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A flute player hears four beats per second when she compares her note to a 523 Hz tuning fork (the note C). She can match the frequency of the tuning fork by pulling out the tuning joint to lengthen her flute slightly. What was her initial frequency?

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 waves. In other words, the beat frequency is the number of beats generated one second at a time. 

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 $A$ is  $f_A = 440Hz$.

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:

\[ =30-25.2 = 4.8cm \]

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