The frequency of a fork A is 3 % more than the frequency o
o more than the frequency of a standard fork whereas the frequency of fork B
is 3% less. The forks A and B produce 6 beats per second
SA and B produce 6 beats per second. The frequency of standard fork will be-
(1) 100 Hz
(2) 106 Hz
(3) 103 Hz
(4) 112
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Answer:
Explanation:
frequency of A be nA=(103/100)nS
nS is frequency if standard fork
frequency of B be nB= (97/100) nS
therefore
nA-nB= 6
6nS/100=6
nS=100 Hz
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