Question

56 tuning forks are arranged in a way such that each fork produces 4 beats per second with its preceding fork. The frequency of the last fork is three times than that of first. The frequency of first fork will be :

• A. 220 Hz
• B. 330 Hz
• C. 110 Hz
• D. 440 Hz

Answer 1

Answer: (C)

110 Hz

Given 56 forks such that when two adjacent forks are struck they produces 4 beats per second

with the last fork having frequency triple of the first fork.

Lets suppose that the first fork has a frequency of $f$.

Then the $2_{nd}$ fork can have a frequency of $f+4orf-4$

but, the frequency of last fork is triple (greater) than the frequency of first

fork which means that the frequencies are gradually increasing as we go from first to the last fork.

So, the frequency of $2_{nd}$ fork is $f_2=f+4$

the $3_{rd}$ fork is $f_3=f+8$ and so on because $f_{beat}=4$ for adjacent elements.

This sequence constitutes an arithmetic progression with $a_1=f\&d=4$

So, frequency of the ${56}_{th}$ fork is $f_{56}=f+(56-1)4=f+220$

$f_{56}=3\times f$

$f+220=3f\Rightarrow f=110Hz$

So C is the correct answer.