A standard tuning fork has frequency 400 Hz. An unknown fork produces 5 beats per second with it. Wax is added to the unknown fork and lowers its frequency by exactly 8 Hz. The new beat frequency becomes 3 beats per second. What was the original frequency of the unknown fork?
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Question

A standard tuning fork has frequency 400 Hz. An unknown fork produces 5 beats per second with it. Wax is added to the unknown fork and lowers its frequency by exactly 8 Hz. The new beat frequency becomes 3 beats per second. What was the original frequency of the unknown fork?

Options

G1 Physics · Accepted Answer

Initially, |f−400| = 5. Therefore, f = 395 Hz or f = 405 Hz. After adding wax, the frequency becomes f−8. If f = 405 Hz, the new frequency is 397 Hz, so the beat frequency is |397−400| = 3 Hz. If f = 395 Hz, the new frequency is 387 Hz and the beat frequency would be 13 Hz. Therefore, f = 405 Hz. Correct option: D.