Question

16 tuning forks are arranged in the order of decreasing frequency. Any two successive forks give 8 beats per second when sounded together. If the first tuning fork gives the octave of the last, then determine the frequency of the last fork

Answer 1

Hey dear,

● Answer -
f16 = 240 Hz

● Explanation -
Let f1, f2, f3...f16 be frequencies of respective forks.

As 8 beats are given for each consecutive pair.
fn = f1 + (n-1)d
f16 = f1 + (16-1)8
f16 = f1 + 120

But f1 gives octave for the last.
f16 = 2f1

Substituting values -
f1 + 120 = 2f1
f1 = 120 Hz

Also,
f16 = 2f1
f16 = 2×120
f16 = 240 Hz

Therefore, frequency of last tuning fork is 240 Hz.

Hope this helps...

Answer 2

Answer:

240 hz is the answer for the question