Question

$64$ tuning forks are arranged such that each fork produces $4$ beats per second with the next one. If the frequency of the last fork is octave of the first, frequency of $16th$ fork is
A. $316\text{ Hz}$
B. $\text{322 Hz}$
C. $\text{312 Hz}$
D. $\text{308 Hz}$

Answer 1

Hint: In this question, we are given each produces a beat frequency of $\text{4 Hz}$ so the difference between the frequencies of two consecutive forks will be $4$ beats per second. This forms an arithmetic progression, and we can find out the frequency of the last fork, but the last fork is the octave of the first. Using this relation, we can find out the frequency of the first fork, and using the A.P we can find out the frequency of the $16th$ fork
Beat frequency, ${\text{f}}_b=\left|{\text{f}}_2-{\text{f}}_1\right|$
Where ${\text{f}}_1$ and ${\text{f}}_2$ are the frequencies of two waves.

Complete answer:
We are given,
There are $64$ tuning forks and each produces $4$ beats per second
Let the frequencies of fork be ${\text{f}}_{{}_1},{\text{f}}_2,{\text{f}}_3,......\text{ , }{\text{f}}_{64}$ respectively.
Since ${\text{f}}_2-{\text{f}}_1={\text{f}}_4-{\text{f}}_3=....={\text{f}}_{64}-{\text{f}}_{63}=4$ they form an arithmetic progression with first term as ${\text{f}}_{{}_1}$ and common difference as $\text{4}$
Therefore, the frequency of last tuning fork, ${\text{f}}_{64}={\text{f}}_1+\left(n-1\right)d$
 $\Rightarrow {\text{f}}_{64}={\text{f}}_1+\left(64-1\right)4$
 $\Rightarrow {\text{f}}_{64}={\text{f}}_1+252$
But it is given the last fork is octave of the first
Thus, ${\text{f}}_{64}=2{\text{f}}_1$
Substituting this in the equation we get,
 $\Rightarrow 2{\text{f}}_1={\text{f}}_1+252$
 $\Rightarrow {\text{f}}_1=252\text{ Hz}$
The Frequency of the first tuning fork is $252\text{ Hz}$
Now to find out the frequency of the $16th$ tuning fork we use the A.P
 ${\text{f}}_{16}={\text{f}}_1+\left(n-1\right)d$
Replacing $n$ with $16$ and $d$ with $4$ we get,
 $\Rightarrow {\text{f}}_{16}={\text{f}}_1+\left(16-1\right)4$
 $\Rightarrow {\text{f}}_{16}={\text{f}}_1+60$
But ${\text{f}}_1=252\text{ Hz}$
Therefore, ${\text{f}}_{16}=252+60$
 $\Rightarrow {\text{f}}_{16}=312\text{ Hz}$
Hene, the frequency of the $16th$ fork is $312\text{ Hz}$
The correct option is option C.

Note:
Beats are produced when the two waves of nearby frequencies are superimposed together. This will occur when the two waves travel in the same path. Beats also cause a periodic variation in the intensity of resultant waves. If the beat frequency is greater than $\text{10 Hz}$ then it cannot be distinguished by human ears.