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

Answer:

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

Answer 2

The initial frequency of the flute player is 527 Hz.

Based on the given information,

• The beats hear by the flute player is 4 beats per second (fbeat), the tuning fork frequency given is 523 Hz.  

• The beat frequency is equivalent to the difference in the initial and final frequency.  

The expression of the beat frequency is,

$fbeat = f1 - f2$

It can be rearranged as,

$f1 = fbeat + f2$

Here f1 is the initial frequency and f2 is the final frequency.

Now putting the values we get,

$f1 = (4beats/s) + 523 Hz\\f1 = 527 Hz$

Thus, the initial frequency is 527 Hz. Therefore, with the increase in the flute length, there is a decrease in the frequency, as the initial frequency is more than the final frequency.

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