Beats

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A piano tuner sounds a reference fork of 256 Hz together with a slightly mistuned string, and hears the loudness rise and fall 4 times every second. What are the possible frequencies of the string?

RankGuru · Accepted Answer

When two sound waves of nearly equal frequency overlap, the resulting amplitude rises and falls periodically, producing beats whose frequency equals the absolute difference of the two source frequencies, $f_{beat} = |f_1 - f_2|$. Here the tuner hears 4 beats per second against a 256 Hz fork, so the string differs from 256 Hz by exactly 4 Hz. Both possibilities satisfy this: $256 + 4 = 260	ext{ Hz}$ or $256 - 4 = 252	ext{ Hz}$. The pair $256	ext{ Hz}$ or $260	ext{ Hz}$ is wrong because 256 Hz would give zero beats. The pair $248	ext{ Hz}$ or $264	ext{ Hz}$ corresponds to a beat frequency of 8 Hz, not 4 Hz. The pair $254	ext{ Hz}$ or $258	ext{ Hz}$ gives only 2 beats per second. This is the NCERT result that beat frequency equals the magnitude of the frequency difference. As a plausibility check, a single beat measurement cannot distinguish whether the string is sharp or flat, so two symmetric answers about the reference frequency are physically expected, which is exactly what $252	ext{ Hz}$ and $260	ext{ Hz}$ represent.

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