Physics · Waves

Beats

Beats are the slow, periodic waxing and waning of loudness heard when two harmonic sound waves of close but unequal frequencies reach the ear together. NCERT Class XI Physics, Section 14.7, treats this as a direct consequence of the superposition principle, and it carries a recurring NEET problem type — the tuning-fork that is loaded with wax or has its tension changed. The single load-bearing result, $\nu_{beat}=|\nu_1-\nu_2|$, looks trivial; the marks are won and lost on the direction logic that the absolute value hides.

What Beats Are

When two harmonic sound waves of close — but not equal — frequencies are heard at the same time, the ear registers a sound at roughly their average frequency, together with a slow, audibly distinct rise and fall in intensity. Each cycle of maximum loudness followed by near-silence is one beat. The frequency at which these loud–soft alternations occur is the beat frequency, and NCERT establishes that it equals the difference between the two source frequencies.

The phenomenon is a special case of interference in time rather than in space. Where two waves of identical frequency redistribute energy across positions to give a fixed pattern, two waves of slightly different frequency drift in and out of phase at a single point, so the loudness at that point cycles between reinforcement and cancellation.

Figure 1 · Two close frequencies superposing

Two harmonic waves of nearly equal frequency (here 11 Hz and 9 Hz, the NCERT illustration) gradually shift from in-phase to out-of-phase and back.

(a) $\nu_1 = 11$ Hz (b) $\nu_2 = 9$ Hz

Superposition of Two Close Frequencies

Consider two waves of equal amplitude $a$ arriving at a point, written as displacements in time with angular frequencies $\omega_1=2\pi\nu_1$ and $\omega_2=2\pi\nu_2$:

$$s_1 = a\cos\omega_1 t, \qquad s_2 = a\cos\omega_2 t$$

By the superposition principle the resultant displacement is the sum $s = s_1 + s_2$. Applying the identity $\cos A + \cos B = 2\cos\!\frac{A-B}{2}\cos\!\frac{A+B}{2}$ gives

$$s = \left[\,2a\cos\!\left(\frac{\omega_1-\omega_2}{2}\,t\right)\right]\cos\!\left(\frac{\omega_1+\omega_2}{2}\,t\right)$$

Because $\nu_1$ and $\nu_2$ are close, $\tfrac{\omega_1+\omega_2}{2}$ is large while $\tfrac{\omega_1-\omega_2}{2}$ is small. The result is therefore a fast oscillation at the average (carrier) frequency $\frac{\nu_1+\nu_2}{2}$ — this sets the pitch we hear — multiplied by a slowly varying amplitude, the term in square brackets. That slow factor is the envelope; it swells and collapses, and its variation is what the ear perceives as the throbbing of loudness.

Figure 2 · The amplitude-modulated resultant

Superposing waves (a) and (b) from Figure 1 gives a fast carrier oscillation whose amplitude is bounded by a slow envelope. The envelope reaches its loud maxima twice per second — a beat frequency of 2 Hz.

loud soft (c) resultant — beats at $|\nu_1-\nu_2| = 2$ Hz

Beat Frequency and the Audibility Limit

Loudness peaks whenever the envelope factor $\cos\!\left(\tfrac{\omega_1-\omega_2}{2}t\right)$ reaches $\pm 1$, because the intensity depends on amplitude regardless of its sign. A maximum at $+1$ and the next at $-1$ are both loud points, so the envelope produces two intensity maxima per cycle of itself. Working this through, the number of beats per second comes out to the plain difference of the frequencies. Since $\omega=2\pi\nu$, NCERT writes the result (Eq. 14.48) as

$$\nu_{beat} = \nu_1 - \nu_2$$

In practice we count beats, never negative numbers, so the working form is the absolute value $\nu_{beat}=|\nu_1-\nu_2|$. The NIOS treatment frames the same idea physically: if fork B completes $n$ more vibrations each second than fork A, then B gains one full vibration over A every $1/n$ second, the two forks pass through one complete in-phase / out-of-phase cycle in that time, and so exactly $n$ beats are heard per second.

NEET Trap

The beat frequency is an absolute difference

A fork of 510 Hz sounded with one of 514 Hz, and the same 510 Hz fork sounded with one of 506 Hz, both give 4 beats per second. The beat count alone never tells you which fork is higher — it only fixes the gap. Two unknown frequencies are always consistent with a single observed beat count.

