic rise and fall in loudness — beats — with beat frequency $|f_1-f_2|$ .

✓Final answer
$(C) Interference.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which property of sound waves is responsible for the formation of beats?$
(a) Amplitude
(b) Wavelength
(c) Frequency
(d) Speed
›Reveal solutionSolution

Beats are produced when two waves of nearly equal but slightly different frequencies superpose, causing periodic constructive/destructive interference — so it's the difference in FREQUENCY between the two waves that creates beats.

When two sound waves of frequencies f1 and f2 (close in value) overlap, their superposition produces a resultant wave whose amplitude rises and falls periodically. This is heard as a throbbing/pulsing loudness variation called 'beats,' occurring at a beat frequency = |f1 - f2|.

If the two waves had the same frequency, no beats would form (just a steady, possibly louder/softer, combined tone depending on phase); beats specifically require a frequency difference between the two source waves.

✓Final answer
$(c) Frequency.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Beats are heard clearly when the frequencies of the two waves are —$
(a) Equal
(b) Nearly equal
(c) Very different
(d) None of these
›Reveal solutionSolution

Beats are clearly heard when the two frequencies are nearly equal.

Beats are the periodic rise and fall of loudness produced when two waves of slightly different frequencies superpose; the beat frequency equals the difference of the two frequencies.

For the ear to follow this waxing and waning, the difference must be small (about ≤ 10 Hz), i.e. the frequencies must be nearly equal. If they differ greatly, the beats are too rapid to be perceived.

✓Final answer
$(b) Nearly equal.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Beats are produced due to ?$
(a) Reflection
(b) Refraction
(c) Interference
(d) Resonance of sound waves
›Reveal solutionSolution

Beats are produced by interference of two sound waves of slightly different frequency.

When two waves of nearly equal frequencies travel in the same direction and superpose, they interfere constructively and destructively alternately, producing a periodic variation in intensity called beats. This is an example of interference of waves (in time), not reflection, refraction or resonance.

✓Final answer
$(c) Interference.$
CBSE 2020Set ANNUAL1 markMCQ
$\frac{1}{4}$
›Reveal solutionSolution

Beat frequency = |f1 − f2| = 4 Hz, so the time between successive intensity maxima is 1/4 s.

Beat frequency $f_{beat} = |f_1 - f_2| = |454 - 450| = 4$ Hz. The time interval between two successive maxima (loudness peaks) equals the beat period:

$T_{beat} = \frac{1}{f_{beat}} = \frac{1}{4}\ \text{s}$

✓Final answer
$\frac{1}{4}$
CBSE 2019Set ANNUAL1 mark
$Q. When two sound sources are sounded together, then 2 beats are produced in .20 second. Find the frequency of the beats.$
›Reveal solutionSolution

Beat frequency = number of beats ÷ time taken = 2 beats / 0.20 s = 10 Hz.

When two sound sources of slightly different frequency are sounded together, the resultant intensity waxes and wanes periodically — this phenomenon is called beats. The beat frequency equals the difference between the two source frequencies, and is measured as the number of beats produced per second:

$f_{beat} = \frac{\text{number of beats}}{\text{time}}$

Here, 2 beats occur in $0.20$ s:

$f_{beat} = \frac{2}{0.20} = 10\text{ Hz}$

✓Final answer
$The frequency of the beats is 10 Hz .$
CBSE 2019Set ANNUAL1 mark
$Q. Write the formula of beat frequency.$
›Reveal solutionSolution

The beat frequency is simply the magnitude of the difference between the two individual frequencies: f_beat = |f1 - f2|.

When two sound waves of nearly equal but slightly different frequencies f1 and f2 travel together, their superposition produces a periodic waxing and waning of loudness called beats. The number of such loud-soft cycles heard per second (the beat frequency) equals the difference between the two frequencies.

✓Final answer
$Beat frequency, f_beat = |f1 - f2|, where f1 and f2 are the frequencies of the two superposing waves.$
CBSE 2017Set ANNUAL1 mark
$Q. Define beats.$
›Reveal solutionSolution

Beats are the periodic loudness variation produced when two waves of nearly equal frequency interfere; beat frequency = |f1 - f2|.

When two sound waves of slightly different frequencies f1 and f2 (with f1 close to f2) travel through the same medium and superpose, at some instants they arrive in phase and reinforce each other (loud sound), and shortly after they arrive out of phase and cancel each other (faint/no sound) — because the phase difference between them keeps changing steadily with time. This periodic waxing and waning of the resultant sound intensity is called beats. The number of times the loudness becomes maximum per second is the beat frequency, and it equals the difference of the two individual frequencies: f_beat = |f1 - f2|. Beats are commonly used by musicians to tune instruments to the same pitch — as the two frequencies get closer, the beat frequency (heard as a slow throbbing) decreases toward zero.

✓Final answer
$Beats are the periodic rise and fall (waxing and waning) in loudness heard when two sound waves of slightly different frequencies superpose; the number of such maxima heard per second is the beat frequency, equal to the difference of the two frequencies (f1 - f2).$

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