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Finding an Unknown Frequency Using Beats and Tuning Forks

Physics · Waves · NEET

A known tuning fork of frequency f gives beats with an unknown source, so the unknown frequency is f + n or f - n (where n = beats per second). To pick the correct sign, you change the unknown source slightly (add wax, or change tension/length) and watch whether the beats go up or down. Memory hook: "beats give you two answers, the change picks the winner."
Known fork f = 256 Hz, beats n = 4/s -> unknown = 260 or 252 HzCombined sound (waxing and waning)Beat envelope: n = |f1 - f2|one beat = one loud-soft cycle
Two nearly-equal frequencies add up so the sound gets loud then soft. Each loud-soft cycle is one beat; counting them per second gives n = |f1 - f2|, which fixes how far the unknown frequency sits from the known fork.

Your doubts, answered

How do I find the unknown frequency from a known tuning fork and the beats I hear?

Beat frequency = |f_known - f_unknown|. So if a tuning fork of 256 Hz gives 4 beats/s with an unknown fork, the unknown = 256 + 4 = 260 Hz OR 256 - 4 = 252 Hz. You always get TWO answers first. That is normal. You then need one extra piece of information (add wax, change tension, etc.) to choose the right one.

Why does the beat method give two possible frequencies, not one?

Beats depend only on the size of the difference, not its direction. |f - x| = 4 is true whether x is 4 above f or 4 below f. The ear cannot tell which fork is higher just from the beat count. That is why NEET always adds a second clue (loading, filing, tension) so you can remove the wrong value.

How does adding wax (loading) change a tuning fork's frequency?

Adding wax adds mass, and more mass means the fork vibrates slower, so its frequency DECREASES. Filing the prongs removes mass, so frequency INCREASES. Use this rule to decide: if you add wax to the unknown fork and beats decrease, the unknown was ABOVE the known one (lowering it brought it closer).

For a stretched string, does higher tension mean higher or lower frequency?

Higher. For a string f is proportional to sqrt(T), so increasing tension raises the frequency and decreasing tension lowers it. This is the exact clue used in the NEET sitar/guitar PYQs: change tension, see if beats grow or shrink, then decide the sign.

If the beat frequency increases after I change the source, is the unknown above or below the fork?

Work out which way you pushed the source. If lowering the unknown frequency makes beats INCREASE, then the two were already far in that direction, so the unknown was BELOW the fork. If lowering it makes beats DECREASE, the unknown was ABOVE the fork. Always trace one candidate value through the change and check it matches.

⚠️ The NEET trap
✗ Beats = 6 Hz with a 530 Hz fork, so the answer is simply 530 + 6 = 536 Hz.
✓ You must test both 536 Hz and 524 Hz against the tension/wax clue; only the value that makes the beats change in the stated direction is correct (524 Hz in NEET 2020).
🧠 Beats always give two candidates; NTA rewards the one that survives the 'what happens when I change it' test, not the first number you write.

Real NEET questions

2020

In a guitar, two strings A and B of the same material are slightly out of tune and produce beats of frequency 6 Hz. When the 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:

A · 536 Hz
B · 537 Hz
C · 523 Hz
D · 524 Hz ✓
Solution: Step 1: Beat frequency = |f_A - f_B| = 6, so f_B = 530 + 6 = 536 Hz OR 530 - 6 = 524 Hz (two candidates). Step 2: For a string f is proportional to sqrt(T), so decreasing tension in B LOWERS f_B. Test f_B = 536: lowering it gives |530 - 535| = 5, beats decrease. That contradicts the question (beats rose to 7). Test f_B = 524: lowering it gives |530 - 523| = 7, beats increase to 7 Hz. This matches. So f_B = 524 Hz. Answer: D.
2016

Three sound waves of equal amplitude have frequencies (n-1), n and (n+1) Hz. They superimpose to give beats. The number of beats produced per second is:

A · 1
B · 4
C · 3
D · 2 ✓
Solution: Pairwise beat frequencies: |n - (n-1)| = 1 Hz and |(n+1) - n| = 1 Hz. The pair (n-1) and (n+1) differs by 2 Hz, which lines up with the maxima of the 1 Hz beats. The resultant produces 2 beats per second. Answer: D. (This is the counting-side of the same idea: beats come from the difference of frequencies.)

Solved Waves NEET PYQs

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Frequently asked

What is the basic formula to find an unknown frequency using beats?

f_unknown = f_known +/- n, where f_known is the tuning fork frequency and n is the number of beats heard per second. The +/- sign is fixed by an extra experiment such as loading with wax or changing tension.

Why do we use a tuning fork of KNOWN frequency?

Because beats only tell you the difference between two frequencies. If neither is known you cannot solve for either. A calibrated fork gives you one fixed value so the unknown becomes the only variable.

What happens if the beat frequency becomes zero?

Zero beats means the two frequencies are exactly equal (f_unknown = f_known). This is how musicians tune instruments: they adjust until the beats disappear, which NCERT describes for players tuning together.

Adding wax to the UNKNOWN fork lowers its frequency. How does that decide the sign?

If lowering the unknown makes beats decrease, the unknown was ABOVE the known fork (it moved closer). If lowering it makes beats increase, the unknown was BELOW the fork (it moved further). Match the observed change to pick +n or -n.

Is this concept important for NEET Physics?

Yes. NEET repeatedly asks beat problems where you must choose the correct sign using tension or loading logic (e.g. NEET 2020 guitar strings). It combines superposition, beats and string frequency in one question, so it is high-yield.

Related concepts:
Beats and Beat Frequency: What Causes ThemBeat Frequency Formula and How to Solve Beat ProblemsSonometer and the Laws of Vibrating StringsHow Tension Changes Wave Speed on a StringPrinciple of Superposition of Waves Explained
Practise Physics for NEET:
Physics MCQs (free, chapter-wise)Waves PYQs