Rule: $\nu_{beat}=|\nu_1-\nu_2|$. From a beat count alone an unknown fork has two possible frequencies, $\nu_{known}\pm\nu_{beat}$.

Beats are only heard distinctly when the frequency difference is small. The human ear cannot resolve two intensity changes separated by less than about one-tenth of a second, so once the difference exceeds roughly 10 Hz, more than ten beats occur each second and the separate waxing and waning blur into a single steady tone. This is why the technique is restricted, in NCERT and in problems, to forks that differ by only a few hertz.

Build the foundation

Beats are interference in time. The general adding-of-waves rule is set out in Superposition of Waves.

The Wax-Loading Tuning-Fork Logic

Because the beat count gives two candidate frequencies, an extra observation is needed to choose. The standard move is to alter one fork by a known amount and watch whether the beats speed up or slow down. The physics of that alteration is fixed: loading a prong with wax adds mass and lowers the fork's natural frequency; filing the prongs removes mass and raises it. For strings the analogue is tension — increasing tension raises frequency.

Figure 3 · Decision table for loading one fork

An unknown fork gives $b$ beats per second with a known fork of frequency $\nu_0$, so the unknown is either $\nu_0+b$ or $\nu_0-b$. Load the unknown fork with wax (lowering it) and read off this table.

Load unknown fork with wax (its frequency drops) Beats INCREASE gap widened Beats DECREASE gap narrowed Unknown was LOWER $\nu = \nu_0 - b$ Unknown was HIGHER $\nu = \nu_0 + b$

The reasoning behind the table is worth saying in words, because the trap is easy to fall into. The unknown fork is either above or below the known one by the beat amount. Loading pushes the unknown down. If it was already the lower fork, pushing it further down widens the gap, so the beats grow more rapid. If it was the higher fork, pushing it down moves it toward the known frequency, narrowing the gap, so the beats slow. The direction of change in the beat count therefore resolves the ambiguity completely.

NEET Trap

Loading lowers; filing raises — and read the change carefully

Two errors recur. First, candidates forget that wax lowers the frequency and filing raises it; reverse this and every conclusion flips. Second, after correctly lowering the loaded fork, they misread the consequence: if the beats increase after loading, the loaded fork was the lower one — not the higher. The string version (NEET 2020) substitutes tension for wax: decreasing tension lowers the string's frequency, exactly like loading.

Rule: loading / lower tension ⇒ frequency falls. Beats rise after this ⇒ the changed source was the lower of the pair.

Worked Example · NCERT 14.6

Two sitar strings A and B are slightly out of tune and produce 5 beats per second. The tension of B is slightly increased and the beat frequency falls to 3 Hz. If A is 427 Hz, find the original frequency of B.

Increasing tension raises B's frequency. Had B been higher than A, raising it further would have increased the beat frequency — but the beats decreased. Therefore B was the lower fork, $\nu_B<\nu_A$. With $\nu_A-\nu_B=5$ Hz and $\nu_A=427$ Hz, the original frequency of B is $427-5=\mathbf{422}$ Hz.

Applications

The chief use of beats is in tuning. A musician sounds an instrument's string or pipe against a reference of the desired pitch and adjusts it until the beats slow to zero; the disappearance of beats signals that the two frequencies have become exactly equal. The same principle, run in reverse, measures an unknown frequency: the beat count gives the gap, and the loading test gives the sign.

UseWhat is knownWhat beats reveal
Tuning an instrumentA reference frequencyAdjust until beats vanish ⇒ frequencies equal
Measuring an unknown frequencyA standard fork of frequency $\nu_0$Beat count $b$ ⇒ unknown is $\nu_0\pm b$
Resolving the $\pm$ signDirection of frequency change (wax / file / tension)How the beat count shifts fixes higher vs lower
Worked Example · NIOS 14.3

A tuning fork of unknown frequency gives 5 beats per second with a fork of 500 Hz. What is the unknown frequency?

The beat count fixes only the gap: $\nu' = 500 \pm 5$, so the unknown is either $\mathbf{495}$ Hz or $\mathbf{505}$ Hz. A second observation — loading or filing one fork and watching the beats — is needed to choose between them.

