Question

A set of 24 tuning forks are arranged so that each gives 6 beats per second with the previous one. If the frequency of the last tuning fork is double the frequency of the first, frequency of the second tuning fork is,
A) 134 Hz
B) 144 Hz
C) 132 Hz
D) 276 Hz

Answer 1

Hint: The number of beats heard per second, depends on the frequency of the two neighbouring tuning forks. The number of beats is the difference in the frequencies of the two tuning forks.

Complete step by step answer:
Beats is a phenomenon that occurs when two sound waves of equal amplitude and phase but with slightly different frequencies interact with each other.
In the problem, it is given that there are 24 tuning forks present, which are arranged in such a way that the beats produced between one fork and the previous forks are 6. Also, the frequency of the last fork is double the frequency of the first.
So if two tuning forks of frequencies ${{\text{f}}_{\text{1}}}\text{ and }{{\text{f}}_{2}}$ are present, then the number of beats per second is given by,
$\text{Beats}={{\text{f}}_{\text{1}}}-{{\text{f}}_{2}}$
So there are 24 tuning forks, so n=24. Each of them has a frequency difference of 6, then d=6.
We know that in an arithmetic progression, the $n^{th}$ term can be written as,
${{a}_{n}}=a+(n-1)d$

Where,
a is the first term of the progression (frequency of the first fork)
d is the common difference. ( In our case the frequency difference)
So the 24th term can be written as,
${{a}_{24}}=a+(24-1)(6)$
We know from the question that the $24^{th}$ frequency is twice that of the first frequency.
${{a}_{24}}=\text{2}a$
Substituting above condition in the equation we get,
$2a=a+(23\times 6)$
$\therefore a=138$
So the frequency of the first tuning fork is 138 Hz. We need to find the frequency of the second tuning fork,
${{\text{f}}_{\text{2}}}=138+6$
$\therefore {{\text{f}}_{\text{2}}}=144\text{Hz}$
So the answer to our question is option (B)- 144 Hz.

Note: We can define beats as the interference of two sound waves with different frequencies, seen as a periodic variation in the medium, whose rate is the difference between two frequencies.
The frequency of the new sound wave formed due to the interference of two similar waves with difference in frequencies is the average of the two frequencies.