This problem shows us the **frequency** of a **vibrating resonator** such as a **tuning fork.** The concept required to solve this problem is related to **frequency** and **wavelength relation, young modulus** to calculate the stress on the resonator, and **beat frequency.**

A **tuning fork** is a **two-string, fork-shaped** acoustical resonator utilized in many areas to create a specified **tone.** The **frequency** of a tuning fork relies on its **measurements** and the **material** it’s created from.

One major aspect is the **beat frequency,** which is equal to the **absolute value** of the change in the **frequency** of the two **successive**Â **waves.** In other words, the beat **frequency** is the number of beats generated **one second** at a time.Â

The **formula** to calculate the **beat frequency** of a tuning **fork** or any other vibrating device is the **difference** in frequency of the **two consecutive** waves:

\[ f_b = |f_2 – f_1| \]

$f_1$ and $f_2$ are the **frequencies** of **two successive waves.**

## Expert Answer

We are given the **initial frequency** of the **flute:**

\[f_{initial} = 527Hertz \]

It is also the **frequency** of the flute.

The **frequency** of **each beat** being produced is $4Hertz$, such that:

\[f_{beat} = 4Hertz \]

The **wavelength** and the **absolute size** of the flute are **directly** proportional. So an increase in the **wavelength** of the flute will result in an **increase** in the **length** of the flute as well. But this is not the **same** in the case of the **frequency.** Since **frequency** and **wavelength** are **inversely proportional** to each other as per the formula:

\[v=\dfrac{f}{\lambda} \]

\[\lambda=\dfrac{f}{v}\]

The **frequency** of the flute will **decrease** when the **wavelength** and the total **length** of the **flute** are increased.

So in order to **calculate** the **frequency** of the flute player, we will equate it to the frequency of the **tuning fork,** such that the **frequency** of the **flute** should be higher than that of the **fork frequency.**

So,

\[f_b=523 + 4 \]

\[f_b=527Hertz\]

## Numerical Result

The **initial frequency** of the **flute** player is $527Hertz$.

## Example

The **length** of a **violin** string is $30cm$. The **musical** note $A$ is $440Hz$. How far should you set your **finger** from the end of the **string** to play the note $C$ having **frequency** $523 Hz$?

Given the **length** of the string $L = 30cm = 0.30m$, and the **frequency** of note $A$ isÂ $f_A = 440Hz$.

We know that a **string** fixed at both ends builds **standing waves.** A straightforward **string** sounds the **fundamental frequency** of:

\[ f_1 = \dfrac{v}{2L} \]

For note $A$ the **frequency** with length $L_A$ becomes:

\[ f_{1A} = \dfrac{v}{2L_A} \]

For a different **length** $L_C$, the **frequency** of note $C$ is:

\[ f_{1C} = \dfrac{v}{2L_C} \]

**Dividing** both equations:

\[ \dfrac{ f_{1A}}{ f_{1C}} = \dfrac{\dfrac{v}{2L_A}}{\dfrac{v}{2L_C}} \]

\[ =\dfrac{L_A}{L_C} \]

\[ L_C = \dfrac{ f_{1A}}{ f_{1C}}L_A \]

**Substituting** the values:

\[ L_C = \dfrac{440}{523}\times 30\]

\[ L_C = 25.2cm\]

Since the **string** is $30cm$ long, the **position** to place the **finger** is:

\[ =30-25.2 = 4.8cm \]