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

Hey dear,

● Answer -

f16 = 312 Hz

● Explanation -

Let f1, f2, f3...f64 be frequencies of respective forks.

As 4 beats are given for each consecutive pair.

fn = f1 + (n-1)d

f64 = f1 + (64-1)4

f64 = f1 + 252

But f64 gives octave for the f1.

f64 = 2f1

f1 + 252 = 2f1

f1 = 252 Hz

Also,

f16 = f1 + (n-1)d

f16 = 252 + (16-1)4

f16 = 252 + 60

f16 = 312 Hz

Therefore, frequency of 16th fork is 312 Hz.

Thanks dear...

Answer 2

Answer:

312 Hz

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