Question

$16$ tuning forks are arranged in the order of increasing frequencies. Any two successive forks gives $8\text{ beats/sec}$ when sounded together. If the last fork gives the octave of the first, then the frequency of the fork is –
$A)\,n=120$
$B)\,n=160$
$C)\,n=180$
$D)\,n=220$

Answer 1

Hint: We can find the difference in the frequencies of the successive frequencies by using the beats frequency given. The last fork is the octave of the first fork means that the frequency of the last one is twice that of the first. Using these two equations and solving them simultaneously, we can get the required answer.

Formula used:
$f=\left| {{f}_{1}}-{{f}_{2}} \right|$

Complete step by step answer:
We will find the difference in the frequencies between two successive tuning forks. For this we will use the information of the beats frequency given.
The beats frequency $f$ that is developed when two close frequencies ${{f}_{1}}$ and ${{f}_{2}}$ superimpose together is given by
$f=\left| {{f}_{1}}-{{f}_{2}} \right|$ --(1)
Now, let us analyze the question.
Let the frequency of the first tuning fork be $n$.
Let the difference between successive tuning forks be $x$.
Hence, the frequency of the second tuning fork will be $n+x$, third will be $n+x+x=n+2x$ and so on.
The general equation for the frequency of the ${{p}^{th}}$ tuning fork in the sequence will be
${{f}_{p}}=n+\left( p-1 \right)x$ --(2)
Now, the given beats frequency when two successive tuning forks are sounded together is $f=8\text{se}{{\text{c}}^{-1}}=8Hz$ $\left( \because 1{{\sec }^{-1}}=1Hz \right)$
Now, we will take the case of the first and the second tuning forks and the beat frequency when they are sounded together.
Hence, using (1), we get
$8=\left| \left( n+x \right)-\left( n \right) \right|=\left| n+x-n \right|=\left| x \right|$
$\therefore x=8Hz$ --(3)
Since, it is given that the frequency is increasing therefore $x=8Hz$ and not $x=-8Hz$.
Using (2), we will get the frequency of the fifteenth tuning fork as
${{f}_{16}}=n+\left( 16-1 \right)x=n+15x$ --(4)
Putting (3) in (4), we get
${{f}_{16}}=n+\left( 15\times 8 \right)=n+120$ --(5)
Now, it is given that the sixteenth tuning fork that is the last one, gives an octave of the first one. What this means is that the frequency of the sixteenth tuning fork is twice the frequency of the first one.
$\therefore {{f}_{16}}=2{{f}_{1}}=2n$ --(6)
Putting (6) in (5), we get
$2n=n+120$
$\therefore 2n-n=120$
$\therefore n=120Hz$
Hence, we have got the frequency of the first tuning fork as $120Hz$.
Therefore, the correct option is $A)\,n=120$.

Note: This problem could also have been solved by realizing that the sequence of the frequency of the tuning forks forms an increasing arithmetic progression, since it has a common difference of frequency between two successive tuning forks and applying the direct formula for the last term of an arithmetic progression as a function of the first term and then solving according to the question. However, essentially, we have also done that and arrived upon the same result in the calculation. We have avoided memorizing the formulae for the arithmetic progression and thus, by proceeding logically have minimized the chances of error that comes while writing memorized formulae.