The frequency of tuning fork A is 2% more than the frequency of a standard tuning fork. The frequency of a tuning fork B is 3% less than the frequency of the same standard tuning fork. If 6 beat/s are heard when the tuning fork A and B are excited , then frequency of A will be

Options

A. 120Hz
B. 122.4Hz
C. 116.Hz
D. 130Hz

Answer

B

Explanation

(i) `n_A = 102/100 n_s` ,
(ii) `n_B = 97/100 n_s` ,
(iii) `n_A - n_B = 6 therefore (102n_s)/(100) - (97n_s)/(100) = 6`
`therefore n_s = 600/5 = 120 Hz`,
(iv) `n_A = 102/100xx120=122.4Hz`