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?

Question

Hickman

Physics

9 November, 20:06

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?

+2

Answers (1)

Know the Answer?

Alisa Rasmussen · Answer 1

Initial Frequency = 527Hertz

Explanation:

Given

Frequency of beat = 4 Hertz

Frequency of Fork = 523 Hertz

Let Fb = Frequency of beat

Ff = Frequency of Fork

The wave length and the length of the flute are directly proportional.

So, when the wavelength gets increased, the length of the flute is also increased.

But it is not so for the frequency.

When the wavelength and the length of the flute are increased, the frequency gets reduced.

For the flute player frequency to equate the frequency of the turning fork, the frequency of the flute must be greater than the frequency of the fork.

So, Frequency = 4 + 523 = 527 Hertz

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?

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?