Question

A set of 25 tuning forks arranged in series of decreasing frequencies. Each gives $3\dfrac{{beats}}{{\sec ond}}$with the succeeding one. The first fork is the octave of the last. The frequency of the sixteenth fork in series is? (Use nth term of A.P)

Answer 1

Hint: A tuning fork is a simple steel device which is used by musicians, which vibrates and produces a particular note of specific pitch. The tuning forks are basically arranged in the decreasing frequency order, so we will assume the frequency of the last tuning fork and then we will obtain mathematical expressions for the first, third and sixteenth forks using the given relation. Then we will solve the formed equations and find the frequency of the sixteenth fork.

Complete step by step answer:
According to question,
Let us consider the frequency of the last tuning fork to be \[a\].
Since, the frequency of the last tuning fork is \[a\]. So, the frequency of the first tuning force will be $2a - 3$ as it produces $3\dfrac{{beats}}{{\sec ond}}$.
In the same way, the frequency of the third tuning fork will be $2a - 6$.
So, the frequency of the last tuning fork will come out to be $2a - 24(3)$.
So, we can say that,
$a = 2a - 72$
$a = 72hz$
Now, we have to find the frequency of the sixteenth fork in the series,
$f = 2a - 15(3)$
On putting the value of \[a\], we get,
$f = 2(72) - 15(3)$
$f = 144 - 45$
$f = 99hz$
So, the frequency of the sixteenth fork in series is, $f = 99hz$.

Note: In tuning forks sound is produced by striking one of its pads, which is made of rubber and then a vibration is produced in it. These vibrations produce longitudinal waves. The longitudinal waves are the waves, which consists of periodic vibrations that take place in the same direction as of the propagation of waves.