A tuning fork produces 4 beats per second when sounded with a 256 Hz standard fork. When a little wax is attached to the tuning fork, the beat frequency drops to 2 beats per second. The original frequency of the tuning fork was:
A. 252 Hz
B. 260 Hz
C. 254 Hz
D. 258 Hz
Answer: Option B
Solution (By JKExamLibrary)
Beat frequency is the absolute difference between two frequencies. Given 4 beats with 256 Hz, the unknown fork is either 256+4=260 Hz or 256-4=252 Hz. Adding wax increases mass, which strictly lowers the frequency of a fork. If it was 252 Hz, lowering it would increase the gap from 256, increasing beats. If it was 260 Hz, lowering it brings it closer to 256, decreasing beats to 2. Thus, it was originally 260 Hz.
This question belongs to:
Science
Physics
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