Question

64 tuning forks are arranged such that each fork
produces 4 beats per second with next one. If
the frequency of the last fork is octave of the
first, the frequency of 16th fork is​

Answer 1

Answer:

Each tuning forck gives 4 beats with the next, so the difference in the frequencies of two consecutive forks is 4

hence

f64=f1+(n-1)*-4f

=2f+(64-1)*-4f=250 hz

Frequency of 30th tuning fork f30=f1+(n-1)*(-4)=2*250+(30-1)*-4=384 hz