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A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
A
508 Hz
B
510 Hz
C
514 Hz
D
516 Hz
Detailed Solution
Beat frequency = difference of the two frequencies, so the string frequency is either 512 + 4 = 516 Hz or 512 − 4 = 508 Hz.
Frequency of a string $f \propto \sqrt{T}$, so increasing the tension increases its frequency.
If the string were at 516 Hz, raising its frequency would increase the difference from 512 Hz, and the beats would increase.
If the string were at 508 Hz, raising its frequency (to 510 Hz) brings it closer to 512 Hz, and the beats decrease to 2 per second, as observed.
Hence the original frequency of the piano string was 508 Hz.
Frequency of a string $f \propto \sqrt{T}$, so increasing the tension increases its frequency.
If the string were at 516 Hz, raising its frequency would increase the difference from 512 Hz, and the beats would increase.
If the string were at 508 Hz, raising its frequency (to 510 Hz) brings it closer to 512 Hz, and the beats decrease to 2 per second, as observed.
Hence the original frequency of the piano string was 508 Hz.