Question

25 tuning forks are arranged in series in the order of decreasing frequency. Any two successive fork produce 3 beat/sec. If the frequency of first tuning fork is the octave of the last fork, the frequency of 21st fork is:
a)72 Hz
b)84 Hz
c)87 Hz
d)144 Hz

Answer 1

84Hz?

Answer 2

No idea.. I don't have the answer.. well can you just say how do you get it?

Answer 3

Difference between two successive frequencies is 3 Hz.
Total is 24 intervals, so overall difference is 3 x 24 = 72 Hz.

f25 doubled yields f1 and f25 + 72 Hz = f1

So f1 = 72 Hz and f25 = 144 Hz.

Answer 4

you mean f1=144 and f25=72..?? That makes f21=84..
Are you sure about the answer? I guess you are..

Answer 5

I am sure that beat frequency between f1 and f2 is: | f1 - f2 |
I am sure that an octave is f2/f1 = 2

It seems the rest follows from that, provided I have not made a mistake in counting the number of intervals!

Answer 6

yeah i guess you are right.. What i didn't know actually was the octave thing.. I didn't know how to relate first and last frequencies..
Thanks!

Answer 7

@ujjwal

For your information:

Ascending octave is ratio 2:1
Asc. fifth is ratio 3:2
Asc. fourth is ratio 4:3
Asc. major third is ratio 5:4
Asc. minor third is ratio 6:5

Answer 8

those are the ratios of frequencies? @Vincent-Lyon.Fr

Answer 9

Yep!

Answer 10

\[\nu1-\nu21=60\]
\[\nu1-\nu25=72\]
are the given conditions

\[\nu1=2*\nu25\]

also the given data and using the above 3 equations



\[so 2*\nu25-\nu25=72\]
\[\nu25=72\]

using this value in the second equation
\[ \nu1=72+72\]
=144