Beats
A piano tuner sounds a reference fork of 256 hertz alongside a slightly mistuned string and hears four beats each second, so what are the possible string frequencies?
Select the correct option:
Solution
252 Hz or 260 Hz
Beats arise, as explained in NCERT Class 11, Chapter 15 (Waves), when two sound waves of slightly different frequencies superpose, producing a periodic rise and fall in loudness at a beat frequency equal to the absolute difference of the two frequencies: fbeat=∣f1−f2∣. Here the fork is 256 Hz and four beats per second are heard, so ∣fstring−256∣=4. This gives two possibilities: fstring=256+4=260 Hz or fstring=256−4=252 Hz. Without extra information both are valid. The option 256 Hz or 264 Hz is wrong because 256 Hz would produce zero beats. The option of only 260 Hz ignores the equally valid lower solution 252 Hz. The option 248 Hz or 264 Hz corresponds to a beat frequency of 8 Hz, not the observed 4 Hz. In practice a tuner resolves this ambiguity by deliberately changing the string tension slightly and listening: if the beats slow down the string frequency is moving toward 256 Hz, and if they speed up it is moving away, which reveals whether the true value was 252 or 260 Hz. The two frequencies produce identical beat counts because the ear responds to the magnitude of the frequency difference, not its sign. A plausibility check: the two answers lie symmetrically 4 Hz above and below the reference, exactly as the absolute-difference formula demands, and 4 beats per second is a slow, easily counted throb, confirming both the magnitude and the two-fold ambiguity.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More beats Practice Questions
About This Question
- Subject
- physics
- Chapter
- oscillations and waves
- Topic
- beats
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
252 Hz or 260 Hz
Beats arise, as explained in NCERT Class 11, Chapter 15 (Waves), when two sound waves of slightly different frequencies superpose, producing a periodic rise and fall in loudness at a beat frequency equal to the absolute difference of the two frequencies: fbeat=∣f1−f2∣. Here the fork is 256 Hz and four beats per second are heard, so ∣fstring−256∣=4. This gives two possibilities: fstring=256+4=260 Hz or fstring=256−4=252 Hz. Without extra information both are valid. The option 256 Hz or 264 Hz is wrong because 256 Hz would produce zero beats. The option of only 260 Hz ignores the equally valid lower solution 252 Hz. The option 248 Hz or 264 Hz corresponds to a beat frequency of 8 Hz, not the observed 4 Hz. In practice a tuner resolves this ambiguity by deliberately changing the string tension slightly and listening: if the beats slow down the string frequency is moving toward 256 Hz, and if they speed up it is moving away, which reveals whether the true value was 252 or 260 Hz. The two frequencies produce identical beat counts because the ear responds to the magnitude of the frequency difference, not its sign. A plausibility check: the two answers lie symmetrically 4 Hz above and below the reference, exactly as the absolute-difference formula demands, and 4 beats per second is a slow, easily counted throb, confirming both the magnitude and the two-fold ambiguity.
This medium difficulty physics question is from the chapter oscillations and waves, covering the topic of beats. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse oscillations and waves questions on RankGuru.