Quick Recap

Beats in one screen

  • Beats = periodic waxing and waning of loudness from superposing two close, unequal frequencies.
  • Resultant $s = 2a\cos\!\left(\tfrac{\omega_1-\omega_2}{2}t\right)\cos\!\left(\tfrac{\omega_1+\omega_2}{2}t\right)$ — a carrier at the average frequency under a slow envelope.
  • Beat frequency $\nu_{beat}=|\nu_1-\nu_2|$; pitch heard is the average $(\nu_1+\nu_2)/2$.
  • Beats are perceived only if the difference is small (under about 10 Hz), since the ear cannot resolve faster alternations.
  • Wax loading lowers a fork's frequency; filing raises it; tension raises a string's frequency.
  • A beat count gives two candidates $\nu_0\pm b$; the way beats change after a known adjustment fixes which.

NEET PYQ Snapshot — Beats

Beats appear most often as the tuning-fork / string direction problem — the load or tension changes and you read the beat count.

NEET 2020

In a guitar, two strings A and B made of the same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B is:

  1. 524 Hz
  2. 536 Hz
  3. 537 Hz
  4. 523 Hz
Answer: (1) 524 Hz

The 6-beat gap makes B either $530+6=536$ Hz or $530-6=524$ Hz. Decreasing tension lowers B's frequency. If B were 536 Hz (higher than A), lowering it would move it toward 530 Hz and the beats would fall — but they rise to 7 Hz. So B must be the lower fork at 524 Hz: lowering it further widens the gap, increasing the beats. Hence $\nu_B = 524$ Hz.

Concept · NCERT 14.6

Two sitar strings A and B are slightly out of tune and give 5 beats per second. The tension of B is slightly increased and the beat frequency falls to 3 Hz. If A is 427 Hz, the original frequency of B is:

  1. 432 Hz
  2. 422 Hz
  3. 424 Hz
  4. 430 Hz
Answer: (2) 422 Hz

Increasing tension raises B. If B were higher than A, the beats would have grown; instead they fell, so B is lower. With $\nu_A-\nu_B=5$ Hz and $\nu_A=427$ Hz, $\nu_B=422$ Hz.

Concept

A tuning fork gives 4 beats per second with a standard fork of 256 Hz. On loading the unknown fork with a little wax, the beat frequency becomes 6 beats per second. The original frequency of the unknown fork is:

  1. 252 Hz
  2. 256 Hz
  3. 260 Hz
  4. 248 Hz
Answer: (1) 252 Hz

The unknown is either $256+4=260$ Hz or $256-4=252$ Hz. Loading lowers the unknown. The beats increased to 6, so the gap widened — which happens only if the unknown was already below the standard. Hence the unknown was the lower fork, $256-4=252$ Hz.

FAQs — Beats

The points examiners turn into traps, answered straight.

What is the beat frequency formula?
The beat frequency equals the magnitude of the difference of the two superposing frequencies, $\nu_{beat}=|\nu_1-\nu_2|$. The number of beats heard per second is this difference. Note the absolute value: a fork of 510 Hz with one of 514 Hz, and a fork of 510 Hz with one of 506 Hz, both give 4 beats per second.
Why must the two frequencies be close for beats to be heard?
The ear cannot resolve two intensity changes that occur in an interval shorter than about one-tenth of a second. If the frequency difference exceeds roughly 10 Hz, more than 10 beats occur per second and the waxing and waning merge into a continuous tone, so distinct beats are no longer perceived.
What is the frequency of the sound actually heard during beats?
The pitch heard is that of the average (carrier) frequency, $(\nu_1+\nu_2)/2$. The slowly varying amplitude — the envelope — oscillates at the beat frequency $|\nu_1-\nu_2|$, producing the waxing and waning of loudness on top of that average tone.
Does loading a tuning fork with wax raise or lower its frequency?
Loading a prong with wax adds mass, which lowers the fork's natural frequency. Filing the prongs removes mass and raises the frequency. This is the standard way to identify an unknown frequency: change one fork in a known direction and watch whether the beat count rises or falls.
A fork gives 5 beats per second with a 500 Hz fork. What is its frequency?
The beat count alone gives two candidates, $500+5=505$ Hz or $500-5=495$ Hz, because the beat frequency is an absolute difference. To decide between them you must change one fork (load or file it) and observe whether the beats increase or decrease.
If beats increase after loading one fork, was that fork the higher or lower one?
Loading lowers the loaded fork's frequency. If the beat frequency increases, the loaded fork must have been the lower of the two — lowering it further widens the gap. If the beat frequency decreases, the loaded fork was the higher one, and lowering it narrowed the gap.