Question

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?

Answer 1

Concepts and reason

The concept used to solve this problem is beat frequency.

Use the expression for beat frequency to find the initial frequency of the flute player.

Fundamentals

Beat can be defined as “the interference pattern between two sounds of slightly different frequencies”.

The expression for the beat frequency is,

${f_{beat}} = {f_1} - {f_2}$

Here, beat frequency is ${f_{beat}}$, final frequency is ${f_2}$, and initial frequency is ${f_1}$.

The expression for the beat frequency is,

${f_{beat}} = {f_1} - {f_2}$

Rearrange the above expression for ${f_1}$.

${f_1} = {f_{beat}} + {f_2}$

The expression for the initial frequency is,

${f_1} = {f_{beat}} + {f_2}$

Substitute $4\,{\rm{beats/s}}$ for ${f_{beat}}$ and $523\,{\rm{Hz}}$ for ${f_2}$.

$\begin{array}{c}\\{f_1} = \left( {4\,{\rm{beats/s}}} \right) + \left( {523\,{\rm{Hz}}} \right)\\\\ = 527\,{\rm{Hz}}\\\end{array}$

Therefore, the initial frequency is $527\,{\rm{Hz}}$.

Ans:

The initial frequency of the flute player is ${\bf{527}}\,{\bf{Hz}}$.